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NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Introduction

An average is which represents the large number of observations in a concise or single numerical data. It is a representative value around which all the values of the variable concentrate. It is called the measure of central tendency.

The word ‘average’ has been defined differently by various authors. According to Dr. Bowley, “Statistics is the science of averages.”

According to Gorton and Cowden, “An average is a single value within the range of the data used to represent all the values in the series.”

It is clear from the above definitions that an average is a single value that represents a group of values.

Read and Learn More Class 10 Maths Solutions Exemplar

Objectives Of Statistical Averages

The statistical average has the following objectives:

1. Provide a Brief Picture of Data: We can present complex data in a simple manner and concise form with the help of averages. An ordinary person can easily remember these averages.
2. Comparative Study: Measures of central value make easier the comparison of two or more than two groups. For example, it is impossible to conclude any result on the basis of marks obtained by all the students of two colleges but we can easily conclude by comparing the average marks obtained by both colleges.
3. Representation of the Group: Averages represent the picture. the whole group and that value enables Us to gel an idea of the entire data.

Properties Of Statistical Averages

Good averages should possess the following properties:

1. An average should be easy to understand and simple to compute.
2. It should be properly defined.
3. It should be based on all the observations.
4. It should be used for further statistical computation.

Types Of Averages

There are different types:

1. Mathematical Averages:
   • Arithmetic mean,
   • Geometric mean,
   • Harmonic mean.
2. Average Related with Position:
   • Median,
   • Mode

Statistics Mean

Mean is that value that can be calculated by dividing the sum of all terms of the series by the number of terms of the series.

Characteristics of Mean

1. It depends on all the terms of the group.
2. It can be calculated easily.
3. Its value is definite and based on calculations.
4. We can use algebraic methods on it.
5. The algebraic sum of deviations measured from the mean is always zero.
6. It can be determined in every condition.
7. Mean can be a value that does not exist in the series.

For example, the average of 5, 9, and 10 is \(\frac{5+9+10}{3}=8\) which is not an observation in the series. Now, we shall discuss the mean of grouped data.

Direct Method for the Mean of a Grouped Frequency Distribution

Step 1: Find the class mark x_i for each class: \(x_i=\frac{\text { lower limit }+ \text { upper limit }}{2}\)

Step 2: Find x_i for each i.

Step 3 : Find the mean using the formula, mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)

Direct Method For The Mean Of A Grouped Frequency Distribution Solved Examples

Question 1. Find the mean of the following frequency distribution by direct method:

Solution:

Now, \(\text { mean } \bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{2170}{74}=29.32\)

Question 2. If the mean of the following data is 26, then find the value of p:

Solution:

Now, arithmetic mean \(\bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}\)

⇒ \(26=\frac{890+15 p}{30+p}\)

⇒ 780 + 26p = 890 + 15p

⇒ 11p = 110

⇒ p = 10

The value of p = 10

Question 3. Find the mean from the following data:

Solution:

First, we will convert the given table into a table containing class intervals and corresponding frequencies:

Now, \(\text { mean } \bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{1485}{80}=18.5625\)

Assumed-Mean Method for the Mean of Grouped Frequency Distribution

Step 1: Find the class marks x_i for each class interval: \(x_i=\frac{\text { lower limit }+ \text { upper limit }}{2}\)

Step 2: Choose a suitable value of x_i from the middle values as the assumed mean and represent it by ‘a’.

Step 3: Find the deviations d_i =X_i – a for each i.

Step 4: Fin d_id_i for each i.

Step 5: Find the mean using the formula: \(\bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i}\)

Question 4. Find the mean for the following distribution table by shortcut method:

Solution:

Let assumed mean a = 35

⇒ \(\text { mean } \bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i}=35+\frac{160}{50}=35+3.2=38.2\)

Question 5. The height of 84 trees is given in the following table. Find their arithmetic mean by shortcut method:

Solution:

Let assumed mean a = 55

∴ \(\text { mean } \bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i}=55+\frac{450}{84}=55+5.36=60.36 \mathrm{~cm}\)

Step Deviation Method for the Mean of Grouped Frequency Distribution

Step 1: Find the class marks x_i for each class. \(x_i=\frac{\text { lower limit }+ \text { upper limit }}{2}\)

Step 2: Choose a suitable value of x_i from the middle values as the assumed mean and represent it by ‘a’.

Step 3: Find h = upper limit – a lower limit which is the same for all the classes.

Step 4: Calculate \(u_i=\frac{x_i-A}{h}\) for each i.

Step 5: Calculate f_iμ_i for each i.

Step 6: Find the mean using the formula \(\bar{x}=a+\frac{\Sigma f_i u_i}{\sum f_i} \times h\).

An Important Result

First, we should know the meanings of step and deviation (used in the step-deviation method for finding the mean).

Deviation

All xi’s are deviated (displaced) from their places by adding or subtracting the same non-zero quantity from each value. This is called the deviation.

Step

Big jump or fall in all xi’s in multiples i.e., all xi’s are deviated from their places by multiplying or by dividing with the same number.

But if we divide by only 1, then the step-deviation method and assumed method are the same.

So, if a student is asked to find the mean by step-deviation method and he solves this by assumed mean method then there is no harm.

Question 6. Find the mean of the following table by step deviation method:

Solution:

Here, h = 3

Let assumed mean a = 13

∴ \(\text { mean } \bar{x}=a+\frac{\Sigma f_i \cdot u_i}{\Sigma f_i} \times h=13+\frac{0}{174} \times 3=13\)

Question 7. The marks obtained by 30 students are given in the following table. Find their mean-by-step deviation method:

Solution:

Let assumed mean a = 25

Here, h = 10

∴ \(\text { mean } \bar{x}=a+\frac{\Sigma f_i u_i}{\Sigma f_i} \times h=25+\frac{2}{30} \times 10=25.67\)

Question 8. Find the mean for the following data by step-deviation method:

Solution:

The given data can be written in the following form:

Now, \(\text { mean }=a+\frac{\Sigma f_i u_i}{\Sigma f_i} \times h=35+\frac{-48 \times 10}{75}=28.6\)

Question 9. Find the mean for the following distribution:

Solution:

First, we will convert the given table into the exclusive form:

Here, h = 5

Let assumed mean a = 42

Now, \(\text { meàn }=a+\frac{\Sigma f_i u_i}{\Sigma f_i} \times h=42+\frac{(-79) \times 5}{70}=36.36\)

Question 10. Find the class limits corresponding to each frequency if the mean of the following distribution is 33 and the assumed mean is 35:

Solution:

Here, we are given \(\bar{x}=33 \text { and } a=35\)

Now, \(\text { mean } \bar{x}=a+\frac{\Sigma f_i u_i}{\Sigma f_i} \times h\)

⇒ \(33=35+\frac{-20}{100} \times h\)

⇒ -200 = -20 h ⇒ h = 10

Now, if \(u_i=-3 \Rightarrow \frac{x_i-35}{10}=-3 \Rightarrow x_i-35=-30 \Rightarrow x_i=5\) and width i.e., h = 10, So, class limits of the first group is 0-10. (subtract and add \(\frac{h}{2}\) from lower and upper boundaries) if \(u_i=-2 ⇒ \frac{x_i-35}{10}=-2 ⇒ x_i-35=-20 x_i=15\) and width i.e., h = 10. So class limits of the second group are 10-20 (subtract and add \(\frac{h}{2}\) from lower and upper boundaries). Similarly, the class limits of other groups are 20-30, 30-40, 40-50 and 50-60.

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Median

The median is the value of the variable which divides the group into two equal parts, one part comprising all values greater and the other all values less than the median.

or

The median of a series is the value of that actual or estimated when a series is arranged in ascending or descending order which divides the distribution into two equal parts.

Properties acre Merits of Median

1. It is easy to calculate and simple to understand. In some cases, it can be located simply by inspection.
2. It is a definite average.
3. It eliminates the effect of extreme values of variable by which it is not affected.
4. It is not capable of further algebraic treatment.
5. It can be determined graphically.
6. Median can be defined for qualitative data where it is possible to rank the observations in some order.
7. Median is especially useful in the case of open-end distributions.

Demerits of Median

1. It requires the data to be arrayed which is difficult if a number of data are large.
2. Our determination is not based on all the observations.
3. The median multiplied by the number of items does arithmetic means.

Median for Ungrouped Data

Method: Arrange the data in ascending or descending order of their magnitudes. Let the total number of observations be N.

1. If N is odd, then median = \(\left(\frac{N+1}{2}\right) \text { th term }\)
2. If N is even, then median = \(\frac{\frac{N}{2} \text { th term }+\left(\frac{N}{2}+1\right) \text { th term }}{2}\)

Median for Grouped Data

Method: We make the cumulative frequency table for the grouped data. Now we will find that class interval by dividing the total number of frequencies by 2 in which the median lies. This class interval is called the median class. Now we will use the following result to find the median:

Median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

where l₁ = lower limit of the median class

l₂ = upper limit of median class

i = l₂ – l₁

f = frequency of median class

N = sum of frequencies

C = cumulative frequency of the class preceding (just before) the median class

Remark:

If an inclusive (discontinuous) series is given then first of all we will prepare an exclusive series from the given inclusive series. To make it we subtract and add the same number i.e.,

⇒ \(\frac{\text { Lower limit of a class – Upper limit of previous class }}{2}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Median Solved Examples

Question 1. Find the median from the following table:

Solution:

Here, N = 65

∴ For median class \(\frac{N}{2}=\frac{65}{2}=32.5\)

∴ Median class = 20-30

Here l₁=20, l₂ = 30

⇒ i = 30-20= 10,

f= 18, C= 19

∴ Median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i=20+\frac{(32.5-19)}{18} \times 10=20+\frac{135}{18}=27.5\)

Question 2. Find the median from the following table:

Solution:

Here, N = 100

⇒ \(\frac{N}{2}=\frac{100}{2}=50\)

Median class = 40-55

Here l₁ = 40, l₂ = 55

⇒ i = 55 – 40 = 15, f = 44, C = 26

Now, Median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i=40+\frac{(50-26)}{44} \times 15\)

= \(40+\frac{15 \times 24}{44}=40+8.18=48.18\)

Question 3. Find the median from the following data:

Solution:

Converting into a simple frequency table :

Here, N = 600

⇒ \(\frac{N}{2}=\frac{600}{2}=300\)

∴ Median class = 40-50

and l₁ = 40, l₂ = 50, i = 50 – 40 = 10, f = 200, C = 236

Now, median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i=40+\frac{(300-236)}{200} \times 10\)

= \(40+\frac{10 \times 64}{200}=40+3.2=43.2\)

Question 4. Find the median for the following frequency distribution:

Solution:

First, we will convert the given data into exclusive form:

Here, N = 400

⇒ \(\frac{N}{2}=\frac{400}{2}=200\)

Median class is 165.5 – 168.5

Now, l₁ = 165.5, l₂ = 168.5, i = 1 68.5 – 165.5 = 3, f= 136, C = 132

and median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

= \(165.5+\frac{(200-132)}{136} \times 3\)

= \(=165.5+\frac{3 \times 68}{136}=167\)

Question 5. Find the median for the following data:

Solution:

Construct the following table from the given data:

Here, N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

The median class is 30 – 40

Now, l₁= 30, l₂ = 40, i = 40 – 30 = 10, f = 13, C = 22

and median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

= \(30+\frac{(25-22)}{13} \times 10=32.31\)

Question 6. Find the missing frequency if the median for the given distribution is 24:

Solution:

Here, N = 62 + p

⇒ Median = 24

⇒ The median class is 20-30

∴ l₁ = 20, l₂ = 30 ⇒ i = 30- 20 = 10

f = 25, C = 30

Now, Median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

⇒ \(24=20+\frac{\left(\frac{62+p}{2}-30\right)}{25} \times 10\)

⇒ \(4=\frac{10}{25}\left(\frac{62+p-60}{2}\right)\)

20 = p + 2 ⇒ p = 18

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Mode

Mode is the value of a variable that occurs most often, i.e., the value of the observation having the maximum frequency (most frequent item of the group).

Modal Class

The class having maximum frequency is called the modal class.

Computation of Mode for a Continuous Frequency Distribution

The formula for Calculating Mode:

Mode = \(l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h\)

Where, l = lower limit of modal class

l = frequency of modal class

h = width of the modal class

f₀ = frequency of the class preceding the modal class

f₂ = frequency of the class following the modal class

Step 1: Determine the class of maximum frequency.

Step 2: Obtain all values required in the formula.

Step 3: Substitute values in the formula and solve.

Mode Solved Examples

Question 1. Compute the mode for the following frequency distribution:

Solution:

The modal class is 20-30 as it has the maximum frequency.

∴ I = 20, f₁ = 28, f₀ = 16, f₂ = 20

and h = 10

Mode \(M=l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

Mode \(M=20+\frac{28-16}{2 \times 28-16-20} \times 10\)

= \(20+\frac{12}{56-16-20} \times 10=20+\frac{12}{20} \times 10=26 \)

Hence, mode = 26

Question 2. Calculate the value of mode for the following frequency distribution:

Solution:

The given data is in inclusive form. So, we convert it into exclusive form, as given below:

The modal class is 12.5 – 16.5 as it has the maximum frequency.

l =12.5, f₁ = 15, f₀ = 12, f₂ = 14, h = 4

Mode = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

Mode = \(12.5+\frac{15-12}{2 \times 15-12-14} \times 4\)

= \(12.5+\frac{3}{4} \times 4=12.5+3=15.5\)

Hence, mode = 15.5

Question 3. The mode of the following series is 17.3. Find the missing frequency:

Solution:

15-20 is the modal class as mode 17.3 lies in this class.

Here, l = 15, f₁ = 24, f₀=x (say), f₂ = 17 and h = 5 and mode = 17.3

∴ Mode(M) = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

⇒ \(17.3=15+\frac{24-x}{2 \times 24-x-17} \times 5\)

⇒ \(17.3=15+\frac{24-x}{31-x} \times 5\)

⇒ 2.3 (31 – x) = 120-5x

⇒ 2.3 x 31 – 2.3 x = 120 – 5x

⇒ 5x – 2.3 x = 120 – 71.3

⇒ 2.7 x = 48.7

∴ \(x=\frac{48.7}{2.7}=18.03\)

Hence, the missing frequency is 18.

Question 4. A survey regarding the heights (in cm) of 50 girls in class 10 of a school was conducted and the following data was obtained:

Find the mean, median, and mode of the above data.

Solution:

1. Mean: Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{7490}{50}=149.80\)

2. Median: Here, \(\frac{N}{2}=\frac{50}{2}=25\)

∴ The median class is 150 – 160

∴ Median = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

= \(150+\frac{25-22}{20} \times 10\)

= 150 + 1.5 = 151.5

3. Mode: The modal class is 150-160 as it has the maximum frequency.

∴ Mode = \(\text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h\)

= \(150+\left(\frac{20-12}{40-12-8}\right) \times 10\)

= \(150+\frac{8}{20} \times 10\)

= 150 + 4 = 154.

Hence, the mean height of the girls = 149.80 cm

the median height = 15 1.5 cm

Empirical Relationship Between Three Measures of Central Tendency

The relationship between mean, median, and mode is.

Mode = 3(Median) – 2(Mean)

and the modal height = 154 cm

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 – Empirical Relationship between Three Measures of Central Tendency Solved Examples

Question 1. Find the mean and median of the following frequency distribution:

Also, find the mode of the following data.

Solution:

We have

Let assumed mean A = 55, h= 10, Σf= 50 and Σfu = -27

Mean \(\bar{x}=A+\left[h \times \frac{\Sigma(f u)}{\Sigma f}\right]\)

⇒ \(\bar{x}=55+\left(10 \times \frac{-27}{50}\right)\)

⇒ \(\bar{x}=55-5.4\)

⇒ \(\bar{x}=49.6\)

Here, N = 50 \(\frac{N}{2}=\frac{50}{2}=25\)

Cumulative frequency greater than 25 is 36 and the corresponding class is 50-60.

∴ l₁=50,/= 12, l₂ = 60, C = 24, i = 60-50= 10

Now, \(\text { median }(M)=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

= \(50+\frac{10}{12}(25-24)=50+\frac{10}{12}\)

= 50 + 0.83 = 50.83

Mode = 3(Median) – 2(Mean)

Mode = 3 x 50.83- 2 x 49.60

= 152.49-99.20 = 53.29

Mode = 53.29

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Cumulative Frequency Curve (Or Ogive)

An ogive is a freehold graph showing the curve of a cumulative frequency distribution.

If we plot the points talcing the upper limit of the class intervals as x coordinates and their corresponding cumulative frequencies asy coordinates and then join these points by a free hand curve, the curve so obtained is called the cumulative frequency curve.

We may follow the following steps:

1. Step 1: Construct a cumulative frequency table.
2. Step 2: Mark the actual class limits along the X-axis.
3. Step 3: Mark the cumulative frequency of respective classes along the Y-axis.
4. Step 4: Plot the points corresponding to cumulative frequency at each upper limit point.
5. Step 5: Join the points plotted by a free-hand curve.

Less Than Series

When we mark the upper-class limits along the X-axis and corresponding cumulative frequency polygon along the 7-axis, then the obtained curve is called cumulative frequency curve or ogive for less than series.

Greater Than Series

When we mark the lower class limits along the X-axis and corresponding cumulative frequency polygon along the Y-axis, then the curve so obtained is called cumulative frequency curve or ogive for greater than series.

To Obtain Median From Cumulative Frequency Curve

Method-1

Step 1: Draw a cumulative frequency curve for the given frequency distribution.

Step 2: Locate \(\frac{N}{2}\) on the Y-axis. Let it be M.

Step 3: From point M draw a line parallel to X-axis cutting the curve at point N.

Step 4: From N, draw NP perpendicular to the X-axis cutting the X-axis at P.

The value of P is the median of the data.

Method-2

Step 1: Draw both ogives (less than series and more than series) on the same axis.

Step 2: From the point of intersection of these curves draw perpendicular X-axis. Let it meet the X-axis at point M. Value of M is the required median.

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 – To Obtain Median From Cumulative Frequency Curve Solved Examples

Question 1. Draw a less than cumulative frequency curve (ogive) for the following distribution:

Solution:

Taking upper-class limits along the X-axis and corresponding cumulative frequencies along the Y-axis mark the points (10, 7) (15, 16), (20, 28), (25, 36), and (30, 42).

Join the points marked by a free-hand curve.

Remark:

Students are advised to leave some points (one on two) without joining, which do not lie on the curve. It is not necessary to join all the points.

This will not affect the median. You can imagine that the given information is not correct, but you have to make a free-hand curve.

Question 2. In a study of the cases of diabetes the following data was obtained:

Draw a less than ogive for the above data.

Solution:

The given frequency distribution is discontinuous, to convert it into a continuous distribution, we subtract 0.5 from the lower limit and add 0.5 to the upper limit respectively. So, the continuous cumulative frequency distribution table is as follows:

Required ogive is given below:

Question 3. Construct a less than ogive and a more than ogive from the following data:

Solution:

Cumulative frequency distribution table according to ages (in years) less than and more than are:

Question 4. During the medical checkup of 35 students of a class, their weights were recorded as follows:

Draw a less than type and a more than type ogive from the given data. Hence obtain the median weight from the graph.

Solution:

1. Less than Series:

We may prepare the less-than series as follows:

Scale:\(\left\{\begin{array}{l}
\text { Along the } X \text {-axis, } 5 \text { small div. }=1 \mathrm{~kg} . \\
\text { Along the } Y \text {-axis, } 10 \text { small div. }=5 \mathrm{~kg} .
\end{array}\right.\)

We plot the points A(40, 3), B(42, 5), C(44, 9), D(46, 14), E(48, 28), F(50, 32) and G(52, 35).

Join AB, BC, CD, DE, EF, and FG with a free hand to get the curve representing ‘Less than series’.

2. More than Series:

We may prepare more than one series as follows:

On the same graph-paper as above we plot the points E(38, 35), Q(40, 32), R(42, 30), S(44, 26), 7(46, 21); (7(48, 7) and 7(50, 3). Join EQ, QR, RS, ST, TV, and UV with a free hand to get the curve representing ‘More than series’.

Scale:\(\left\{\begin{array}{l}
\text { Along the } X \text {-axis, } 5 \text { small div. }=1 \mathrm{~kg} \\
\text { Along the } Y \text {-axis, } 10 \text { small div. }=5 \text { students }
\end{array}\right.\)

The two curves intersect at point L. Draw LM ⊥ OX.

∴ Median weight = OM = 46.5 kg

Question 5. Find the median for the following distribution:

Solution:

Using the given data, first of all, construct a cumulative frequency table and then draw ogive (cumulative frequency curve).

Since, the no. of terms, n = 35

∴ Median = \(\left(\frac{n+1}{2}\right) \text { th term }\)

= \(\left(\frac{35+1}{2}\right) \text { th term }=18 \text { th term }\)

Through mark 18 on the Y-axis, draw a horizontal line that meets the curve at point A.

Through point A, on the curve, the draw meets the X-axis at point B. The value of point B on the X-axis is the median.

It is clear from the cumulative frequency curve drawn that the die median is 26.

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Excercise 14.1

Question 1. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?

Solution:

Mean \(\bar{x}=\frac{\sum f_i \cdot x_i}{\sum f_i}=\frac{162}{20}=8.1\)

∴ The mean number of plants per house = 8.1.

The direct method is used here because the values of X_i and f_i are very small.

Question 2. Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

Mean \(\bar{x}=\frac{\sum f_i \cdot x_i}{\sum f_i}=\frac{27260}{50}=545.20\)

∴ Mean wages of workers = ₹ 545.20

Question 3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f:

Solution:

Now, mean pocket allowance = \(\frac{\sum f_i \cdot x_i}{\sum f_i}\)

⇒ \(18=\frac{752+20 f}{44+f}\)

⇒ 792 + 18f = 752 + 20f

⇒ 792 – 752 = 20f – 18f

⇒ 40 = 2f ⇒ f = 20

The missing frequency f = 20

Question 4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Solution:

Let assumed mean A = 75.6

∴ Mean heartbeats per minute

⇒ \(\bar{x}=A+\frac{\sum f_i \cdot d_i}{\sum f_i}=75.5+\frac{12}{30}\)

= 75.5 + 0.4 = 75.9

The mean heartbeats per minute for these women = 75.9

Question 5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following was the distribution of mangoes according to the number of boxes:

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution:

Here, h = 3

Let assumed mean A = 57

∴ Mean \(\bar{x}=A+h \cdot \frac{\sum f_i \cdot u_i}{\sum f_i}\)

= \(57+\frac{3 \times 25}{400}=57+0.1875\)

= 57.1875

∴ Mean number of mangoes = 57.1875

≈ 57.19

Here, the assumed mean method is used.

Question 6. The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.

Solution:

Here, h = 50

Let assumed mean A = ₹ 225

∴ Mean \(\bar{x}=A+h \cdot \frac{\sum f_i \cdot u_i}{\sum f_i}\)

= \(225+50 \times \frac{(-7)}{25}\)

= 225 – 14 = ₹ 221

Therefore, the mean daily expenditure on food per family = ₹ 211.

Question 7. To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.

Solution:

∴ Mean \(\bar{x}=\frac{\sum f_i \cdot x_i}{\sum f_i}=\frac{2}{30}\)

= 0.0986 parts per million.

The mean concentration of SO₂ in the air = 0.0986 parts per million.

Question 8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:

Mean \(\bar{x}=\frac{\sum f_i \cdot x_i}{\sum f_i}=\frac{499}{40}=12.475\)

∴ The mean number of absences of students = 12.475 days per student

Question 9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:

Mean \(\bar{x}=\frac{\sum f_i \cdot x_i}{\sum f_i}=\frac{2430}{35}=69.43\)

∴ The mean percentage rate of literacy = 69.43

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.2

Question 1. The following table shows the ages of the patients admitted to a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:

Lot assumed mean A = 40

∴ Mean \(\bar{x}=A+\frac{\sum f_1 \cdot d_i}{\sum f_i}=40+\frac{(-370)}{80}\)

= 40 – 4.625

= 35.375 years

For more,

Highest frequency = 23

∴ Modal class: 35 – 45

Now, l₁= 35, l₂ = 45

f = 23, f₁ =21, f₂ = 14

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(35+\frac{(23-21)(45-35)}{46-21-14}\)

= \(35+\frac{2 \times 10}{11}\)

= 35. 1.8 = 36.8 years.

Mode = 36.8 years.

Question 2. The following data gives information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetime of the components.

Solution:

Highest frequency = 61

∴ Modal class: 60 – 80

Now, l₁ = 60, l₂ = 80

f = 61, f₁ = 52, f₂ = 38

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(60+\frac{(61-52)(80-60)}{2 \times 61-52-38}\)

= \(\begin{equation}
60+\frac{9 \times 20}{32}=60+5.625
\end{equation}\)

= 65.625 hours

Question 3. The following data gives the distribution of the total monthly household expenditure of 200 families in a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:

Highest frequency = 40

∴ Modal class: 1500-2000

Now, l₁ = 1500, l₂ = 2000

f = 40, f₁ = 24, f₂= 33

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(1500+\frac{(40-24)(2000-1500)}{2 \times 40-24-33}\)

= \(1500+\frac{16 \times 500}{23}\)

= 1500 + 347.83 = ₹ 1847.83

Let assumed A = ₹ 2750

Here, h = 500

Mean Monthly expenditure

⇒ \(\bar{x}=A+h \cdot \frac{\sum f_i \cdot u_i}{\sum f_i}\)

⇒ \(2750+\frac{500 \times(-35)}{200}\)

= 2750 – 87.50 = ₹ 2662.50

Mean Monthly expenditure = ₹ 2662.50

Question 4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools in India. Find the mode and mean of this data. Interpret the two measures.

Solution:

Here, maximum frequency = 10

∴ Model class: 30-35

Now, l₁ = 30, l₂ = 35

f = 10, f₁ = 9, f_i2 = 3

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(30+\frac{(10-9)(35-30)}{2 \times 10-9-3}\)

= \(30+\frac{5}{8}=30+0.625\)

= 30.625 students per teacher

For mean,

let assumed mean A = 27.5

Now, mean = \(A+\frac{\sum f_i \cdot d_i}{\sum f_i}=27.5+\frac{60}{35}\)

= 27.5 + 1.7

= 29.2 students per teacher

So, in the higher secondary schools of India, the mean number of students per teacher is 29.2 states while the number of students per teacher in maximum states is 30.625.

Question 5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.

Solution:

Here, maximum frequency = 18

∴ Modal class: 4000 – 5000

Now, l₁ = 4000, l₂ = 5000

f = 18, f₁ = 4, f₂ = 9

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(4000+\frac{(18-4)(5000-4000)}{2 \times 18-4-9}\)

= \(4000+\frac{14 \times 1000}{23}\)

= 4000 + 608.7 = 4608.7

≈ 4608 run (approximately)

The mode of the data = 4608 run (approximately)

Question 6. A student noted the number of cars passing through a spot on the road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Solution:

Here, maximum frequency = 20

∴ Model class: 40-50

Now, l₁ = 40, l₂ = 50

f = 20, f₁ = 12, f₂ = 11

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(40+\frac{(20-12)(50-40)}{2 \times 20-12-11}\)

= \(40+\frac{8 \times 10}{17}=40+4.7\)

= 44.7

The mode of the data = 44.7

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.3

Question 1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean, and mode of the data and compare them.

Solution:

Let assumed mean A = 115

Here, h = 20

Here, N = 68

⇒ \(\frac{N+1}{2}=\frac{68+1}{2}=34.5\)

∴ Median class = 125 – 145

Now, l₁ = 125, l₂ = 145

f = 20, C = 22

and median (M) = \(l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

= \(125+\frac{\left(\frac{68}{2}-22\right)(145-125)}{20}\)

= \(125+\frac{12 \times 20}{20}\)

= 125 + 12 = 137

Mean \(\bar{x}=A+h \cdot \frac{\sum f_i \cdot d_i}{\sum f_i}\)

= \(115+\frac{20 \times 75}{68}\)

= 115 + 22.06 = 137.06

Again, maximum frequency = 20

∴ Modal class: 125-145

Now, l₁ = 125, l₂= 145

f = 20, f₁ = 13, f₂ = 14

∴ Mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(125+\frac{(20-13)(145-125)}{2 \times 20-13-14}\)

= \(125+\frac{7 \times 20}{13}\)

= 125 + 10.77 = 135.77

The values of median, mean and mode are equal approximately.

Question 2. If the median of the distribution given below is 28.5. find the value of x and y:

Solution:

Now, 45 + x + y = 60

⇒ x + y = 15 → (1)

Given that, median = 28.5

∴ Median class: 20 – 30

Now, l₁ = 20, l₂ = 30

f = 20, c = 5 + x, N = 60

Median = \(l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

⇒ \(28.5=20+\frac{\left(\frac{60}{2}-5-x\right)(30-20)}{20}\)

⇒ \(28.5-20=\frac{(25-x)(10)}{20}\)

⇒ \(8.5=\frac{25-x}{2} \quad ⇒ \quad 17=25-x\)

⇒ x = 25 – 17 = 8

From equation (1),

8 + y = 15

⇒ y = 15 – 8 = 7

∴ x = 8, y = 7

The value of x and y is 8 and 7.

Question 3. A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons aged 18 years onwards but less than 60 years.

Solution:

N = 100

∴ \(\frac{N+1}{2}=\frac{100+1}{2}=50.5\)

⇒ Median class: 35 – 40

∴ l₁ = 35, l₂ = 40

f = 33, C = 45

Median = \(l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

= \(35+\frac{\left(\frac{100}{2}-45\right)(40-35)}{33}\)

= \(35+\frac{5 \times 5}{33}=35+0.76\)

= 35.76 years.

Median = 35.76 years.

Question 4. The lengths of40 the leaves of a plant are measured correctly to the nearest millimeter and the data obtained is represented in the following table:

Find the median length of the leaves.

[Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to 117.5-126.5, 126.5-135.5, …, 171.5 – 180.5.]

Solution:

∴ \(N=40 \Rightarrow \frac{N+1}{2}=\frac{40+1}{2}=20.5\)

⇒ Median class: 144.5 – 153.5

l₁ = 144.5, l₂ = 153.5

f = 12, C = 17

and median = \(l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

= \(144.5+\frac{\left(\frac{40}{2}-17\right)(153.5-144.5)}{12}\)

= \(144.5+\frac{3 \times 9}{12}=144.5+2.25\)

= 146.75 mm

Therefore, the median length of leaves = 146.75 mm

Question 5. The following table gives the distribution of the lifetime of 400 neon lamps:

Find the median lifetime of a lamp.

Solution:

Here, N = 400

∴ \(\frac{N+1}{2}=\frac{400+1}{2}=200.5\)

Median class: 3000 – 3500

Now, l₁ = 3000, l₂ = 3500

f = 86, C = 130

and median = \(=l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

= \(=3000+\frac{\left(\frac{400}{2}-130\right)(3500-3000)}{86}\)

= \(3000+\frac{70 \times 500}{86}\)

= 3000 + 406.98 = 3406.98 hours

Therefore, the mean lifetime of lamps = 3406.98 hours.

Question 6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.

Solution:

N = 100

∴ \(\frac{N+1}{2}=\frac{100+1}{2}=50.5\)

⇒ Median class: 7 – 10

Now, l₁ = 7, l₂ = 10

f = 40, C = 36

∴ Median = \(l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

= \(7+\frac{\left(\frac{100}{2}-36\right)(10-7)}{40}\)

= \(7+\frac{14 \times 3}{40}\)

= 7 + 1.05 = 8.05

Mean = \(\frac{\sum f_i \cdot x_i}{\sum f_i}=\frac{832}{100}=8.32\)

Fir mode,

Maximum frequency = 40

∴ Modal class 7 – 10

Now, l₁ = 7, l₂ = 10

f = 40, f₁ = 30, f₂ = 16

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(7+\frac{(40-30)(10-7)}{2 \times 40-30-16}\)

= \(7+\frac{10 \times 3}{34}\)

= 7 + 0.88 = 7.88

Mode = 7.88

Question 7. The distribution below gives the weights of 30 students in a class. Find the median weight of the students.

Solution:

Here, N = 30

∴ \(\frac{N+1}{2}=\frac{30+1}{2}=15.5\)

⇒ Median class: 55 – 60

Now, l₁ = 55, l₂ = 60, f = 6, C= 13

and median = \(l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

= \(55+\frac{\left(\frac{30}{2}-13\right)(60-55)}{6}\)

= \(55+\frac{2 \times 5}{6}\)

= 55 + 1.67

= 56.67 kg

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.4

Question 1. The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less type cumulative frequency distribution, and draw its ogive.

Solution:

Cumulative frequency curve (ogive)

Question 2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution:

To find the median:

1. Take the point \(\left(0, \frac{N}{2}\right) \equiv\left(0, \frac{35}{2}\right)\) on Y-axis.
2. Draw a line parallel to the X-axis from points \(\left(0, \frac{35}{2}\right)\) which is the curve at point P.
3. Find the abscissa of point P (read). It is 46.8.
4. Therefore, the required median = 46.8 kg.

Question 3. The following table gives the production yield per hectare of the heat of wheat of 100 farms in a village.

Change the distribution to a more than type distribution, and draw its ogive.

Solution:

Convert the given distribution to the ‘more than’ type distribution.

Distribution of “More than” type

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Multiple Choice Questions And Answers

Question 1. In the formula \(\bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i}\) to determine the mean of grouped data, di is the deviation from and the following:

1. The lower limit of classes
2. The upper limit of classes
3. Mid-point of classes
4. Frequencies of classes

Answer: 3. Mid-point of classes

Question 2. If x_i is the mid-points of class intervals of grouped data and f_i are their corresponding frequencies, then the value of Σ(f_i – x_i – \(\bar{x}\)) is:

1. -1
2. 0
3. 1
4. None of these

Answer: 2. 0

Question 3. For the distribution

the sum of the upper limits of the median class and modal class is:

1. 70
2. 50
3. 60
4. 30

Answer: 1. 70

Question 4. The abscissa of the point of intersection of cumulative frequency curves of ‘less than type’ and ‘more than type’ gives the ______ of the data:

1. Mean
2. Median
3. Mode
4. All of these

Answer: 2. Median

Question 5. A cumulative frequency curve is necessary in

1. Mean
2. Median
3. Mode
4. All of these

Answer: 2. Median

Question 6. For the distribution

the upper limit of the median class is:

1. 19
2. 19.5
3. 29
4. 29.5

Answer: 4. 29.5

Question 7. In the distribution

the modal class is:

1. 20 – 30
2. 30 – 40
3. 40 – 50
4. 50 – 60

Answer: 3. 40 – 50

Question 8. The modal class of the following frequency distribution will be

1. 1 – 3
2. 3 – 5
3. 5 – 7
4. 7 – 9

Answer: 2. 3 – 5

Question 9. The arithmetic mean of 3, 4, 6, and x is 5. The value of x will be:

1. 5
2. 2
3. 7
4. 3

Answer: 3. 7

Question 10. The arithmetic mean of positive even numbers from 1 to 10 will be:

1. 2
2. 4
3. 6
4. 5

Solution: 3. 6

Question 11. 1 hejnean ol n observations is \(\bar{x}\). If each observation is increased by a, then the new mean will be:

1. \(\bar{x}\)+ a
2. \(\bar{x}\)+ a/2
3. \(\bar{x}\)– a
4. \(\bar{x}\)– a/2

Answer: 1. \(\bar{x}\)+ a

Question 12. For a frequency distribution, the relation between mean, median, and mode is:

1. Mode = 3 mean – 2 median
2. Mode = 2 median – 3 mean
3. Mode = 3 median – 2 mean
4. Mode = 3 median + 2 mean

Answer: 3. Mode = 3 median – 2 mean

Question 13. The mean of natural number from 1 to 9 will be:

1. 3
2. 5
3. 8
4. 9

Answer: 2. 5

Question 14. For a frequency distribution, the mean is 20.5 and the median is 20. Its mode will be:

1. 21.5
2. 17
3. 19
4. 20.5

Answer: 3. 19

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Introduction

In our daily lives, we come across some situations, for which we do not have any definite answer, although we know what the possible results (outcomes) are.

Probability For Example :

1. When a coin is tossed, it is certain that it will come down but it may turn up a head or may not turn up a head.
2. When a die is thrown, it may turn up a particular number 3 or not.
3. It may rain today.
4. The above situations predict die uncertainty (chance), therefore all are referred to a phenomenon which may or may not occur. The probability estimates the degree of uncertainty regarding the happening of a given phenomenon.

Read and Learn More Class 10 Maths Solutions Exemplar

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Experiment

An operation which gives some outcomes (results) under some given conditions is known as the ‘experiment’.

For Example:

1. “Tossing a coin” is an experiment with two outcomes: head and tail.
2. “Throwing a die” is an experiment with six outcomes: 1,2, 3, 4, 5 and 6. The plural of die is dice.

Probability Trial

Experimenting repeatedly under similar conditions is called a trial.

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Sample Space

The set (collection) of all possible outcomes of usually denoted by S.

For Example:

(1). In tossing a coin ← experiment

Sample space denoted by S=\(\left\{\begin{array}{cc}
H & T \\
(\text { Head ) } & \text { (Tail) }
\end{array}\right\}\)

(2). In throwing a die ← experiment

S = {1,2, 3, 4, 5, 6}

(3). In tossing a coin 3 times ← experiment

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

(4). In throwing two dice simultaneously ← experiment

S=\(\left\{\begin{array}{l}
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
\end{array}\right\}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Event

Any outcome or combination of some outcomes constitutes an event. Each event is a subset (part) of the sample space.

For Example:

(1). Let S = {1, 2, 3, 4, 5, 6}, when a die is thrown and let E₁ is the event of “getting an odd number”.

∴ E₁ = {1, 3, 5}

(2). Let E₂ be the event of “getting an even number”.

∴ E₂ = {2, 4, 6}

(3) Let E₃ be the event of “getting a prime number”.

∴ E₃ = {2,3, 5}

All above events are part of a subset of a sample space.

Remark: Total number of elements in a sample space = (Number of outcomes of an event)No‘of times

For Example:

(1). In tossing a coin 3 times,

The total number of elements in a sample space denoted by

n(S) = (2)^{3 → No.of times}

Possible outcomes of a coin = 8

(2). In throwing 2 dice,

n(S) = (6)^{2 → No.of times}

Possible outcomes of an experiment = 36

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Simple Event Or Elementary Event

An event is called a simple event if it contains only one element. In terms of set, it is a singleton set.

For Example:

(1) In “tossing a coin”, sample space S = {H, T}

Let A = event of getting a head = {H} and B = event of getting a tail = {T}

Here A and B are simple events.

(2). In “tossing a coin 2 times”, S = {(H, H), (H, T), (T, H), (T, T)}

Let A = event of getting a head on the first coin and getting a tail on the second coin = {(H, T)}

B = event of getting a head on the first coin = {(H, H), (H, T)}

Here, A is a simple event but B is not a simple event as B has two sample points.

To understand the experiment, sample space and event, are given below.

It is necessary that event is a part of the sample space

About the Playing Cards

1. A pack of playing cards has 52 cards in all.

2. It has 2 Colours – [Red (26 cards); Black (26 cards)]

3. It has 4 suites

Numbering on each card of each suit as :

Actually, A, J, Q and IC are written in place of 1, 11, 12 and 13.

So, in a pack of 52 cards, 4 cards of the same number are there.

Note: Kings, Queens and Jacks are called the face cards while the aces together with face cards are called the honour cards.

For Example: Number 2 will be in spade, club, heart as well as in diamond. Similarly, the number IC will be in spade, dub, heart and diamond also.

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Equally Likely Events

The events are said to be equally likely when we have no reason to believe that one is more likely to occur than the other. Each of the possible outcomes has an equal chance of occurring.

For Example:

1. In “rolling a die”. There are six possible outcomes in the sample space, each is equally likely to occur.
2. In “tossing a coin”, getting a tail or a head are equally likely events.

Now, let us try to understand the meaning of probability.

(Probability: probable, perhaps, possibly, chances etc.)

For this, first, we see some experiments.

(1 ). In tossing a coin, S = {H, T}

What are the chances of getting a head—only {H}?

You will say 50% or fifty-fifty—Correct.

∴ \(50 \% \text { means } \frac{50}{100}, \text { i.e., } \frac{1}{2}=\frac{n(\text { favourable cases })}{n(S)}\)

Our favourable case is getting a head ⇒ n(E) = 1

and the total number of outcomes in a sample space ⇒ n(S)= 2

(2). In throwing a die, S = {l, 2, 3, 4, 5, 6}

What are the chances of getting an even number, i.e., {2,4, 6}?

Again, you will say 50% or fifty-fifty—Correct.

∴ Probability of an event =\(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)

Now, tell me O’ girl!

What is the probability of getting a head exactly 2 times when we toss a coin 3 times simultaneously?

Now, you see example (1) of sample space and give the correct answer.

Yes! Your answer is correct. It is \(\frac{3}{8}\)

Now, you can make understand to your friend easily that
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

In these 8 outcomes, only 3 underlined contain exactly two heads.

∴ Our favourable cases = 3 and total number of outcomes = 8

∴ Required probability = \(\frac{3}{8}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Minimum And Maximum Values Of Probability Of An Event

The value of the probability of any event always lies between 0 and 1. When the probability of an event is 0, it is said to be an impossible event and when the probability of an event is l, it is said to be a sure event.

Explanation: We know that a set (collection) of favourable outcomes is a part of the whole sample space.

∴ Number of favourable cases \(\leq\) Total number of cases.

⇒ \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }} \leq 1\) → (1)

Now, see the following example :

“In throwing a die numbered l to 6”, find the probability of getting a number greater than 6”.

Here there is no such possibility to get a number which is greater than 6.

Favourable number of cases ≥ 0

So. always a number of favourable cases \(\leq\) 0

∴ \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }} \leq 0\)

Combining Eqs. (1) and (2), we get

∴0 \(\leq\) Probability of an event \(\leq\) 1

Remember: In each experiment, we assume that the chance of each outcome is the same.

• Die or die must be fair.
• The coin must be unbiased (We do not consider the possibility that the coin can land on its side). We will consider that coin will land on any one of its faces, i.e., either head or tail.

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Theorems Of Addition Of Probabilities

If an event £ can happen in any one of the n different ways with the probabilities P₁, P₂, P₃ …. respectively, then the probability for the event £ is equal to the sum of the individual probabilities P₁, P₂, P₃, …Pn

For Example:

When a die is rolled, then any one of the numbers,1,2, 3, 4, 5 or 6 can be shown up.

Total number of outcomes = 6

Number of even numbers (2, 4 or 6)= 3

∴ Probability of getting an even number = \(\frac{\text { Favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{6}=\frac{1}{2}\)

Also, Probability of getting 2 = \(\frac{1}{6}\)

Probability of getting 4 = \(\frac{1}{6}\)

Probability of getting 6 = \(\frac{1}{6}\)

∴ Probability of getting an even number = \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) = \(\frac{1}{2}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Complementary Events

If E is an event, then the event ‘not £’ is a complementary event of E.

For Example:

When we throw a die, let E be the event getting a number less than or equal to 2, then the event ‘not E’, i.e., getting a number greater than 2 is a complementary event of E. Complementary of E is denoted by E or \((\bar{E})\)

∴ \(P(E)+P(\bar{E})\)=1

Let E be an event, then the number of outcomes favourable to E is greater than or equal to zero and is less than or equal to the total number of outcomes. It follows than

∴ \(0 \leq P(E) \leq 1\).

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Exhaustive Events

All the possible outcomes of any trial taken together are called exhaustive events.

For Example:

From a group of 2 boys and 3 girls, two children are selected at random. Describe the events:

S = {B₁B₂, B₁G₁, B₁G₂, B₂G₂, B₂G₁,B₂G₂, B₂G₂, G₁G₂, G₂G₃, G₁G₃}

E₁ = event that both the selected children are girls

= {G₁G₂, G₂G₃, G₁G₃}

E₂ = event that the selected children consist of 1 boy and 1 girl

= {B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₂}

E₃ = event that at least one boy is selected

= {B₁B₂, B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₃}

Here E₁ and E₂ together do not constitute S.

∴ E₁ and E₂ are not exhaustive events,

and E₂ and E₃ also do not form the whole sample space S.

∴ E₂ and E₃ are not exhaustive events.

But E₁ and E₃ together form a whole sample space.

∴ E₁ and E₃ are exhaustive events.

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Solved Examples

Question 1. In any situation that has only two possible outcomes, each outcome will have probability \(\frac{1}{2}\). Find whether it is true or false.

Solution:

False,

The probability of each outcome will be \(\frac{1}{2}\) only when the two outcomes are equally likely.

For Example:

On the eve of Deepawali, we are lighting the candles in night on the roof.

The candles may be “lightening off due to heavy rain” or “they may not be lightening off” are not equally likely events because they do have not equal chances of occurring.

The possibility of falling rain at the time of the festival Deepawali is very less.

Question 2. A marble is chosen at random from 6 marbles numbered 1 to 6. Find the probability of getting a marble having number 2 and 6 on it.

Solution:

Given

A marble is chosen at random from 6 marbles numbered 1 to 6.

The favourable case is to get a marble on which both numbers 2 and 6 are written. But there is no such marble.

So, n(E) = 0 and n(S) = 6

∴ Required probability =\(\frac{n(E)}{n(S)}=\frac{0}{6}\)=0

Question 3. A marble is chosen at random from 6 marbles numbered 1 to 6. Find the probability of getting a marble having the number 2 or 6 on it.

Solution:

Given

A marble is chosen at random from 6 marbles numbered 1 to 6.

Here n(E) = 2   [either 2 or 6]

and n(S) = 6

∴ Required probability = \(\frac{2}{6}=\frac{1}{3}\)

Question 4. A fair coin is tossed

1. List the sample space.
2. What is P(tail)?

Solution:

(1). When a fair coin is tossed, there are two possible outcomes. Either the head appears on the uppermost face or the tail appears on the uppermost face.

S = {H, T}; n {S} = 2

(2). \(P(\text { tail })=\frac{\text { No. of favourable outcomes }}{\text { No. of possible outcomes }}=\frac{1}{2}\)

Question 5. A fair die is tossed. List the sample space. State the probability of the event :

1. A number less than 3 appears
2. A number greater than 3 appears
3. A number greater than or equal to 3 appears

Solution:

When a fair die is tossed, the number of possible outcomes is 1, 2, 3, 4, 5 and 6.

Sample space S = {1,2, 3,.4, 5, 6); n (S) = 6

(1). Let A be the event which denotes “a number less than 3 appears”.

A= {1,2}, n(A) = 2;

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)

(2). Let E be the event which denotes “a number greater than 3” appears.

E= {4,5,6}; n{E) =3

⇒ P(E)=\(\frac{n(E)}{n(S)}=\frac{3}{6}=\frac{1}{2}\)

(3). Let B be the event in which “a number greater than or equal to 3” appears.

B = {3, 4, 5, 6}; n(B) = 4

⇒ P(B)=\(\frac{n(B)}{n(S)}=\frac{4}{6}=\frac{2}{3}\)

Question 6. Two dice are thrown simultaneously. Find the probability of getting:

1. An even number is the sum
2. An even number as the product
3. The sum as a prime number
4. A total of at least 10
5. A doublet.

Solution:

Elementary events associated with the random experiment of throwing two dice are :

{1,1} {1,2} {1,3} {1,4} {1,5} {1,6}

{2,1} {2,2} {2,3} {2,4} {2,5} {2,6} .

{3,1} {3,2} {3,3} {3,4} {3,5} {3,6}

{4,1} {4,2} {4,3} {4,4} {4,5} {4,6}

{5,1} {5,2} {5,3} {5,4} {5,5} {5,6}

{6,1} {6,2} {6,3} {6,4} ‘{6,5} {6,6}

∴ Total number of elementary events = 6 x 6 = 36.

(1). Let A be the event of getting an even number as the sum, i.e., 2, 4, 6, 8, 10, 12.

The sum is even.

∴ Sum may be 2 or 4 or 6 or 8 10 or 12.

So, elementary events favourable to events are (1,1), (1,3), (3,1),(2,2), (1,5), (5, 1), (2, 4), (4, 2),'(3, 3), (2, 6), (6, 2), (4, 4), (5, 3), (3, 5), (5, 5), (6, 4), (4, 6) and (6, 6). C

clearly, a favourable number of outcomes =18

Hence, required probability = \(\frac{18}{36}=\frac{1}{2}\)

(2). Let B be the event of getting an even number as the product, i.e., 2, 4, 6, 8, 10, 12, 16, 18,20,24,30,36.

∴ Elementary events favourable to event £ are : (1, 2), (2, 1), (1, 4), (4, 1), (2, 2), (1, 6), (6, 1), (2, 3), (3, 2), (2, 4), (4, 2), (2, 5), (5, 2), (2, 6), (6, 2), (3, 4), (4, 3), (4, 4), (3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 6), (6, 5), (6, 6).

∴ Favourable number of outcomes = 27

Hence, required probability = \(\frac{27}{36}=\frac{3}{4}\)

(3). Let C be the event of getting the sum as a prime number, i.e., 2, 3, 5, 7, 11.

Elementary events favourable to event C are :(1, 1), (1 2), (2, 1),(1, 4), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (6, 5), (5, 6)

∴ Favourable number of outcomes =15

Hence, required probability = \(\frac{15}{36}=\frac{5}{12}\)

(4). Let D be the event of getting a total of at least 10, i.e., 10, 11, 12.

Then, the elementary events favourable to D are (6, 4), (4, 6), (5, 5), (6, 5), (5, 6) and (6, 6).

∴ Favourable number of outcomes = 6

Hence, required probability = \(\frac{6}{36}=\frac{1}{6}\)

(5). Let E be the event of getting a Doublet, i.e., (1,1), (2,2), (3,3), (4, 4), (5,5), (6,6).

Clearly, a favourable number of outcomes = 6

∴ \(P(E)=\frac{6}{36}=\frac{1}{6}\)

Question 7.

• A die is thrown once. Find the probability of getting :
  1. A prime number
  2. A number between 3 and 6
  3. A number greater than 4
  4. A number at most 4
  5. A factor of 6.
• When two dice are thrown find the probability of obtaining :
  1. A total of 6
  2. A total of 10
  3. Same number on both dice
  4. A total of 9.
• Two different dice are thrown together. Find the probability that the numbers obtained :
  1. Have a sum of less than 7
  2. Have a product of less than 16
  3. Is a doublet of odd numbers.

Solution:

1. As a die is rolled once, therefore there are six possible outcomes, i.c., 1,2,3, 4, 5, 6.

(1). Let A be an event “getting a prime number”.

Favourable cases for a prime number are 2, 3, 5,

i.e., n(A) = 3

Hence, P(A)=\(\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}\)

(2). Let A be an event “getting a number between 3 and 6”.
Favourable cases for events A are 4 or 5.

i.e., n(A) = 2

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)

(3). Let A be an event “a number greater than 4”.

Favourable cases for events A are 5 and 6.

i.e., n(A) = 2

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)

(4). Let A be the event of getting a number at most 4.

∴ A = {1,2, 3, 4} ⇒ n(A) = 4, n(S) = 6

∴ Required probability = \(\frac{n(A)}{n(S)}=\frac{4}{6}=\frac{2}{3}\)

(5). Let A be the event of getting a factor of 6.

∴ A = {1,2, 3, 6} r ⇒ n(A) = 4, n(S) = 6

∴ Required probability = \(\frac{n(A)}{n(S)}=\frac{4}{6}=\frac{2}{3}\)

2. Since a pair of dice is thrown once, so there are 36 possible outcomes.

i.e., n(S) = 36

(1). Let A be an event “a total 6”. Favourable cases for a total of 6 are (2, 4), (4, 2), (3, 3),(5, 1), (1,5).

i.e., n(A) = 5

Hence, P(A)=\(\frac{n(A)}{n(S)}=\frac{5}{36}\)

(2). Let A be an event “a total of 10”. Favourable cases for a total of 10 are (6, 4), (4, 6),(5,5).

i.e., n(A) = 3

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{3}{36}=\frac{1}{12}\)

(3). Let A be an event “the same number of both the dice”. Favourable cases for the same number on both the dice are (1, 2), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).

i.e.. n(A)= 6

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

(4). Let A be an event “of getting a total of 9”. Favourable cases for a total of 9 are (3,6), (6,3), (4,5), (5,4). ”

i.e., n(A) = 4

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

3. We have, n(S) = 36

(1). Let A be an event “a sum less than 7” i.e., 2, 3, 4, 5, 6.

Favourable cases for a sum less than 7 are : (1, 1), (1, 2), (2, l),(l,3), (3, 1), (2, 2), (1,4), (4, 1), (2, 3), (3, 2), (1,5), (5, 1), (2, 4), (4, 2), (3, 3)

i.e., n(A) = 15

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)

(2). Let/1 be an event “product less than 16” i.e., 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15.

Favourable cases for a product less than 16 are : (1, 1), (1, 2), (2, 1),(1, 3), (3, 1), (1, 4), (4, 1), (2, 2), (1, 5), (5, 1), (1, 6), (6, 1), (2, 3), (3, 2), (2, 4), (4, 2), (3, 3), (2, 5),(5, 2), (2, 6), (6, 2), (3, 4), (4, 3), (3, 5), (5, 3).

n(A) = 25

Hence, P(A)=\(\frac{n(A)}{n(S)}=\frac{25}{36}\)

(3). Let A be an event “a doublet of odd numbers”.

Favourable cases for a doublet of odd numbers are (1, 1), (3, 3), (5, 5)

i.e., n(A) = 3

Hence P(A)=\(\frac{n(A)}{n(S)}=\frac{3}{36}=\frac{1}{12}\)

Question 8. A standard deck of 52 cards is shuffled. Ritu draws a single card from the deck at random. What is the probability that the card is a jack?

Solution:

Given

A standard deck of 52 cards is shuffled. Ritu draws a single card from the deck at random.

S = Sample space of all possible outcomes, or 52 cards.

Thus, n(S) = 52

E = Event of selecting a jack. There are four jacks in the deck; jack of hearts, of diamonds, of spades, and of clubs.

Thus n(E) = 4 .

∴ P(E)=\(\frac{\text { Number of possible jacks }}{\text { Total number of possible cards }}=\frac{4}{52}=\frac{1}{13}\)

Hence, the probability that the card is jack is \(\frac{1}{13}\)

Question 9. A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:

1. Red
2. Black or white
3. Not black.

Solution:

Given

A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random.

Total number of balls = 5 + 7 + 3= 15

1. Number of red balls = 7

∴ P(drawing a red ball) = \(\frac{7}{15}\)

2. Number of black or white balls = 5 + 3 = 8

∴ P(drawing a black or white ball) =\(\frac{8}{15}\)

3. Number of balls which are not black =15-5 = 10

∴ P(drawing a ball that is not black) = \(\frac{10}{15}\) = \(\frac{2}{3}\)

Hence, the probability of getting a red ball, a black or white ball and not a black ball are \(\frac{7}{15}, \frac{8}{15}\) and \(\frac{2}{3}\) respectively.

Question 10. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?

Solution:

Given

The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18

Total number of apples in a heap = 900

Let the number of rotten apples in a heap = x

Number of rotten apples = \(\frac{\text { Number of rotten apples }}{\text { Total number of apples in a heap }}\)

Now, probability (rotten apple) = \(0.18=\frac{x}{900}\)

∴ x = 900 x 0.18 = 162

∴ Number of rotten apples in a heap of apples =162

Question 11. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing:

1. An ace
2. ‘2’ of spades
3. ‘10’ of a black suit.

Solution:

1. The number of possible outcomes of drawing an ace is 4 as the number of aces in the deck is 4.

The total number of outcomes relating to drawing a card from a deck of 52 cards is 52.

∴ P(drawing an ace) = \(\frac{4}{52}=\frac{1}{13}\)

2. As there is only one card ‘2’ of spades, there is one favourable outcome only.

∴ Probability of drawing ‘2’ of spades = \(\frac{1}{52}\)

3. Since, there are two cards of TO’ of a black suit, one 10 of spade and the other 10 of the club, the probability of drawing 10 of a black suit = \(\frac{2}{52}=\frac{1}{26}\)

Question 12. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

1. A two-digit number,
2. A number divisible by 5.

Solution:

Given

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box

We have, n(S) = 90

(1) Let A be the event of getting “a two-digit number”.

∴ Favourable cases are 10, 11, 12, 13, 14, …, 90.

∴ n(A)=81

∴ P(A)=\(\frac{n(A)}{n(S)}=\frac{81}{90}=\frac{9}{10}\)

(2) Let B be the event of getting “a number divisible by 5 “.

∴ Favourable cases are 10,15,20,25,30, …, 90.

Let there be n in numbers.

T_n =90

∴ 10+(n-1) 5 =90 [\(a_n\) of A.P.=a+(n-1)d]

⇒ (n-1) 5 =80 ⇒ n-1 = 16 ⇒ n=17

∴ n(B) =17

∴ P(B) =\(\frac{n(B)}{n(S)}=\frac{17}{90}\)

Question 13. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to all of the numbers 1,2, 3, …,12. What is the probability that it will point to

1. 10
2. An odd number
3. A number which is a multiple of 3?

Solution:

Given

A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to all of the numbers 1,2, 3, …,12.

n(S) = Total number of elementary events = 12.

1. Favourable number of elementary events that point to 10 = 1

∴ Required probability = \(\frac{1}{12}\)

2. Let A be the event “point of the arrowhead towards an odd number”

∴ n(A) = {1,3, 5, 7, 9, 11} = 6

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{6}{12}=\frac{1}{2}\)

3. Let E be the event “pointing towards a number which is a multiple of 3”

∴ n(E) = {3, 6, 9, 12} = 4

∴ P(E) = \(\frac{n(A)}{n(S)}=\frac{4}{12}=\frac{1}{3}\)

Question 14. A bag contains 15 white balls and some black balls. If the probability of drawing a black ball is thrice that of a white ball, find the number of black balls in the bag.

Solution:

Given

A bag contains 15 white balls and some black balls. If the probability of drawing a black ball is thrice that of a white ball

Number of white balls in a bag =15

Suppose the number of black balls in the bag = x

Total number of balls in the bag = 15 + x

P(white ball) =\(\frac{15}{15+x}\)

P(black ball) =\(\frac{x}{15+x}\)

Now, P(black ball ) =3 P( white ball )

⇒ \(\frac{x}{15+x} =3 \times \frac{15}{15+x} \quad \Rightarrow \quad x=45\)

Therefore, the number of black balls in the bag is 45

Question 15. Find the probability of having 53 Tuesdays in a:

1. Non-leap year
2. Leap year.

Solution:

1. A non-leap year contains 365 days. So, by dividing it by 7, we get 52 weeks and 1 more day.

So, since 52 weeks are there, it means 52 Tuesdays will also be there necessarily with probability 1 and 1 more day may be either Sun. or Mon. or Tue. or Wed. or Thur. or Fri. or Sat.

So, to get 53 Tuesdays, we have to select one more Tuesday from these 7 possibilities with probability \({1}{7}\)

Therefore, probability of having 53 Tuesdays in a non-leap year = \(1 \times \frac{1}{7}=\frac{1}{7}\)

2. In a leap year, there are 366 days and 364 days make 52 weeks and therefore 52 Tuesdays. So, the probability of getting 52 Tuesdays till now is l(sure event). The remaining two days can be

Sunday – Monday
Monday – Tuesday
Tuesday – Wednesday
Wednesday – Thursday
Thursday – Friday
Friday – Saturday
Saturday – Sunday

Hence, favourable outcomes = 2 and total outcomes = 7.

Therefore, the probability of having 53 Tuesdays in a leap year is \(1 \times \frac{2}{7}=\frac{2}{7}\)

Question16. It is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb?

Solution:

Given

It is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box.

The number of non-defective bulbs in the box = 600 – 12 = 588

So, probability of taking out a non-defective bulb = \(\frac{588}{600}=\frac{49}{50}\)=0.98

Question 17. Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on

1. The same day
2. Different days
3. Consecutive days?

Solution:

Given

Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any day as on another day.

1. Both the customers visit the shop on the same day, i.e., there are 6 days (Monday to Saturday) and each is equally likely.

∴ Probability of both will visit on the same day

= First will go on Monday and the other will also go on Monday or the first will go on Tuesday and the other will go on Tuesday:

or first will go on Saturday and other will also go on Saturday

=\(\frac{1}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{6}+\ldots+\frac{1}{6} \times \frac{1}{6}=6 \times \frac{1}{6} \times \frac{1}{6}=\frac{1}{6}\)

Alternative Method:

The first person will definitely go to the market on any of the six days with probability 1 (sure event). If both go to the market the same day then the second person has only 1 choice (only that 1 day when the first goes). So, its probability of going to the market = \(\frac{1}{6}\)

∴ Required probability = \(1 \times \frac{1}{6}=\frac{1}{6}\)

2. The first person will definitely go to the market on any of the six days with probability 1 (sure event). Since both will not go on the same day, so second person has now only 5 choices (except this 1 day when first go to the market). So, the probability of the second man go to the other day = \(\frac{5}{6}\)

∴ Required probability = \(1 \times \frac{5}{6}=\frac{5}{6}\)

3. There are 5 consecutive days, i.e., Monday Tuesday, Tuesday Wednesday, Wednesday Thursday, Thursday Friday, Friday Saturday.

∴ Required probability = 5 \times \frac{1}{6} \times \frac{1}{6}=\frac{5}{36}

Question 18. Find the probability of getting 52 Sundays in a leap year.

Solution:

A leap year has 366 days, so by dividing it by 7, we get 52 weeks and 2 days more. 52 weeks means 52 Sundays surely. Now, what will you say?

Perhaps you will say that the probability of getting 52 Sundays in a leap year is 1. Your answer is not correct. Why?

Think about the two remaining days. If from the remaining 2 days, 1 day is the Sunday, then there are 53 Sundays in a leap year.

So, the question does not end at the stage that probability is 1. We have to consider necessarily the two remaining days.

The remaining 2 days may be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday).

For only 52 Sundays we want the combination of (Sunday, and Monday) or (Saturday, and Sunday not to occur. So, to get the required probability.

= (52 Sundays with probability 1) and (5 other possibilities out of 7 with probability \(\left(\frac{5}{7}\right)\))

= \(1 \times \frac{5}{7}=\frac{5}{7}\)

Question 19. If a number x is chosen at random from the numbers -2, -1, 0, 1, 2. What is the probability that \(x^2<2\)?

Solution:

We are presenting a table for x and x.

Clearly, x can take any one of the five values.

∴ Total number of cases = 5

Now, x³ < 2 when x = -1 or 0 or 1

∴ Favourable cases = 3

∴ Required probability =\(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}=\frac{3}{5}\)

Question 20. A circle with a radius of 10 cm is drawn somewhere on a rectangular piece of paper of dimensions 10 cm x 30 cm, This paper is kept horizontally on the tabletop and a point is marked on the rectangular paper without seeing It. bind the probability that it will marked outside the circle. {Take \(\pi=\frac{22}{7}\)

Solution:

Given

A circle with a radius of 10 cm is drawn somewhere on a rectangular piece of paper of dimensions 10 cm x 30 cm, This paper is kept horizontally on the tabletop and a point is marked on the rectangular paper without seeing It.

Area of rectangle =40 \(\times 30 \mathrm{~cm}^2=1200 \mathrm{~cm}^2 \)

Area of circle =\(\pi(10)^2=100 \pi \mathrm{cm}^2\)

∴ Area of remaining portion =(1200-100 \(\pi) \mathrm{cm}^2\)

∴ Required probability =\(\frac{\text { Favourable portion }}{\text { Total portion }}=\frac{100(12-\pi)}{1200}\)

= \(\frac{12-\frac{22}{7}}{12}=\frac{84-22}{7 \times 12}=\frac{62}{7 \times 12}=\frac{31}{42}\)

Question 21. A square dart board is placed in the first quadrant from x = 0 to,x = 6 and = 0 to x = 6. A triangular region on the dartboard is enclosed y = 2, x = 6 and y = x. Find the probability that a dart that randomly hits the dartboard will land in the triangular region formed by the three lines.

Solution:

Given

A square dart board is placed in the first quadrant from x = 0 to,x = 6 and = 0 to x = 6. A triangular region on the dartboard is enclosed y = 2, x = 6 and y = x.

Area of square board =\((6)^2=36 \mathrm{sq}\). units

Now, DE = 6 – 2 = 4 units

BD = 6 – 2 = 4 units

∴ Area of \(\triangle B D E =\frac{1}{2} \times D E \times B D=\frac{1}{2} \times 4 \times 4\)

= 8 sq units

Hence, required probability =\(\frac{Area of \triangle B D E}{{ Area of square } O A B C}\)

= \(\frac{8}{6 \times 6}=\frac{2}{9}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.1

Question 1. Complete the following statements:

1. Probability of an event E + Probability of the event ‘not E’ = ________.
2. The probability of an event that cannot happen is ________ Such an event is called ___________.
3. The probability of an event that is certain to happen is _______ Such an event is called _______.
4. The sum of the probabilities of all the elementary events of an experiment is _________.
5. The probability of an event is greater than or equal to _______ and less than or equal to ________.

Solution:

1. Probability of an event E + Probability of the event ‘not F = 1.
2. The probability of an event that cannot happen is zero. Such an event is called an impossible event.
3. The probability of an event that is certain to happen is 1. Such an event is called a certain event.
4. The sum of the probabilities of all the elementary events of an experiment is 1.
5. The probability of an event is greater than or equal to zero and less than or equal to one.

Question 2. Which of the following experiments have equally likely outcomes? Explain.

1. A driver attempts to start a car. The car starts or does not start.
2. A player attempts to shoot a basketball. She/he shoots or misses the shot.
3. A trial is made to answer a true-false question. The answer is right or wrong.
4. A baby is born. It is a boy or a girl?

Solution:

1. A driver attempts to start a car. Here the possibility of starting a car is more than that of not starting the car. So this experiment is not equally likely.
2. A player attempts to shoot a basketball.
   • In the same condition, the possibilities of shoots or misses the shoots are not the same. So this experiment is not equally likely.
3. A trial is made to answer a true-false question. The possibilities of right or wrong answers are the same. So this experiment is equally likely.
4. A baby is born. The possibilities of this baby being a boy or a girl are equal. So this experiment is equally likely.

Question 3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Solution:

Tossing a coin is considered to be a fair way because the coin is symmetric and the possibilities of getting head or tail are equal.

Question 4. Which of the following cannot be the probability of an event?

1. \(frac{2}{3}\)
2. -1.5
3. 15%
4. 0.7

Solution: 2. -1.5

Because the probability of an event cannot be smaller than zero.

Question 5. If P(E) = 0.05, what is the probability of ‘not E’?

Solution :

P(E) = 0.05

Probability of ‘E-not’ = 1 – P(E)

= 1 – 0.05 = 0.95

Question 6. A bag contains lemon-flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out:

1. An orange-flavoured candy?
2. A lemon-flavoured candy?

Solution :

There are only lemon-flavoured candies in the bag.

If a candy is drawn at random from the bag, then

1. The probability of drawing an orange-flavoured candy = 0.

So. the probability of drawing an orange-flavoured candy = 0.

2. The event of drawing a lemon-flavoured candle is a certain event.

So. the probability of drawing a lemon-flavoured candy = 1

Question 7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Solution:

The probability of 2 students not having the same birthday = 0.992

∴ The probability of 2 students having the same birthday = 1 – 0.992 = 0.008

Question 8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

1. Red?
2. Not red?

Solution:

Total balls 3 + 5 = 8

Total possible outcomes of drawing a ball at random from the bag = 8

1. Favourable outcomes of drawing a red ball = 3

∴ The probability of drawing a red ball

Favourable outcomes of = \(\frac{\text { drawing a red ball }}{\text { Total possible outcomes }}\)

=\(\frac{3}{8}\)

2. Probability that the ball drawn is not red = 1 – probability that the ball drawn is red.

Question 9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be

1. Red?
2. White?
3. Not green?

Solution:

Total marbles = 5 + 8 + 4=17 Total possible outcomes of drawing a marble at random from the box = 17

1. Favourable outcomes of drawing a red marble = 5.

∴ The probability that the marble drawn is red

Favourable outcomes of = \(\frac{\text { drawing a red ball }}{\text { Total possible outcomes }}\)

=\(\frac{5}{17}\)

2. Favourable outcomes of drawing a white marble = S

∴ Possibility of drawing a white marble

Favourable outcomes of \(\frac{\text { drawing a red ball }}{\text { Total possible outcomes }}\)

= \(\frac{8}{17}\)

3. Favourable outcomes of drawing a green marble = 4

⇒ Favourable outcomes that the marble drawn is not green = 17 – 4 =13

∴ The probability that the marble drawn is not green

Favourable outcomes = \(\frac{\text { marble drawn is not green }}{\text { Total possible outcomes }}\)

= \(\frac{13}{17}\)

Question 10. A piggy bank contains a hundred 50 p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the tyre coins will fall out when the bank is turned upside down, what is the probability that the coin?

1. Will be a 50 p coin?
2. Will not be a ₹ 5 coin?

Solution:

No. of coins of 50 paise =100

No. of coins of ₹ 1 =50

No. of coins of ₹ 2 =20

No. of coins of ₹ 5 = 10

Total coins = 100 + 50 + 20 + 10 = 180

Total outcomes that one coin will fall out = 180

1. The favourable outcomes that the coin fall out is 50 paise = 100

∴ Probability that coin fall out is of 50 paise =\(\frac{100}{180}=\frac{5}{9}\)

2. The favourable outcome that the coin falls out is ₹ 5 = 10

∴ Probability that coin fall out is of Rs 5 = =\(\frac{10}{180}=\frac{1}{18}\)

Therefore, the probability that the coin falls out is not ₹ 5

= \(1-\frac{1}{18}=\frac{17}{18}\)

Question 11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Solution :

Total no. of fish = 5 + 8 = 13

Total possible outcomes of taking a fish = 13

Favourable outcomes of taking a male fish = 5

Probability of a male fish

Favourable outcomes

= \(\frac{\text { drawing a male fish }}{\text { Total no. of possible outcomes of drawing a fish }}\)

= \(\frac{5}{3}\)

Question 12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1,2,3,4,5,6,7,8. and these are equally likely outcomes.

What is the probability that it will point at:

1. 8?
2. An odd number?
3. A number greater than 2?
4. A number less than 9?

Solution :

Total no. of outcomes of pointing by an arrow =8

1. Favourable outcomes that the arrow indicates the number 8 = 1

∴ Probability of this event = \(\frac{1}{8}\)

Favourable outcomes that the arrow indicates an odd number = {1,3,5, 7} = 4

∴ Probability of this event = \(\frac{4}{8}=\frac{1}{2}\)

3. Favourable outcomes that the arrow indicates a number greater than 2 = {3,4,5,6,7,8} = 6

∴ Probability of this event = \(\frac{6}{8}\) =\(\frac{3}{4}\)

4. Favourable outcomes that the arrow indicates a number less than 9. = {1,2, 3, 4, 5, 6, 7, 8} = 8

∴ Probability of this event = \(\frac{8}{8}\)=1

Question 13. A die is thrown once. Find the probability of getting:

1. A prime number
2. A number lying between 2 and 6
3. An odd number.

Solution:

Possible outcomes in one throw of a die = {1,2,3,4,5,6}

Total possible outcomes = 6

1. Prime numbers = {2, 3, 5} = 3

∴ Probability of getting a prime number = \(\frac{3}{6}=\frac{1}{2}\)

2. Numbers lying between 2 to 6 = {3,4,5} = 3

∴ Probability of this event = \(\frac{3}{6}=\frac{1}{2}\)

3. Odd numbers = {1, 3, 5} = 3

∴ Probability of this event =\(\frac{3}{6}=\frac{1}{2}\)

Question 14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:

1. A king of red colour
2. A face card
3. A red-face card
4. The jack of hearts
5. A spade
6. The Queen of diamonds

Solution:

Total possible outcomes of drawing a card from the pack of cards = 52.

1. Favourable outcomes of getting a king of red colour = 2.

∴ Probability of drawing a icing of red colour = =\(\frac{2}{52}=\frac{1}{26}\) .

2. Favourable outcomes of getting a face card = 12

∴ Probability of drawing a face card = \(\frac{12}{52}=\frac{3}{13}\)

3. Favourable outcomes of getting a face card of red colour = 6

∴ Probability of drawing a face card of red colour =\(\frac{6}{52}=\frac{3}{26}\)

4. Favourable outcomes of drawing a jack of heart = 1

∴ Probability of drawing a jack of heart = \(\frac{1}{52}\)

5. Favourable outcomes of drawing a card of spade = 13

∴ Probability of drawing a card of spade = \(\frac{13}{52}={1}{4}\)

6. Favourable outcomes of drawing a queen of diamonds = 1

∴ Probability of drawing a queen of diamond = \(\frac{1}{52}\)

Question 15. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

1. What is the probability that the card is the queen?
2. If the queen is drawn and put aside, what is the probability that the second card picked up is
   1. An ace?
   2. A queen?

Solution:

Total favourable outcomes of drawing a card from the five playing cards = 5

1. Favourable outcomes of drawing a queen = 1

∴ Probability of drawing a queen = \(\frac{1}{5}\)

2. Remaining cards when ‘queen’ is drawn and put aside = 4

(1). Favourable outcomes of drawing a card of ace = 1

∴ Probability of drawing the second card ace = \(\frac{1}{4}\)

(2). Favourable outcomes of drawing a queen = 0

∴ Probability of drawing the second card of queen = \(\frac{0}{4}\)=0

Question 16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Solution :

No. of good pens = 132

No. of defective pens = 12

Total pens = 132 + 12 = 144

Total favourable outcomes of drawing a pen = 144

Favourable outcomes of drawing a good pen = 132

∴ Probability of drawing a good pen = \(\frac{132}{144}\) = \(\frac{11}{12}\)

Question 17.

1. A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
2. Suppose the bulb drawn in (1) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Solution:

Total bulbs = 20

Defective bulbs = 4

1. If a bulb is drawn at random then total possible outcomes = 20

Favourable outcomes of a defective bulb = 4.

∴ Probability of a defective bulb = \(\frac{4}{20}=\frac{1}{5}\)

2. If the bulb drawn is not defective and it is not replaced. Now a bulb is drawn then

Total possible outcomes =19

Favourable outcomes of a defective bulb = 4

∴ Probability of a defective bulb = \(\frac{4}{19}\)

⇒ Probability of a non-defective bulb =\(1-\frac{4}{19}=\frac{15}{19}\)

Question 18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

1. A two-digit number
2. A perfect square number
3. A number divisible by 5.

Solution:

Total no. of discs = 90

Possible outcomes of drawing a disc = 90

1. Two digit numbers = {10, 11, 12, …90}

Favourable outcomes of drawing a two’ digit number = 81

Now, the probability of getting a two-digit number on the disc =\(\frac{81}{90}=\frac{9}{10}\)

2. Perfect square numbers in given numbers
= {1,4, 9, …,81}

⇒ Favourable outcomes of drawing a perfect square number = 9

Now, the probability that the number marked on the disc is a perfect square =\(\frac{9}{90}=\frac{1}{10}\)

3. Numbers divisible by 5 in the given numbers = (5, 10, 15,…, 90}

⇒ Favourable outcomes of a number drawn divisible by 5 = 18

Now, the probability that the number marked on the disc is divisible by 5 = \(\frac{18}{90}=\frac{1}{5}\)

Question 19. A child has a die whose six faces show the letters as given below:

A B C D E A

The die is thrown once. What is the probability of getting

1. A?
2. D?

Solution:

Total possible outcomes in a throw of die = 6

1. Favourable outcome of getting A = 2

∴ Probability of getting A = \(\frac{2}{6}=\frac{1}{3}\)

2. Favourable outcomes of getting D = 1

∴ Probability of getting D = \(\frac{1}{6}\)

Question 20. Suppose you drop a die at random on the rectangular region shown. What is the probability that it will land inside the circle with a diameter of 1 m?

Solution:

Area of rectangle = \(3 \times 2=6 \mathrm{~m}^2\)

Diameter of circle =1 m

⇒ Radius of circle =\(\frac{1}{2} \mathrm{~m}\)

⇒ Area of circle =\(\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4} \mathrm{~m}^2\)

Now, the probability that in one throw of a die, the die lands inside the circle

= \(\frac{\text { Area of circle }}{\text { Area of rectangle }}\)

= \(\frac{\pi / 4}{6}=\frac{\pi}{4 \times 6}=\frac{\pi}{24}\)

Question 21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that:

1. She will buy it?
2. She will not buy it?

Solution:

Total number of ball pens = 144

No. of defective ball pens = 20

∴ No. of good ball pens = 144 – 20 = 124

No. of favourable outcomes of drawing a pen randomly = 144

1. To buy the pen, it must be a good one.

∴ Favourable outcomes of drawing a good ball pen=124

⇒ Probability of buying a good ball pen =\(\frac{124}{144}=\frac{31}{36}\)

2. If the ball pen is defective then she will not buy it.

∴ Favourable outcomes of drawing a defective ball pen = 20

⇒ Probability of drawing a defective ball pen =\(\frac{20}{144}=\frac{5}{36}\)

Question 22. Two dice, one blue and one grey, are thrown at the same time,

1. Complete the following table:

2. A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac{1}{11}\). Do you agree with this argument? Justify your answer.

Solution :

1. In a throw of two dice, possible outcomes are

⇒ \(\left\{\begin{array}{llllll}
(1,1), & (1,2), & (1,3), & (1,4), & (1,5), & (1,6) \\
(2,1), & (2,2), & (2,3), & (2,4), & (2,5), & (2,6) \\
(3,1), & (3,2), & (3,3), & (3,4), & (3,5), & (3,6) \\
(4,1), & (4,2), & (4,3), & (4,4), & (4,5), & (4,6) \\
(5,1), & (5,2), & (5,3), & (5,4), & (5,5), & (5,6) \\
(6,1), & (6,2), & (6,3), & (6,4), & (6,5), & (6,6)
\end{array}\right\}\)

Total possible outcomes = 36

Outcome of getting the sum 2 = (1, 1)

Favourable outcomes of getting a sum 2 = 1

∴ Probability of getting the sum 2 = \(\frac{1}{36}\)

Outcomes of getting the sum 3 = (1, 2), (2, 1)

Favourable outcomes of getting the sum 3 = 2

∴ Probability of getting the sum 3 = \(\frac{2}{36}={1}{18}\)

Outcomes of getting the sum 4 = (1,3), (2, 2), (3, 1)

Favourable outcomes of getting the sum 4 = 3

∴ Probability of getting the sum 4 = \(\frac{3}{36}={1}{12}\)

Outcomes of getting the sum 5 = (1,4), (2, 3), (3, 2), (4, 1)

Favourable outcomes of getting the sum 5=4

∴ Probability of getting the sum 5 = \(\frac{4}{36}={1}{9}\)

Outcomes of getting the sum 6 = (1,5), (2, 4), (3, 3), (4, 2), (5, 1)

Favourable outcomes of getting the sum 6 = 5

∴ Probability of getting the sum 6 = \(\frac{5}{36}\)

Outcomes of getting the sum 7 = (1,6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

Favourable outcomes of getting the sum 7 = 6

∴ Probability of getting the sum 7 = \(\frac{6}{36}={1}{6}\)

Outcomes of getting the sum 8 = (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)

Favourable outcomes of getting the sum 8 = 5

∴ Probability of getting the sum 8 = \(\frac{5}{36}\)

Outcomes of getting the sum 9 = (3,6), (4, 5), (5, 4), (6, 3)

Favourable outcomes of getting the sum 9 = 4

∴ Probability of getting the sum 9 = \(\frac{4}{36}={1}{9}\)

Outcomes of getting the sum 10 = (4, 6),(5, 5), (6, 4)

Favourable outcomes of getting the sum 10 = 3

∴ Probability of getting the sum 10 = \(\frac{3}{36}={1}{12}\)

Outcomes of getting the sum 11 = (5, 6), (6, 5)

Favourable outcomes of getting the sum 11 = 2

∴ Probability of getting the sum 11= \(\frac{2}{36}={1}{18}\)

Outcomes of getting the sum 12 = (6, 6)

Favourable outcomes of getting the sum 12 = 1

∴ Probability of getting the sum 12 = \(\frac{1}{36}\)

Now, the given table will be as follows:

2. The student’s argument is wrong because the favourable outcomes of every event are different.

Question 23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanifwins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Solution :

Given

A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanifwins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise.

All possible outcomes in three throws of a coin of ₹ 1.

HHH, HHT, HTH, THH, HTT, THT, TTH,

TTT → Total possible outcomes = 8

Favourable outcomes of 3 heads = 1

Favourable outcomes of 3 tails = 1

∴ Favourable outcomes to win = 2

Now, favourable outcomes to lose = 8 – 2 = 6

Therefore, the probability to lose

= \(\frac{\text { Favourable outcomes to lose }}{\text { Total possible outcomes }}\)

= [/latex]\frac{6}{8}=\frac{3}{4}[/latex]

Question 24. A die is thrown twice. What is the probability that:

1. 5 will not come up either time?
2. 5 will come up at least once?

[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Solution:

In two throws of a die, the all possible outcomes are :

(1.1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2.1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3.1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4.1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5.1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6.1), (6,2), (6,3), (6,4), (6,5), (6,6)

Total possible outcomes = 36

Outcomes in which 5 comes up = 11

Outcomes in which 5 does not come up = 36 – 11 = 25

1. Favourable outcomes that 5 does not come up = 25

∴ Probability of not getting 5 = \(\frac{25}{36}\)

2. Favourable outcomes that 5 will come up at least once = 11

∴ Probability that 5 will come up at least once = \(\frac{11}{36}\)

Question 25. Which of the following arguments are correct and which are not correct? Clive reasons for your answer.

1. If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\)
2. If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\)

Solution:

1. When two coins are tossed together, then there are not three outcomes, but the following four outcomes are obtained:

HH, HT, TH, TT

So, the .ugumom of the student is false.

2. In one throw of a die

All possible outcomes {1, 2, 3, 4, 5, 6} 6

Outcomes of getting an even number = {2. 4, 6} = 3

Outcomes of gelling an odd number = (1,3,5} =3

∴ Possibility of getting an odd number= \(\frac{3}{6}=\frac{1}{2}\)

Therefore, the argument of the student is true.

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.2

Question 1. Two customers Shvam and Ehta visit a particular shop in the same week (Tuesday to Saturday), Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on

1. The same day?
2. Consecutive days?
3. Different days?

Solution:

Hence, the total possible outcomes are as follows:

Toral possible outcomes = 25

1. Outcomes that both customers visit on the same day

= (T. T), (W, W), (Th. Th), (F, F), (S, S)

Total favourable outcomes = 5

So. is probability Drat Bodi will visit on the same day

= \(\frac{5}{25}=\frac{1}{5}\)

2. Outcomes that both customers visit on consecutive days

= (T, W), (W; Th), (Th, F), (F, S), (W, T), (Th, W), (F, Th), (S, F)

Total favourable outcomes = 8

So, the probability that both visit on consecutive days the shop

= \(\frac{8}{25}\)

3. Probability that both visit the shop on the same day = \(\frac{1}{5}\) [from part (1)]

∴ The probability that both visit the shop on different days

1= \(\frac{1}{5}= {4}{5}\)

Question 2. A die is numbered in such a way that its faces show the numbers 1,2,2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a tew values of the total score on the two throws:

Number in the first throw What is the probability that the total score is

1. Even?
2. 6?
3. At least 6?

Solution:

Total possible outcomes = 36

1. Favourable outcomes of getting the sum as an even number = 18

∴ Probability of getting the sum as even number = \(\frac{18}{36}=\frac{1}{2}\)

2. Favourable outcomes of getting a sum 6 = 4

∴ Probability of getting a sum 6= \(\frac{4}{36}=\frac{1}{9}\)

3. Favourable outcomes of getting a sum less than 6=15.

∴ Probability of getting a sum less than 6 =\(\frac{15}{36}=\frac{5}{12}\)

Question 3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Solution:

Given

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball

Let the number of blue balls =x

Given, the number of red balls = 5

Total balls = x + 5

∴ The probability of drawing a blue ball

⇒ P(B)=\(\frac{x}{x+5}\)

and the probability of drawing a red ball

⇒ P(R)=\(\frac{5}{x+5}\)

Given that, P(B) =2 \(\times P(R)\)

⇒ \(\frac{x}{x+5} =\frac{2 \times 5}{x+5}\)

⇒ x =10

⇒ Number of blue balls =10

Question 4. A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?

If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.

Solution:

Given

A box contains 12 balls out of which x are black. If one ball is drawn at random from the box,

Total balls in the box = 12

Black balls = x

∴ Probability of drawing a black ball = \(\frac{x}{12}\)

On putting 6 more black balls in the box, Total balls = 12 + 6= 18 Black balls = x + 6

Now, probability of drawing a black ball = \(\frac{x+6}{12}\)

Given that, \(\frac{x+6}{18} =2 \times \frac{x}{12}\)

⇒ \(\frac{x+6}{18} =\frac{x}{6}\)

⇒ x + 6 = 3 x

⇒ 2 x = 6 ⇒ x = 3

Question 5. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green 2 is \(\frac{2}{3}\). Find the number of blue balls in the jar.

Solution:

Given

A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green 2 is \(\frac{2}{3}\).

Total marbles in jar = 24

Let green marbles in jar = x

∴ Probability of drawing 1 green marble from jar = \(\frac{x}{24}\)

Given that, \(\frac{x}{24}=\frac{2}{3}\)

x = \(\frac{2}{3} \times 24\)

⇒ x = 16

∴ Green marbles = 16

⇒ Blue marbles = 24 – 16 = 8

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Multiple Choice Question And Answers

Question 1. The probability of a sure event is:

1. 0
2. 1
3. -l
4. \(\frac{1}{2}\)

Answer: 2. 1

Question 2. The sum of the probabilities ofan event and its complementary event is:

1. -1
2. 0
3. 1
4. None of these

Answer: 3. 1

Question 3. The probability of winning a match by Ravi is \(\frac{2}{5}\). The probability of his losing the match is

1. \(\frac{3}{5}\)
2. \(\frac{2}{5}\)
3. 1
4. 0

Answer: 1. \(\frac{3}{5}\)

Question 4. If P(A) represents the probability of an event, A, then:

1. P(A)<0
2. P(A)>1
3. 0 ≤ P(A) ≤1
4. -1 ≤ P(A)≤ 1

Answer: 3. 0 ≤ P(A) ≤1

Question 5. When a dice is thrown, the probability of getting an even number less than 3, is:

1. 0
2. \(\frac{1}{2}\)
3. \(\frac{1}{3}\)
4. \(\frac{1}{6}\)

Answer: 4. \(\frac{1}{6}\)

Question 6. The probability of selecting a prime number from the numbers 1 to 20, is:

1. \(\frac{7}{20}\)
2. \(\frac{2}{5}\)
3. \(\frac{9}{20}\)
4. \(\frac{1}{2}\)

Answer: 2. \(\frac{2}{5}\)

Question 7. In a year, which is not a leap year, the probability of 53 Monday is:

1. \(\frac{6}{7}\)
2. \(\frac{3}{7}\)
3. \(\frac{2}{7}\)
4. \(\frac{1}{7}\)

Answer: 4. \(\frac{1}{7}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids

The objects which occupy space have three dimensions) arc called solids.

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids Cuboid And Cube

A cuboid is a rectangular solid having six faces all of which are rectangles. The cuboid has six faces. The opposite faces are all congruent. Two adjacent faces meet in a line.

This line is called the edge of the cuboid. There are 12 edges of a cuboid. The point where three adjacent edges meet is called the vertex of the cuboid. There are 8 vertices of a cuboid.

The three edges that meet on the vertex of a cuboid are called its length, breadth and height. A cube is a cuboid in which all the edges are of equal length and every one of the six faces is a square.

Read and Learn More Class 10 Maths Solutions Exemplar

Total Surface (Surface Area) of a Cuboid and a Cube

The sum of the area of the faces of the cuboid is called its surface. The areas of opposite faces are equal, so the total surface of the A cuboid is

2lb + 2bh + 2hl = 2 (bh+ hl +lb)

For a cube, l = b = h = a (say)

Surface of a cube = 2(a.a + a.a + a.a) = 6a²

The Length of the Diagonal of a Cuboid and a Cube

Diagonal of a floor \(D^{\prime} B^{\prime}=\sqrt{l^2+b^2}\). So, if we want to find the length of the longest rod DB’, then we make a new right-angled A treating, DB’ as hypotenuse, base as D’B’ and height as DD’ which is h. So, the length of the diagonal of a cuboid DB’.

= \(\sqrt{\left(D^{\prime} B^{\prime}\right)^2+\left(D D^{\prime}\right)^2}=\sqrt{l^2+b^2+h^2}\)

For a cube, l = b = h = a (say)

∴ The length of the diagonal of a cube = \(\sqrt{a^2+a^2+a^2}=a \sqrt{3}\)

Lateral Surface Area of a Cuboid and a Cube

• Lateral surface area of a cuboid = 2(l + b)h. [CSA = TSA- (area of top and area of bottom)]
• Lateral surface area of a cube = 4a², where a = edge of the cube. It is also called the curved surface area or area of 4 walls.

The volume of a Cuboid and a Cube

The volume of any solid figure is the amount of space enclosed within its bounding faces.

The volume of a cuboid of length l units, breadth b units and height h units are lbh cubic units.

= (length x breadth x height) cubic units

= (area of the base x height) cubic units.

For a cube,

Volume of cube = a³ cubic units,

where, the length of the edge of the cube is a unit.

Note: The measurements of volume are as follows:

1 cm³ = 10³ cubic mm

1 m³ = 1000 litre = 1 kl

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids Cylinder

If a rectangle is revolved about one of its sides as its axis, the solid so formed is called a right circular cylinder.

The side AD about which the rectangle ABCD revolves is called the height of the cylinder.

The line CD is called the generating line because when it revolves round AB, it generates the cylinder.

If you cut the hollow cylinder along AD and spread the piece, it becomes a rectangle whose one side is AD and the other side is AB which is equal to the circumference of a circle whose radius is the radius of the cylinder.

Hence, Side AB = 2πr

If h is the height of the cylinder, the area of ABCD

= AB x AD = 2πrh

Thus, the area of the curved surface of the cylinder

= 2πrh = perimeter of base x height

Whole Surface of a Cylinder

The whole surface area of a cylinder

= Curved surface + Area of the base + Area of the top

= 2πrh + πr2 + nr² = 2πrh + 2πr² = 2πr(h + r)

Volume of a Cylinder

The volume of a cylinder = Area of the base x Height

= nr2 x h = nr2h

Note: If the top and bottom are removed from the total surface area, then only the curved surface area remains.

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids Hollow Cylinder

Solids like iron pipes, rubber tubes, etc., are in the shape of hollow cylinders.

For a hollow cylinder of height h and with external and internal radii R and r respectively, we have

Volume of the material = Exterior volume- Interior volume
= πR²h- πr²h = πh (R²– r²)

The curved surface of the hollow cylinder
= External surface + Internal surface = 2πRh + 2πrh = 2πh(R + r)

The total surface area of the hollow cylinder
= Curved surface + 2(Area of base rings)
= (2πRh + 2πrh) + 2(πR²– πr²) = 2πh(R + r) + 2π(R² – r²)
= 2π(R + r) (h+R-r)

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids Right Circular Cone

A right circular cone is a solid generated by the revolution of a right-angled triangle about sides containing the right angle as the axis.

Let CAB be a right-angled triangle, right-angled at A. The hypotenuse CB revolves around the side CA as the fixed axis. The hypotenuse CB will generate the curved surface of a cone.

The radius of the circular base is called the radius of the cone. It is usually denoted by ‘r’.

The point C is called the vertex of the cone.

The length CA of the axis is called the height of the cone. It is usually ‘denoted by ‘h’.

The hypotenuse CB is called the generating line of the cone and its length is called the slant height. It is usually denoted by l.

The volume of Right Circular Cone

Volume of a cone = \(\frac{1}{3}\) (area of the base) x height’

i.e., \(V=\frac{1}{3} \pi r^2 h\)

where r = radius of the base and h = height.

Slant height of right circular cone = \(l=\sqrt{h^2+r^2}\)

The curved surface of the Right Circular Cone

Curved surface of a cone = \(\pi r l=\pi r \sqrt{h^2+r^2}\)

where r = radius of the base, l = slant height of the cone.

Total Surface of Right Circular Cone

Total surface of the cone = Area of the base + Area of the curved surface

i.e., total surface of the cone = πr² + πrl = πr(r + l).

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids Sphere

When a circle is revolved about its diameter, the solid thus formed is a sphere. Let AB be the diameter of a circle ADB and 0 be its centre.

If the circle ADB revolves around AB, point D takes different positions and a closed solid is formed.

A sphere may also be defined as a solid bounded by a closed surface, all points on which are at a constant distance from a fixed point.

Curved Surface Area and Volume of a Sphere

The curved surface area of a sphere = 4πr², where r is the radius of the sphere.

The volume of sphere = \(\frac{4}{3} \pi r^3\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids Hemisphere

A plane through the centre of a sphere divides the sphere into two equal parts. Each part is called a hemisphere.

If r is the radius of the hemisphere, then

Volume of hemisphere = \(\frac{2}{3} \pi r^3\)

Curved surface area = 2πr²

Total surface area = 2πr² + πr² = 3πr².

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids Spherical Shell

A spherical shell having external radius R and internal radius r, then we have

Volume of material = \(\frac{4}{3} \pi R^3-\frac{4}{3} \pi r^3=\frac{4}{3} \pi\left(R^3-r^3\right)\)

Volume And Surface Area Of Solids Solved Examples

Question 1. A tent of cloth is cylindrical upto 1 m in height and conical above it of the same radius of base. If the diameter of the tent is 6 m and the slant height of the conical part is 5 m, find the cloth required to make this tent.

Solution:

Given

A tent of cloth is cylindrical upto 1 m in height and conical above it of the same radius of base. If the diameter of the tent is 6 m and the slant height of the conical part is 5 m,

Diameter of base 2r = 6 m

⇒ \(r=\frac{6}{2}=3 \mathrm{~m}\)

Height of cylindrical path h = 1 m

The slant height of conical part l = 5 m

Cloth required in tent = 2πrh + πrl

= πr (2h + l)

= \(\frac{22}{7} \times 3(2 \times 1+5)=66 \mathrm{~m}^2\)

The cloth required to make this tent =66 m²

Question 2. The base and top of a right circular cylindrical drum are hemispherical. The diameter of the cylindrical part is 14 cm and the total height is 30 cm. Find the total surface area of the drum.

Solution:

Given

The base and top of a right circular cylindrical drum are hemispherical. The diameter of the cylindrical part is 14 cm and the total height is 30 cm.

Here 2r = 14

⇒ r = 7 cm

Height of cylindrical part h = 30 – 2 x 7 = 16 cm

The total surface area of the drum = 2 x the Curved surface of the hemisphere + the Curved surface of the cylinder

= 2 x 2πr² + 2πrh

= 2πr(2r + h)

= \(2 \times \frac{22}{7} \times 7(2 \times 7+16)\)

= 44 x 30 = 1320 cm²

The total surface area of the drum = 1320 cm²

Question 3. Three cubes, each with an 8 cm edge, are joined end to end. Find the total surface area of the resulting cuboid.

Solution:

Given

Three cubes, each with an 8 cm edge, are joined end to end.

As is clear from the adjoining figure;

the length of the resulting cuboid = 3 x 8 cm = 24 cm

Its width = 8 cm and its height = 8 cm

i.e., l = 24 cm, b = 8 cm and h = 8 cm

∴ The total surface area of the resulting cuboid

= 2(1 x b + b x h + h x l)

= 2 (24 x 8 + 8 x 8 + 8 x 24) cm² = 896 cm²

The total surface area of the resulting cuboid = 896 cm²

Question 4. A rectangular sheet of tin 58 cm x 44 cm is to be made into an open box by cutting off equal squares from the corners and folding up the flaps. What should be the volume of box if the surface area of box is 2452 cm²?

Solution:

Given

A rectangular sheet of tin 58 cm x 44 cm is to be made into an open box by cutting off equal squares from the corners and folding up the flaps.

Let a square of x cm from each corner be removed from a rectangular sheet.

So, length of box = (58 – 2x) cm

breadth of box = (44 – 2x) cm

and height of box = x cm

∴ Surface area of open box = 2 (lb + bh + hl)- lb

⇒ 2[(58 – 2x) (44 – 2x) + x (44 – 2x) + x (58 – 2x)]- (58 – 2x)(44 – lx) = 2452

⇒ (58- 2x) (44- 2x) + 2x(44- 2x) + 2x (58 – 2x) = 2452

⇒ (29 – x) (22 – x) + x (22 – x) + x (29 – x) = 613

⇒ 638 – 51x + x2 + 22x – x2 + 29x – x2 = 613

⇒ x² = 25

⇒ x = 5 cm (x = – 5 is not admissible)

∴ Volume of box = x(58 – 2x) (44- 2x) = 5(58- 10)(44 – 10)

= 5 x 48 x 34 cm³ = 8160 cm³

The volume of box = 8160 cm³

Question 5. The radii of two right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5:4. Calculate the ratio of their curved surface areas and also the ratio of their volumes.

Solution:

Given

The radii of two right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5:4.

Let the radii of the two cylinders be 2r and 3r respectively and their heights be 5h and 4h.

∴ \(\frac{\text { Curved surface area of 1st cylinder }\left(S_1\right)}{\text { Curved surface area of 2nd cylinder }\left(S_2\right)}=\frac{2 \pi \times 2 r \times 5 h}{2 \pi \times 3 r \times 4 h}\)

i.e., \(\frac{S_1}{S_2}=\frac{5}{6}\) or S1: S2 = 5:6

⇒ \(\frac{\text { Volume of 1st cylinder }\left(V_1\right)}{\text { Volume of 2nd cylinder }\left(V_2\right)}=\frac{\pi \times(2 r)^2 \times 5 h}{\pi \times(3 r)^2 \times 4 h}\)

i.e., \(\frac{V_1}{V_2}=\frac{5}{9}\) or V₁:V₂= 5:9

The ratio of their volumes = 5:9

Question 6. The volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere?

Solution:

Given

The volume and surface area of a solid hemisphere are numerically equal.

We have,

Volume of hemisphere = Surface area of hemisphere

⇒ \(\frac{2}{3} \pi r^3=3 \pi r^2\) = 2r = 9

Hence, the diameter of the hemisphere = 9 units

Question 7. The curved surface area of a cone of height 8 m is 1 88.4 m². Find the volume of a cone.

Solution:

Given

The curved surface area of a cone of height 8 m is 1 88.4 m².

πrl = 188.4

⇒ \(r l=\frac{188.4}{3.14}=60\)

⇒ r²l² = 3600

⇒ r²(h² + r²) = 3600

⇒ r²(64 + r²) = 3600

⇒ r⁴ + 64r² – 3600 = 0

⇒ (r²+ 100) (r²– 36) = 0

∴ r² = -100 or r² = 36

⇒ r = 6 (∵ r² = -100 is not possible)

∴ Volume of a cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times 3.14 \times 36 \times 8=301.44 \mathrm{~m}^3\)

Question 8. A conical tent is required to accommodate 157 persons, each person must have 2 m² of space on the ground and 15 m³ of air to breathe. Find the height of the tent; also calculate the slant height.

Solution:

Given

A conical tent is required to accommodate 157 persons, each person must have 2 m² of space on the ground and 15 m³ of air to breathe.

1 person needs 2 m² of space.

∴ 157 persons needs 2 x 157 m space on the ground

∴ r² = 2 x 157

∴ \(r^2=\frac{2 \times 157}{3.14}=100\) = r = 100

Also, 1 person needs 15 m of air.

∴ 157 persons need 15 x 157 m3 of air.

∴ \(\frac{1}{3} \pi r^2 h=15 \times 157\)

⇒ \(h=\frac{15 \times 157 \times 3}{3.14 \times 100}=22.5 \mathrm{~m}\)

∴ l² =h² + r² = (22.5)² + (10)² = 606.25

∴ \(l=\sqrt{606.25}=24.62 \mathrm{~m}\)

The height of the tent =22.5 m

The slant height= 24.62 m

 

Question 9. The radius and height of a solid right circular cone are in the ratio of 5:12. If its volume is 314 cm find its total surface area. [Take π = 3.14]

Solution:

Given

The radius and height of a solid right circular cone are in the ratio of 5:12. If its volume is 314 cm

Let the radius of the cone = 5x

∴ Height of cone = 12r

∴ l² = (5r)² + (12x)² = 169 x²

∴ \(l=\sqrt{169 x^2}=13 x\)

It is given that volume = 314 cm2

∴ \(\frac{1}{3} \pi(5 x)^2(12 x)=314\)

⇒ \(\frac{1}{3} \times 3.14 \times 25 \times 12 \times x^3=314\)

⇒ \(x^3=\frac{314 \times 3}{3.14 \times 25 \times 12}=1\)

∴ x = 1

∴ Radius r = 5 x 1 = 5 cm

Height h = 1 2 x 1 = 1 2 cm

and slant height l= 13 x I = 13 cm

Now, total surface area of cone = πr(l + r) = 3.14 x 5 (13 + 5) = 3.14 x 5 x 18

= 282.60 cm²

Hence, the total surface area of a cone is 282.60 cm.

Question 10. If h, C, and V respectively are the height, the curved surface area and the volume of a cone. Prove that 3πVh³ – C²h² + 9V² = 0.

Solution:

Given

h, C, and V respectively are the height, the curved surface area and the volume of a cone.

⇒ \(C=\pi r l, V=\frac{1}{3} \pi r^2 h, l^2=h^2+r^2\)

L.H.S. = 3πVh³ – C²h² + 9V²

= \(3 \pi\left(\frac{1}{3} \pi r^2 h\right) h^3-(\pi r l)^2 h^2+9\left(\frac{1}{3} \pi r^2 h\right)^2\)

= \(\pi^2 r^2 h^4-\pi^2 r^2 h^2\left(h^2+r^2\right)+9 \times \frac{1}{9} \pi^2 r^4 h^2\)

= π²r²h⁴ – π²r²h⁴ – π²r⁴h² + π²r⁴h² = 0 = R.H.S. Hence Proved.

Question 11. A cone of equal height and equal base is cut off from a cylinder of height 24 cm and base radius 7 cm. Find the total surface and volume of the remaining solid.

Solution:

Given

A cone of equal height and equal base is cut off from a cylinder of height 24 cm and base radius 7 cm.

Here r = 7 cm, h = 24 cm

The volume of remaining solid = Volume of the cylinder – Volume of a cone

= \(\pi r^2 h-\frac{1}{3} \pi r^2 h=\frac{2}{3} \pi r^2 h\)

= \(\frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 24=2464 \mathrm{~cm}^3\)

Now, l² = h² + r² = (24)² + (7)²

= 576 + 49 = 625

⇒ l = 25 cm

∴ Total surface area of remaining solid

= Curved surface of cylinder + Curved surface of cone + Area of top

= 2πrh +πrI + πr²

= πr(2h + l + r)

= \(\frac{22}{7} \times 7(2 \times 24+25+7)=22 \times 80=1760 \mathrm{~cm}^2\)

Total surface area of remaining solid = 1760 cm²

Question 12: From a wooden cubical block of edge 7 cm, the largest possible right conical piece is cut out whose base is on one of the faces of the cube. Calculate:

1. The volume of the wood left in the block, and
2. the total surface area of the block left. (\(\text { Take } \pi=\frac{22}{7}\))

Solution:

Initial volume of the block = a³ = 7³ cm³

The base of the largest cone will touch the sides of the base of the cube, and the height will be equal to the length of the edge of the cube.

∴ For the cone, r = 3.5 cm and h = 7 cm.

1. The volume of the cone = \(\frac{1}{3} \pi r^2 h=\frac{\pi}{3} \cdot\left(\frac{7}{2}\right)^2 \cdot 7 \mathrm{~cm}^3\)

∴ volume of the wood left = \(\left\{7^3-\frac{1}{3} \cdot \frac{22}{7} \cdot\left(\frac{7}{2}\right)^2 \cdot 7\right\} \mathrm{cm}^3\)

= \(\left(7^3-\frac{11 \times 7^2}{3 \times 2}\right) \mathrm{cm}^3=7^2\left(7-\frac{11}{6}\right) \mathrm{cm}^3\)

= \(\frac{49 \times 31}{6} \mathrm{~cm}^3=\frac{1519}{6} \mathrm{~cm}^3=253 \frac{1}{6} \mathrm{~cm}^3\)

2. The total surface area of the wood left

= Total surface area of the cube- Area of the base of the cone + Curved surface area of the cone.

= 6a² – πr² + πrl

= \(\left\{6 \times 7^2-\frac{22}{7} \cdot\left(\frac{7}{2}\right)^2+\frac{22}{7} \cdot \frac{7}{2} \cdot \sqrt{7^2+\left(\frac{7}{2}\right)^2}\right\} \mathrm{cm}^2\)

= \(\left(6 \times 49-\frac{11 \times 7}{2}+11 \cdot 7 \cdot \frac{\sqrt{5}}{2}\right) \mathrm{cm}^2=\left\{294+\frac{77}{2}(\sqrt{5}-1)\right\} \mathrm{cm}^2\)

= \(\left(294+\frac{77}{2} \times 1.24\right) \mathrm{cm}^2=341.74 \mathrm{~cm}^2\)

Question 13. A sphere is inscribed in a cylinder such that the sphere touches the cylinder. Show that the curved surface is equal to the curved surface of the cylinder.

Solution:

Given

A sphere is inscribed in a cylinder such that the sphere touches the cylinder.

A cylinder circumscribed a sphere is shown.

Here, the radius of the sphere = radius of the cylinder = r

Height of cylinder h = 2r

Now, the curved surface of the sphere = 4πr²

and the curved surface of sphere of cylinder = 2πrh = 2πr(2r) = 4πr²

Therefore, the curved surface of the sphere and the curved surface of the cylinder are equal.

Hence proved.

Question 14. The largest possible cube Is made from a wooden sphere of radius 6√3 cm. Find the surface area of the cube.

Solution:

Given

The largest possible cube Is made from a wooden sphere of radius 6√3 cm

Here, a diagonal of the cube will be the diameter of the sphere

∴ length of a diagonal of the cube = 2 x 6√3 cm = 12√3 cm

If an edge of the cube is a then a diagonal of the cube = √3 a

= 12√3 cm

∴ a = 12 cm.

∴ The surface area of the cube = 6a² = 6.(12)² cm² = 864 cm²

Question 15. A hemisphere is inscribed in a cylinder and a cone is inscribed in the hemisphere. The cone vertex lies in the centre of the upper circular part of the cylinder. Show that: \(\frac{1}{3}\) x Volume of cylinder = \(\frac{1}{2}\)x Volume of hemisphere = Volume of cone

Solution:

Given

A hemisphere is inscribed in a cylinder and a cone is inscribed in the hemisphere. The cone vertex lies in the centre of the upper circular part of the cylinder.

Let radius of cone = radius of cylinder = radius of hemisphere = r

∴ Height of cone = Height of cylinder = r

Now, the volume of cone = \(\frac{1}{3} \pi r^2 \cdot(r)=\frac{1}{3} \pi r^3\) → (1)

Volume of hemisphere = \(\frac{2}{3} \pi r^3\)

⇒ \(\frac{1}{2}\)x Volume of hemisphere = \(\frac{1}{3} \pi r^3\) → (2)

and Volume of cylinder = πr²(r) = r³

⇒ \(\frac{1}{3}\) x Volume of cylinder = \(\frac{1}{3} \pi r^3\) → (3)

From equations (1), (2) and (3)

⇒ \(\frac{1}{3}\) x Volume of cylinder = \(\frac{1}{2}\)x Volume of hemisphere

= Volume of cone

Hence proved.

Question 16. Water in a canal, 5.4 m wide and 1.8 m deep is flowing at a speed of 25 km/hr. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation?

Solution:

Given

Water in a canal, 5.4 m wide and 1.8 m deep is flowing at a speed of 25 km/hr.

Width of canal = 5.4 m

Height of canal = 1.8 m

In one hour, water is moving upto 25 km = 25000 m

∴ We treat it as the length of the canal.

∴ Volume of water in canal in 1 hour = (25000 x 5.4 x 1.8)m³

⇒ Volume of water in the canal in 40 minutes = 243000 m³

∴ Volume of water in canal in 60 minutes = \(\frac{243000}{60} \times 40 \mathrm{~m}^3\)

= 162000 m³

This water can irrigate a field upto the height of 10 cm = 0. 1 m

∴ Volume = Area = Area x height

⇒ 162000 = Area x 0.1

⇒ Area of field = \(\frac{162000}{0.1} \mathrm{~m}^2=1620000 \mathrm{~m}^2\) (∵Note: 1 Hectare = 1000 m)

= \(\frac{1620000}{10000} \text { hectare }\)

= 162 hectare.

So, 162 hectare area can be irrigated.

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Conversion Of Solid Form One Shape To Another And Mixed Problems

When a solid is converted into another without any loss of material then its volume remains the same.

If a larger solid is converted into smaller solids then the number of smaller solids

= \(\frac{\text { Volume of larger solid }}{\text { Volume of } 1 \text { smaller solid }}\)

Conversion Of Solid Form One Shape To Another And Mixed Problems Solved Examples

Question 1. A solid metallic cuboid of dimensions 9 m x 8 m x 2 m is melted and recast into solid cubes of edge 2m. Find the number of cubes so formed.

Solution:

Given

A solid metallic cuboid of dimensions 9 m x 8 m x 2 m is melted and recast into solid cubes of edge 2m.

Let x cubes each of edge 2 m are formed to melt a cuboid of dimensions 9 x 8 x 2.

So, \(\text { number of cubes }=\frac{\text { Volume of cuboid }}{\text { Volume of } 1 \text { cube }}\)

∴ \(x=\frac{9 \times 8 \times 2}{2}=72\)

So, 72 cubes will be formed

Question 2. Three cubes of metal whose edges are in the ratio 3:4 are melted down into a single cube whose diagonal is 12√3 cm, Find the edges of the three cubes.

Solution:

Given

Three cubes of metal whose edges are in the ratio 3:4 are melted down into a single cube whose diagonal is 12√3 cm,

The ratio in the edges = 3:4:5

Let edges be 3x, 4x and 5x respectively.

∴ Volumes of three cubes will be 27x³, 64x³ and 125x³ in cm³ respectively.

Now, sum of the volumes of these three cubes = 27x³ + 64x³ + 125 x³ = 216 x³ cm³

Let the edge of the new cube be a cm.

Diagonal of new cube = a√3 cm

a√3 = 12√3

Volume of new cube = (12)³ = 1728 cm³

Now by the given condition

216x³ = 1728

⇒ x³ = 8

⇒ x = 2

∴ \(\left.\begin{array}{rl}
\text { Edge of 1 cube } & =3 \times 2=6 \mathrm{~cm} \\
\text { Edge of 2 cube } & =4 \times 2=8 \mathrm{~cm} \\
\text { Edge of 3 cube } & =5 \times 2=10 \mathrm{~cm}
\end{array}\right\}\).

Question 3. A cube of metal with a 2.5 cm edge is melted and cast into a rectangular solid whose base is 1.25 cm by 0.25 cm. Assuming no loss in melting find the height of the solid. Also, find the gain in the surface area.

Solution:

Given

A cube of metal with a 2.5 cm edge is melted and cast into a rectangular solid whose base is 1.25 cm by 0.25 cm.

The volume of the cube = (edge)³ = (2.5)³ cu. cm

Area of the base of rectangular solid = 1 .25 x 0.25 sq. cm

∴ Height of solid = \(\frac{\text { Volume }}{\text { Area of base }}=\frac{(2.5)^3}{1.25 \times 0.25}=50 \mathrm{~cm}\)

Surface area of the cube = 6 x (edge)² = 6 x (2.5)² = 37.5 sq. cm

Surface area of the solid = 2(lb + bh + hl) = 2(50 x 1.25 + 1.25 x 0.25 + 50 x 0.25)

= 2(62.5 + 0.3125 + 12.5) = 150.625 sq. cm

∴ Gain in surface area = 150.625 – 37.50 = 113.125 sq. cm

Question 4. A granary is in the shape of a cuboid of size 8m x 6m x 3m. If a bag of grain occupies a space of 0.65 m3, how many bags can be stored in the granary?

Solution:

Given

A granary is in the shape of a cuboid of size 8m x 6m x 3m. If a bag of grain occupies a space of 0.65 m³

The size of the granary is 8 m x 6 m x 3 m,

Volume of granary = 8 x 6 x 3 = 144m³

The volume of one bag of grain = 0.65 m³

The number of bags which can be stored in the granary

= \(\frac{\text { Volume of granary }}{\text { Volume of each bag }}=\frac{144}{0.65}=221.54 \text { or } 221 \text { bags. }\)

Question 5. A cylinderical bucket 28 cm in diameter and 72 cm high is full of water. The water is emptied into a rectangular tank 66 cm long and 28 cm wide. Find the height of the water level in the tank.

Solution:

Given

A cylinderical bucket 28 cm in diameter and 72 cm high is full of water. The water is emptied into a rectangular tank 66 cm long and 28 cm wide.

Let the height of the water level in the tank = x m,

then according to the problem

πr²h = l x b x x

or \(\frac{22}{7} \times 14 \times 14 \times 72=66 \times 28 \times x\)

or \(x=\frac{\frac{22}{7} \times 14 \times 14 \times 72}{66 \times 28}=24 \mathrm{~cm}\)

Question 6. A rectangular container whose base is a square of side 15 cm stands on a horizontal table and holds water upto 3 cm from the top. When a cube is placed in the water and is completely submerged, the water rises to the top and 54 cm of water overflows. Calculate the volume of the cube and its surface area.

Solution:

Given

A rectangular container whose base is a square of side 15 cm stands on a horizontal table and holds water upto 3 cm from the top. When a cube is placed in the water and is completely submerged, the water rises to the top and 54 cm of water overflows.

Volume of the cube submerged = Volume of water that fills 3 cm height of the container + Volume of water that overflows

= 15 x 15 x 3 + 54 = 729 cm³

If the side of the cube submerged = x cm

Its volume = x³ cm³

∴ x³ = 729 = 9 x 9 x 9

⇒ x = 9 cm.

∴ The side of the cube = 9 cm

And its surface area = 6 x (side) =6 x 9 x 9 = 486 cm²

Question 7. A solid spherical ball of iron with a radius of 6 cm is melted and recast into three solid spherical balls. The radii of the two of balls are 3 cm and 4 cm respectively, determine the diameter of the third ball.

Solution:

Given

A solid spherical ball of iron with a radius of 6 cm is melted and recast into three solid spherical balls. The radii of the two of balls are 3 cm and 4 cm respectively,

Let the radius of the third ball = r cm

∴ The volume of three balls formed = Volume of the ball melted

⇒ \(\frac{4}{3} \pi(3)^3+\frac{4}{3} \pi(4)^3+\frac{4}{3} \pi(r)^3=\frac{4}{3} \pi(6)^3\)

⇒ 27 + 64 + r³ = 216

⇒ r³ = 125, i.e., r = 5 cm

∴ The diameter of the third ball = 2 x 5 cm = 10 cm

Question 8. 50 circular plates each of radius 7 cm and thickness 0.5 cm are placed one above the other to form a solid right circular cylinder. Find

1. The total surface area and
2. The volume of the cylinder so formed

Solution:

The height of the cylinder formed by placing 50 plates = 50 x 0.5 = 25 cm

Radius of cylinder formed = Radius of plate = 7 cm

1. Total surface area of cylinder = 2πrh + 2πr²

= \(\left[2 \times \frac{22}{7} \times 7 \times 25+2 \times \frac{22}{7} \times(7)^2\right] \mathrm{cm}^2\)

= (1100 + 308) cm² = 1408 cm²

2. Volume of cylinder = \(\pi r^2 h=\frac{22}{7} \times(7)^2 \times 25=3850 \mathrm{~cm}^3\)

Question 9. A rectangular paper of 22 cm x 12 cm is folded in two different ways and forms two cylinders.

1. Find the ratio of the volumes of two cylinders.
2. Find the difference in the volumes of the two cylinders.

Solution:

When the paper is folded along a 12 cm side,

then height of cylinder h₁ = 22 cm and 2πr₁ = 12

⇒ \(r_1=\frac{12}{2 \pi}=\frac{6}{\pi} \mathrm{cm}\)

∴ Volume = \(V_1=\pi r_1^2 h_1=\pi \times\left(\frac{6}{\pi}\right)^2 \times 22=\frac{792}{\pi}=\frac{792 \times 7}{22}=252 \mathrm{~cm}^3\)

When the paper is folded along a 22 cm side,

then height of cylinder h₂ = 12 cm and 2πr₂ = 22

⇒ \(2 \times \frac{22}{7} \times r_2=22\)

⇒ \(r_2=\frac{7}{2} \mathrm{~cm}\)

∴ Volume V₂ = r_2²h₂

= \(\frac{22}{7} \times\left(\frac{7}{2}\right)^2 \times 12\)

= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12=462 \mathrm{~cm}^3\)

1. Ratio of volumes, V₁:V₂ = 252 : 462 = 6: 11
2. Difference in volume = (462 – 252) cm³ = 210 cm³

Question 10. A semicircle of radius 17.5 cm is rotated about its diameter. Find the curved surface of the generated solid.

Solution:

Given

A semicircle of radius 17.5 cm is rotated about its diameter.

The solid generated by a circle rotated about its diameter is a sphere.

Now, the radius of the sphere r = 17.5 cm.

and its curved surface = \(4 \pi r^2=4 \times \frac{22}{7} \times 17.5 \times 17.5=3850 \mathrm{~cm}^2\)

Question 11. A sphere of radius 6 cm is melted and recast into a cone of height 6 cm. Find the radius of the cone.

Solution:

Given

A sphere of radius 6 cm is melted and recast into a cone of height 6 cm.

Radius of sphere = 6 cm

∴ Volume of sphere = \(\frac{4}{3} \pi(6)^3=288 \pi \mathrm{cm}^3\)

Let the radius of cone = r

Height of cone = 6 cm

∴ Volume of cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi r^2 \times 6=2 \pi r^2\)

Given that, Volume of cone = Volume of a sphere

⇒ 2nr² – 288 7r

⇒ r²= 144

⇒ r = 12

Therefore, the radius of the cone = 12 cm

Question 12. The height and radius of the base of a metallic cone are 27 cm and 16 cm respectively. It is melted and recast into a sphere. Find the radius and curved surface of the sphere.

Solution:

Given

The height and radius of the base of a metallic cone are 27 cm and 16 cm respectively. It is melted and recast into a sphere.

Height of cone h = 27 cm

The radius of cone r = 16 cm

∴ Volume of cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi(16)^2 \times 27 \mathrm{~cm}^3\)

Let the radius of the sphere = R

∴ Volume of sphere = \(\frac{4}{3} \pi R^3\)

Now, Volume of sphere = Volume of cone

⇒ \(\frac{4}{3} \pi R^3=\frac{1}{3} \pi(16)^2 \times 27\)

⇒ \(R^3=\frac{16^2 \times 27}{4}=4^3 \times 3^3\)

R = 4 x 3 = 12cm

and curved surface of sphere = 4R² = 4(12)² = 576 cm²

Question 13. The radius of a metallic sphere is 60 mm. It is melted and recast into a wire of diameter 0.8 mm. Find the length of the wire.

Solution:

Given

The radius of a metallic sphere is 60 mm. It is melted and recast into a wire of diameter 0.8 mm.

Radius of sphere = 60 mm = 6 cm

∴ Volume of sphere = \(\frac{4}{3} \pi(6)^3=288 \pi \mathrm{cm}^3\)

Let, the length of wire = l cm

Radius of wire \(r=\frac{0.8}{2} \mathrm{~mm}=0.4 \mathrm{~mm}=\frac{0.4}{10} \mathrm{~cm}=\frac{4}{100} \mathrm{~cm}\)

∴ Volume of wire = \(\pi r^2 l=\pi\left(\frac{4}{100}\right)^2 \cdot l\)

Now, Volume of wire = Volume of a sphere

⇒ \(\pi \times \frac{4}{100} \times \frac{4}{100} \times l=288 \pi\)

⇒ \(l=\frac{288 \times 100 \times 100}{4 \times 4}=180000 \mathrm{~cm}=1800 \mathrm{~m}\)

Question 14. A wire of diameter 3 mm is wound around a cylinder whose height is 1 2 cm and radius 5 cm so as to cover the curved surface of the cylinder completely. Find

1. Length of the wire.
2. Mass of the wire, assuming the density of copper to be 8.88 g per cm³.

Solution:

Given

A wire of diameter 3 mm is wound around a cylinder whose height is 1 2 cm and radius 5 cm so as to cover the curved surface of the cylinder completely.

1. Let the wire wounded around the cylinder complete n revolutions.

Diameter (width) of wire = 3 m- 0.3 cm

So, the whole height of the cylinder = 0.3 n cm

But the whole height of the cylinder = 12 cm

∴ \(0.3 n=12 \Rightarrow n=\frac{12}{0.3}=\frac{120}{3}=40\)

So, 40 revolutions are completed to wound the wire completely on the cylinder.

In 1 revolution, the length of the wire = 2 πr

∴ In 40 revolutions, the length of the wire = 40 x 2πr

= \(80 \times \frac{22}{7} \times 5 \mathrm{~cm}=1257.14 \mathrm{~cm}\)

= 12.57 m.

2. Now, radius of wire = \(\frac{0.3}{2}=0.15 \mathrm{~cm}\)

Volume of wire = area of cross section x length of wire

= π(0.15)² x 1257.14 = 88.898 cm³

∴ Mass of wire = volume x density

= 88.898 x 8.88 g= 789.41 g

Hence, the length of the wire = 12.57 m

and mass of wire = 789.41 g

Question 15. A metallic cylinder of diameter 16 cm and height 9 cm is melted and recast into a sphere of diameter 6 cm. How many such spheres can be formed?

Solution:

Given

A metallic cylinder of diameter 16 cm and height 9 cm is melted and recast into a sphere of diameter 6 cm.

For the cylinder,

Radius = \(\frac{16}{2} = 8cm[latex]

Height = 9cm

∴ Volume of cylinder = JI(8)2(9) = 576K cm3

Diameter of sphere = 6 cm

∴ Radius of sphere = [latex]\frac{6}{2}=3 cm\)

Now, volume of one sphere = \(\frac{4}{3} \pi(3)^3=36 \pi \mathrm{cm}^3\)

∴ Number of spheres formed = \(\frac{\text { Volume of cylinder }}{\text { Volume of one sphere }}=\frac{576 \pi}{36 \pi}=16\)

Question 16. A solid metallic right circular cone of height 6.75 cm and radius of the base 12 cm is melted and two solid spheres formed from it. If the volume of one of the spheres is 8 times that of the other, find the radius of the smaller sphere.

Solution:

Given

A solid metallic right circular cone of height 6.75 cm and radius of the base 12 cm is melted and two solid spheres formed from it. If the volume of one of the spheres is 8 times that of the other

Let the radius of the smaller sphere be r and the radius of the larger sphere be R.

image

So, \(\frac{1}{3} \pi(12)^2 \times 6.75=\frac{4}{3} \pi r^3+\frac{4}{3} \pi R^3\)

⇒ (12)² x 6.75 = 4(r³ + R³) → (1)

Now, the volume of the larger sphere = 8 x the Volume of the smaller sphere

⇒ \(\frac{4}{3} \pi R^3=8 \times \frac{4}{3} \pi r^3\)

R³ = 8r³ = R = 2r → (2)

From equations (1) and (2), we get

12 x 12 x 6.75 = 4(r³ + 8r³)

⇒ \(9 r^3=\frac{12 \times 12 \times 6.75}{4}\)

⇒ \(r^3=\frac{12 \times 12 \times 6.15}{4 \times 9}=27=(3)^3\)

r = 3 cm.

Hence, the radius of a smaller sphere = 3 cm.

Question 17. Water is flowing at the rate of 3 km an hour through a circular pipe of 20 cm internal diameter into a circular cistern of diameter 10 m and depth 2m. In how much time will the cistern be filled?

Solution:

Given

Water is flowing at the rate of 3 km an hour through a circular pipe of 20 cm internal diameter into a circular cistern of diameter 10 m and depth 2m.

Internal diameter of pipe = 20 cm

∴ Internal radius of pipe = 10 cm = \(\frac{1}{10}m\)

Rate of flow of water = 3 km/h

⇒ Rate of flow of water = \(\frac{3 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{sec}\)

Diameter of cistern = 10 m

Radius of cistern = 5 m

Height of cistern = 2 m

Water discharge by pipe in 1 sec

= π x r² x flow of water

= \(\pi \times\left(\frac{1}{10}\right)^2 \times \frac{3 \times 1000}{60 \times 60} \mathrm{~m}^3 / \mathrm{sec}\)

Volume of cistern = π x 5² x 2 m³

Time taken to fill the cistern = \(\frac{\text { Volume of cistern }}{\text { Volume of water discharge in }1 \mathrm{sec}}\)

= \(\frac{\pi \times 25 \times 2}{\pi \times \frac{1}{100} \times \frac{3 \times 1000}{60 \times 60}} \mathrm{sec}=\frac{25 \times 2 \times 100 \times 60 \times 60}{3 \times 1000} \mathrm{sec}\)

= \(\frac{5 \times 60 \times 60}{3} \sec =\frac{5 \times 60 \times 60}{3 \times 60 \times 60} \text { hours }=\frac{5}{3} \text { hours }\)

Time taken to fill the cistern =\(\frac{5}{3} \text { hours }\)

Question 18. The volume of a sphere is 288 π cm³. 27 small spheres can be formed with this sphere. Find the radius of a small sphere.

Solution:

Given

The volume of a sphere is 288 π cm³. 27 small spheres can be formed with this sphere.

Volume of 27 small spheres = Volume of one big sphere = 288π

Volume of 1 small sphere = \(\frac{288}{27} \pi\)

\(\frac{4}{3} \pi r^3=\frac{32}{3} \pi\) where r = radius of small sphere

r³ = 8

r = 2 cm

The radius of a small sphere = 2 cm

Question 19. Find:

1. The lateral or curved surface area of a closed cylindrical petrol storage tank is 4.2 m in diameter and 4.5 m high.
2. How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank?

Solution:

1. Curved surface area = \(2 \pi r h=2 \times \frac{22}{7} \times \frac{4.2}{2} \times 4.5=59.4 \mathrm{~m}^2\)

2. Total steel used = Total surface area (Assuming thickness = 0 m)

= 2πr(r + h)

= \(2 \times \frac{22}{7} \times \frac{4.2}{2}\left(\frac{4.2}{2}+4.5\right)\)

= \(\frac{44}{7} \times 2.1 \times \frac{13.2}{2}=22 \times 0.3 \times 13.2=87.12 \mathrm{~m}^2\)

Let the actual steel used be x m²

= \(\frac{11}{12} x=87.12\)

= \(x=\frac{87.12 \times 12}{11}=95.04 \mathrm{~m}^2\)

Question 20. In one fortnight (15 days) of a given month, there was a rainfall of 11.10 cm in a river valley. If the area of the valley is 97280 km², show that the total rainfall of a day was approximately equivalent to the addition (sum) to the normal water of three rivers earth 1072 km long, 7.5 m wide and 3m deep. (Round off the volumes upto one place of decimal).

Solution:

Given

In one fortnight (15 days) of a given month, there was a rainfall of 11.10 cm in a river valley. If the area of the valley is 97280 km²

Area of the valley = 97280 km² = 9.728 x 10⁴ km²

= 9.728 x 10¹⁰ m²

∴ Volume (amount) of rainfall in the valley in one fortnight (15 days)

= \(9.728 \times 10^{10} \times \frac{11.10}{100} \mathrm{~m}^3=107.98 \times 10^8 \mathrm{~m}^3\)

∴ Amount of rainfall in 1 day = \(\frac{107.98 \times 10^8}{15} \mathrm{~m}^3\)

= 7.198 x 10⁸ m³ = 7.2 x 10⁸m³ → (1)

Now, length of each river = 1072 km = 1072000 m

breadth of each river = 75 m

and depth of each river = 3 m

∴ Amount (volume) of water in each river = 1072000 x 75 x 3 m³

∴ Volume of water in 3 such rivers = 3 x 1.072 x 10⁶ x 75 x 3 m³

= 7.236 x 10⁸ m³ = 7.2 x 10⁸ m³ → (2)

From (1 ) and (2), we can say that total rainfall in a day is approximately the same as the volume of 3 rivers.

Question 21. The dimensions of a solid iron cuboid are 4.4 m x 2.6 m x 1.0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness of 5 cm. Find the length of the pipe.

Answer:

Given

The dimensions of a solid iron cuboid are 4.4 m x 2.6 m x 1.0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness of 5 cm.

Length of solid iron cuboid = 4.4 m

Breadth of solid iron cuboid = 2.6 m

Height of solid iron cuboid = 1.0 m

∴ Volume of solid cuboid = 4.4 x 2.6 x 1 = 11.44 m³

= 11.44 x 100 x 100 x 100 cm³

This cuboid is melted and recast into a hollow cylinder.

∴ The volume of a cuboid = Volume of a hollow cylinder

⇒ 11.44 x 10⁶ = πR²h – πr²h

⇒ 11.44 x 10⁶ = πh(R²– r²) [∵ r = 30 cm, width = 5 cm, R = 35 cm]

⇒ \(11.44 \times 10^6=\frac{22}{7} \times h(R-r)(R+r)\)

⇒ 11 .44 x 106 x 7 = 22 x h (35 – 30) (35 + 30)

⇒ \(h=\frac{11.44 \times 10^6 \times 7}{22 \times 5 \times 65} \mathrm{~cm}=\frac{11440000 \times 7}{22 \times 5 \times 65} \mathrm{~cm}\)

= 11200 cm = 112 m

Hence, the length of the pipe = 112 m

Question 22. A tank measures 2 m long, 1.6 m wide and 1 m depth. Water is there upto 0.4 m in height. Brides measuring 25 cm x 14 cm x 10 cm are put into the tank so that water may come upto the top. Each brick absorbs water equal \(\frac{1}{7} \text { th }\) to that of its own volume. How many bricks will be needed?

Solution:

Let x bricks be needed

∴ Volume of x bricks = x X 0.25 X 0.1 4 X 0.1 m³

The volume of water absorbed by bricks

= \(=\frac{1}{7} x \times 0.25 \times 0.14 \times 0.1\)

∴ Remaining water in the tank = \(2 \times 1.6 \times 0.4-\frac{1}{7} \times x \times 0.25 \times 0.14 \times 0.1\)

Now, the Volume of water in the tank + Volume of x bricks

= Volume of tank

⇒ \(\left(2 \times 1.6 \times 0.4-\frac{x}{7} \times 0.25 \times 0.14 \times 0.1\right)+x \times 0.25 \times 0.14 \times 0.1\) = 2 x 1.6 x 1

⇒ \(\frac{6 x}{7} \times 0.25 \times 0.14 \times 0.1=2 \times 1.6 \times 0.6\)

⇒ \(x=\frac{2 \times 1.6 \times 0.6 \times 7}{6 \times 0.25 \times 0.14 \times 0.1}=\frac{2 \times 16 \times 6 \times 7 \times 1000}{6 \times 25 \times 14 \times 1}=640\)

Question 23. A metallic solid sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm. How many cones will be made?

Solution:

Given

A metallic solid sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm.

Let n cones be recast from the sphere.

∴ The sum of volumes of n cones = volume of a sphere

⇒ \(n\left[\frac{1}{3} \pi(3.5)^2 \times 3\right]=\frac{4}{3} \pi(10.5)^3\)

⇒ \(n=\frac{4 \times 10.5 \times 10.5 \times 10.5}{3.5 \times 3.5 \times 3}=126\)

Hence, 126 cones will be made.

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Frustum Of A Right Circular Cone

Frustum: If a right circular cone is cut by a plane parallel to the base of the cone then the portion between the plane and base is called the frustum of the cone.

How To Find The Volume And Surface Area Of A Bucket

Let a bucket of height h and radii of upper and lower ends be r₁ and r₂ respectively.

Now we shall find three parts:

1. The slant is the height of the bucket.
2. Curved surface area and total surface area of the bucket.
3. The volume of the bucket or capacity in litres.

Proof :

1. Let the slant height of the bucket be l

Now, draw DC ⊥ AB

l²=h^{2 +} (r₁ – r₂)²

⇒ \(l=\sqrt{h^2+\left(r_1-r_2\right)^2}\)

2. Let EO = H and DO = L

ΔABO ~ ΔDEO

∴ \(\frac{A B}{D E}=\frac{B O}{E O}=\frac{A O}{D O}\)

⇒ \(\frac{r_1}{r_2}=\frac{h+H}{H}=\frac{l+L}{L}\)

⇒ \(\frac{r_1}{r_2}=\frac{h+H}{H} \quad \text { and } \quad \frac{r_1}{r_2}=\frac{l+L}{L}\)

⇒ Hr₁ = hr₂ + Hr₂ and Lr₁ = lr₂ + Lr₂

⇒ H(r₁– r₂) = hr₂ ⇒ L(r₁ – r₂) = lr₂

⇒ \(H=\frac{h r_2}{r_1-r_2} \quad \text { and } \quad L=\frac{l r_2}{r_1-r_2}\)

Curved surface area of bucket = C.S.A of larger cone- C.S.A of smaller cone

= \(\pi r_1(L+l)-\pi r_2 L=\pi r_1\left[\frac{l r_2}{r_1-r_2}+l\right]-\pi r_2\left(\frac{l r_2}{r_1-r_2}\right)\)

= \(\pi r_1 l\left(\frac{r_2+r_1-r_2}{r_1-r_2}\right)-\frac{\pi r_2^2 l}{r_1-r_2}\)

= \(\frac{\pi l}{r_1-r_2}\left(r_1^2-r_2^2\right)=\frac{\pi l\left(r_1+r_2\right)\left(r_1-r_2\right)}{\left(r_1-r_2\right)}\)

= \(\pi l\left(r_1+r_2\right) \text { where } l=\sqrt{h^2+\left(r_1-r_2\right)^2}\)

Total surface area = C.S.A. of bucket + Area of the smaller circle

= πl(r₁ + r₂) + πr₂²

3. Volume of bucket = Volume of larger cone – Volume of smaller cone

= \(\frac{1}{3} \pi l_l\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \pi r_1^2(H+h)-\frac{1}{3} \pi r_2^2 \cdot H\)

= \(\frac{1}{3} \pi r_1^2\left[\frac{h r_2}{r_1-r_2}+h\right]-\frac{1}{3} \pi r_2^2\left(\frac{h r_2}{r_1-r_2}\right)\)

= \(\frac{1}{3} \pi r_1^2 h\left(\frac{r_2+r_1-r_2}{r_1-r_2}\right)-\frac{1}{3} \frac{\pi h r_2^3}{r_1-r_2}\)

= \(\frac{1}{3} \frac{\pi h}{r_1-r_2}\left(r_1^3-r_2^3\right)\)

= \(\frac{1}{3} \frac{\pi h}{r_1-r_2}\left(r_1-r_2\right)\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \pi h\left(r_1^2+r_2^2+r_1 r_2\right)\)

Note: Actually, the bucket is a frustum cone made of cutting the bucket by a plane parallel to the base.

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Frustum Of A Right Circular Cone Solved Examples

Question 1. A cone is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base, Compare the volume of the two parts.

Solution:

Given

A cone is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base

We can solve this using similarity.

Let r and li be the radius and height of a cone OAB

Let OE = \(\frac{h}{2}\)

As OED and OFB are similar

∴ \(\frac{O E}{O F}=\frac{E D}{F B} \frac{h / 2}{h}=\frac{E D}{r}\)

⇒ \(E D=\frac{r}{2}\)

Now volume of cone OCD = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \pi \times\left(\frac{r}{2}\right)^2 \times \frac{h}{2}=\frac{\pi r^2 h}{24}\)

and Volume of cone OAB = \(\frac{1}{3} \times \pi \times r^2 \times h=\frac{\pi r^2 h}{3}\)

∴ \(\frac{\text { Volume of part } O C D}{\text { Volume of part } C D A B}=\frac{\frac{\pi r^2 h}{24}}{\frac{\pi r^2 h}{3}-\frac{\pi r^2 h}{24}}=\frac{\frac{1}{24}}{\frac{1}{3}-\frac{1}{24}}=\frac{\frac{1}{24}}{\frac{8-1}{24}}=\frac{1}{7}\)

Question 2. The height of a right circular cone is trisected by two planes drawn parallel to the base. Show that the ratio of volumes of the three portions starting from the top is in the ratio 1:7:19.

Solution:

Given

The height of a right circular cone is trisected by two planes drawn parallel to the base.

Since height is trisected, therefore by basic proportionality theorem, base radii of three cones VCD, VA’B’ and VAB are also in the ratio 1: 2 : 3.

( ∵ VL : VM : VN = r₂ : r₁ : r = 1:2:3)

Let volume of cone VCD = V = \(\frac{1}{3} \pi r_2{ }^2 h\)

∴ Volume of cone VA’B’ = \(\frac{1}{3} \pi r_1^2(2 h)=\frac{1}{3} \pi\left(2 r_2\right)^2(2 h)=8 \times \frac{1}{3} \pi r_2{ }^2 h=8 V\)

and volume of cone VAB = \(\frac{1}{3} \pi r^2(3 h)=\frac{1}{3} \pi\left(3 r_2\right)^2(3 h)=27 \times \frac{1}{3} \pi r_2^2 h=27 \mathrm{~V}\)

∴ Ratio of volumes of 3 portions

= Volume(VCD) : Volume(CD£’A’) : Volume(A’B’BA)

= V: 8V – V: 27V – 8V = 1: 7: 19

Hence Proved.

Question 3. The radii of the faces of a frustum of a cone are 3 cm and 4 cm and its height is 5cm. Find its volume.

Solution:

Given

The radii of the faces of a frustum of a cone are 3 cm and 4 cm and its height is 5cm.

Here r = 3 cm, R = 4 cm and h = 5 cm

Volume of frustum of cone = \(\frac{1}{3} \pi l\left(R^2+r^2+R r\right) \text { cu. units }\)

= \(\frac{1}{3} \times \frac{22}{7} \times 5\left(4^2+3^2+4 \times 3\right)\)

= \(\frac{1}{3} \times \frac{110}{7}(16+9+12)\)

= \(\frac{110}{21} \times 37=\frac{4070}{21} \mathrm{~cm}^3\)

Volume of frustum of cone =\(\frac{4070}{21} \mathrm{~cm}^3\)

Question 4. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume is \(\frac{1}{27}\) of the volume of the given cone, at what height above the base, the section has been made?

Solution:

Given

The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume is \(\frac{1}{27}\) of the volume of the given cone

Let OA – h ⇒ AB = 30- h

and let AC = r, BD = R

ΔOAC ~ ΔOBD,

∴ \(\frac{h}{30}=\frac{r}{R} \Rightarrow r=\frac{h R}{30}\)

Now, Volume of smaller cone = \(\frac{1}{27}\) x Volume of larger cone

⇒ \(\frac{1}{3} \pi r^2 h=\frac{1}{27} \times \frac{1}{3} \pi R^2 \times 30\)

⇒ \(\frac{1}{3} \pi\left(\frac{h R}{30}\right)^2 \cdot h=\frac{1}{27} \times \frac{1}{3} \pi R^2 \times 30\)

⇒ \(\frac{h^3}{30^2}=\frac{30}{27} \Rightarrow h^3=\frac{30^3}{3^3} \Rightarrow h=10\)

∴ Reqired height = 30 – 10 = 20 cm.

1 kl = 1 m³

1000 l = 100 x 100 x 100 cm³

1 litre = 1000 cm³

= \(1 \mathrm{~cm}^3=\frac{1}{1000} \text { litre }\)

Question 5. The radii of the ends of a bucket 30 cm high are 21 cm and 7 cm. Find its capacity in litres and the amount of sheets required to make this bucket. (\(\text { Take } \pi=\frac{22}{7}\))

Solution:

Given

The radii of the ends of a bucket 30 cm high are 21 cm and 7 cm.

Although we can solve this problem by using similar triangles, here we are using the direct formula.

Volume of bucket = \(\frac{1}{3} \pi h\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \times \frac{22}{7} \times 30\left(21^2+21 \times 7+7^2\right)\)

= \(\frac{1}{3} \times \frac{22}{7} \times 30 \times 637=20020 \mathrm{~cm}^3=\frac{20020}{1000} \text { lit. }\)

⇒ Capacity = 20.02 lit

Area of sheet = C.S.A. of bucket + Area of base

= πl(r₁ + r₂) + r₂² = π[l(r₁ + r₂) + r2² (∵ \(l=\sqrt{h^2+\left(n_1-r_2\right)^2}\) = \(\sqrt{900+14^2}\) = \(\sqrt{1096}\) = 33.106)

= \(\frac{22}{7}[33.106(28)+49]\)

= 3067.32 = 3067cm² (approx.)

Area of sheet = 3067cm² (approx.)

Question 6. A bucket is 32 cm in diameter at the top and 20 cm in diameter at the bottom. Find the capacity of the bucket in litres if it is 21 cm deep. Also, find the cost of the tin sheet used in making the bucket at the rate of ₹ 1.50 per sq dm.

Solution:

Given

A bucket is 32 cm in diameter at the top and 20 cm in diameter at the bottom.

Here R = 16 cm, r = 10 cm and h = 21 cm

Volume of frustum of a cone = \(\frac{\pi h}{3}\left(R^2+r^2+R r\right)\)

= \(\frac{22}{7} \times \frac{1}{3} \times 21\left(16^2+10^2+16 \times 10\right)\)

= 22(256 + 100 + 160)

= 22 x 516 = 11352cm³

= \(\frac{11352}{1000} \text { litres }=11.352 \text { litres }\)

Now for slant height l of frustum

l² = h² + (R-r)²

l² = 212 + (16 – 10)²

l² = 441 + 36

l² = 477

∴ \(l=\sqrt{477}=21.84 \mathrm{~cm}\)

Now S.A of bucket = πl(R + r) + πr²

= \(\frac{22}{7} \times 21.84 \times(16+10)+\frac{22}{7} \times 10^2\)

= \(\frac{22}{7}(21.84 \times 26+100)\)

= \(\frac{22}{7} \times 667.84=2098.92 \mathrm{~cm}^2=20.99 \mathrm{dm}^2\)

Cost of sheet @ ₹ 1 .50 per sq. dm. = 20.99 x 1.50 = ₹ 31.49

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids Exercise 13.1

Question 1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Given,

volume of cube = 64 cm³

(side)³ = 64

(side)³ = 4³

side = 4 cm

Side of cube = 4 cm

A cuboid is formed by joining two cubes together as shown.

∴ For cuboid

length l = 4 + 4 = 8 cm

height h = 4 cm

Now, the total surface area of a cuboid

= 2(l.b + b.h + l.h)

= 2 (8 x 4 + 4 x 4 + 8 x 4)

= 2 (32 + 16 + 32) = 160 cm²

The surface area of the resulting cuboid = 160 cm²

Question 2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Solution:

Given

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm.

In the adjoining figure, a cylinder is surmounted on the hemisphere.

Diameter of hemisphere 2r = 14 cm

⇒ Radius of hemisphere r = 7 cm

The radius of cylinder r = 1 cm

Now, the total height of the vessel = 13 cm

⇒ h + r = 13 cm

⇒ h = 13 – 7 = 6 cm

Height of cylinder, h = 6 cm

The inner surface area of the cylinder = 2πrh

Inner curved surface area of hemisphere = 2πr²

Inner surface area of vessel = 2πrh + 2πr²

= 2πr(h + r)

= \(2 \times \frac{22}{7} \times 7 \times(6+7)\)

= 44 x 13 = 572 cm²

The inner surface area of the vessel = 572 cm²

Question 3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution:

Given

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm.

In the adjoining figure, a cone of the same cross-section is surmounted on the hemisphere.

The radius of base of cone r = 3.5 cm

⇒ Radius of hemisphere r = 3.5 cm

Total height of toy = 15.5 cm

⇒ h + r = 15.5

⇒ h = 15.5 – 3.5 = 12 cm

∴ Height of cone h = 12 cm

Now, from l² = h² + r²

l² = (12)² + (3.5)²

= 144 + 12.25

= 156.25

⇒ \(l=\sqrt{156.25}=12.5 \mathrm{~cm}\)

∴ The curved surface area of the cone = πrl

and curved surface area of hemisphere = 2πr²

So, the total surface area of the toy = πrl + 2πr²

= πr (l + 2r)

= \(\frac{22}{7} \times 3.5 \times(12.5+2 \times 3.5)\)

= 11 x 19.5 = 214.5cm²

The total surface area of the toy = 214.5cm²

Question 4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution:

Given

A cubical block of side 7 cm is surmounted by a hemisphere.

The base of the hemisphere is on the upper face of a cube of edge 7 cm,

∴ Maximum diameter of hemisphere = edge of the cube

= 7 cm

⇒ 2r = 7 ⇒ \(r=\frac{7}{2} \mathrm{~cm}\)

Now, the total surface area of the solid

= total surface area of cube + curved surface of the hemisphere – an area of the base of the hemisphere

= 6 x (side)² + 2πr²– πr²

= 6 x (side)² + πr²

= \(=6 \times 7 \times 7+\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)

= 294 + 38.5 = 332.5 cm²

The total surface area of the solid = 332.5 cm²

Question 5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution:

Given

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube.

Let the side of the cube = a

Diameter of hemisphere = side of the cube

⇒ 2r = a = \(r=\frac{a}{2}\)

Now, the surface area of remaining solid = total surface of the hemisphere – an area of the base of the hemisphere

= 6a² + 2πr²– πr²

= 6a² + πr²

= \(6 a^2+\pi\left(\frac{a}{2}\right)^2=\frac{24 a^2+\pi a^2}{4}\)

= \(\frac{a^2(24+\pi)}{4} \text { square units }\)

The surface area of the remaining solid = \(\frac{a^2(24+\pi)}{4} \text { square units }\)

Question 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:

Given

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm

Diameter of capsule = 5 mm

2r = 5 mm

r = 2.5 mm

∴ The radius of the cylindrical part = radius of the hemisphere = 2.5 mm

Length of capsule = 14 mm

⇒ h + 2r = 14 mm

⇒ h = 14 – 2r = 14 – 5 = 9 mm

Now the surface area of the capsule = 2 x curved surface of hemisphere + curved surface of the cylinder

= 2 x 2πr² + 2πrh = 2πr (2r + h)

= \(2 \times \frac{22}{7} \times 2.5 \times(5+9)\)

= \(\frac{110}{7} \times 14=220 \mathrm{~mm}^2\)

The surface area of the capsule = 220 mm²

Question 7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m².

Solution:

Given

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m,

In the adjoining, a tent is shown in which a cone is surmounted by a cylinder.

Diameter of cylindrical part 2r = 2. 1 m

⇒ \(r=\frac{2.1}{2} \mathrm{~m}\)

Height of cylindrical part h = 4 m

Radius of conical part \(r=\frac{2.1}{2} \mathrm{~m}\)

The slant height of conical part l = 2.8 m

Now, the area of canvas required to form a tent = curved surface of the cylindrical part + curved surface of the conical part

= 2πrh + πrl

= πr (2h + l)

= \(\frac{22}{7} \times \frac{2.1}{2} \times(2 \times 4+2.8)\)

= 3.3 x 10.8 = 35.64 m²

the area of canvas required to form a tent

Question 8. From a solid cylinder whose height is 2.4 cm and diameter is 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

Solution:

Diameter of cylinder 2r = 1.4 cm

⇒ r = 0.7 cm

∴ The radius of the cylinder = radius of the cone = r = 0.7 cm

Height of cylinder = height of cone = h = 2.4 cm

If the slant height of the cone is l, then

l² = h² + r²= (2.4)² + (0.7)²

= 5.76 + 0.49 = 6.25

⇒ \(l=\sqrt{6.25}=2.5 \mathrm{~cm}\)

Surface area of remaining solid

= area of base of cylinder + curved surface of cylinder + curved surface of cone

= πr² + 2πrh + πrl

= πr (r + 2h + l)

= \(\frac{22}{7} \times 0.7 \times(0.7+2 \times 2.4+2.5)\)

= 2.2 x 8 = 17.6 cm²

Question 9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Solution:

The total surface area of the article

= curved surface of cylinder + 2 x curved surface of a hemisphere

= 2πrh + 2 x 2πr²

= 2πr (h+2r)

= \(2 \times \frac{22}{7} \times 3.5 \times(10+2 \times 3.5)\)

= 22 x 17 = 374 cm²

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids Exercise 13.2

Question 1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Solution:

Radius of hemisphere = radius of cone = r = 1 cm

Height of cone h = radius of cone = 1 cm

Volume of hemisphere = \(\frac{2}{3} \pi r^3\)

Volume of cone = \(\frac{1}{3} \pi r^2 h\)

∴ The volume of solid = volume of hemisphere + volume of a cone

= \(\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h\)

= \(\frac{2}{3} \pi(1)^3+\frac{1}{3} \pi(1)^2(1)\)

= \(\frac{2}{3} \pi+\frac{1}{3} \pi=\pi \mathrm{cm}^3\)

Question 2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution:

The model is shown in the given below.

Diameter 2r = 3 cm

⇒ \(r=\frac{3}{2} \mathrm{~cm}\)

Height of each cone h = 2 cm

Let the height of the cylinder = H

∴ H + h + h = 12 cm

⇒ H + 2 + 2 = 12

⇒ H = 12 – 4 = 8 cm

The volume of model = volume of cylinder + 2 x volume of a cone

= \(\pi r^2 H+2 \times \frac{1}{3} \pi r^2 h\)

= \(\pi r^2\left(H+\frac{2 h}{3}\right)\)

= \(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times\left(8+\frac{2 \times 2}{3}\right)\)

= \(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{28}{3}=66 \mathrm{~cm}^3\)

Air contained in model = 66 cm³

Question 3. A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with a length of 5 cm and a diameter of 2.8 cm.

Solution:

For one gulab jamun,

Diameter 2r = 2.8 cm

⇒ r = 1.4 cm

Height of cylindrical part h = 5 – r – r

= 5-1.4 – 1.4 = 2.2 cm

Volume of one gulab jamun

= volume of cylindrical part + 2 x volume of hemispherical part

= \(\pi r^2 h+2 \times \frac{2}{3} \pi r^3\)

= \(\pi r^2\left(h+\frac{4 r}{3}\right)\)

= \(\frac{22}{7} \times 1.4 \times 1.4 \times\left(2.2+\frac{4 \times 1.4}{3}\right)\)

= 25.051 cm³

⇒ Volume of 45 gulab jamuns = 45 x 25.051

= 1127.295 cm³

Volume of sugar syrup in 45 gulab jamuns

= 30% of 1127.295

= \(1127.295 \times \frac{30}{100} \mathrm{~cm}^3\)

= 338.1885 cm3 ≈ 338 cm³

Question 4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution:

Volume of cuboid = 15 x 10 x 3.5 cm³

= 525 cm³

For conical depression

r = 0.5 cm and h = 1.4 cm

∴ The volume of one cavity

= \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4\)

= \(\frac{1.1}{3} \mathrm{~cm}^3\)

⇒ Volume of four depression= \(4 \times \frac{1.1}{3} \mathrm{~cm}^3\)

= 1.467 cm³

Now, the volume of wood used in the pen stand

= volume of cuboid – volume of four depression

= (525 – 1 .467) cm³ = 523.533 cm³

Question 5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution:

The radius of cone r = 5 cm

and height h = 8 cm

⇒ Volume of cone = \(\frac{1}{3} \pi r^2 h\)

The volume of water filled in the cone

= \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi \times(5)^2 \times 8\)

= \(\frac{200}{3} \pi \mathrm{cm}^3\)

The volume of water flows out on dropping lead shots in it.

= \(\frac{1}{4} \times \frac{200 \pi}{3}=\frac{50 \pi}{3} \mathrm{~cm}^3\)

The radius of shot R = 0.5 cm

Volume of one lead shot = \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi(0.5)^3 \mathrm{~cm}^3\)

= \(\frac{\pi}{6} \mathrm{~cm}^3\)

Now, number of shots

= \(\frac{\text { volume of water flows out }}{\text { volume of one shot }}\)

= \(\frac{50 \pi / 3}{\pi / 6}=\frac{50 \pi}{3} \times \frac{6}{\pi}=100\)

Question 6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use π = 3.14)

Solution:

For the first cylinder,

Diameter 2r = 24 cm

⇒ r = 12 cm

Height h = 220 cm

∴ Volume = πr²h = π x 12 x 12 x 220 cm³

= 31680 π cm³

For the second cylinder,

Radius R = 8 cm

Height H = 60 cm

Volume = πr²H = π x 8 x 8 x 60

= 3840π cm³

Volume of the pole = (31680π + 3840π) cm³

= 35520π cm³

∴ Weight of pole = 35520π x 8 g

= 35520 x 3.14 x 8 g

= 892262.4 g = 892.2624 kg

= 892.26 kg

Question 7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Solution:

The radius of cylinder r = 60 cm and

height h = 1 80 cm

∴ Volume of cylinder = πr²h

= π x 60 x 60 x 180 cm³

= 648000π cm3

The radius of cone R = 60 cm

and height H = 120 cm

∴ Volume of cone = \(\frac{1}{3} \pi R^2 H\)

= \(\frac{1}{3} \pi \times 60 \times 60 \times 120 \mathrm{~cm}^3\)

= 144000 cm³

The volume of solid formed from the cone and hemisphere

= (144000π + 144000π) cm³

= 288000π cm³

⇒ Volume of water displaced by this solid

= 288000π cm³

∴ The volume of the remaining water in a cylinder

= (648000π – 288000π) cm³

= 360000π cm³

= \(360000 \times \frac{22}{7} \mathrm{~cm}^3\)

= \(1131428.57 \mathrm{~cm}^3=\frac{1131428.57}{1000000} \mathrm{~m}^3\)

= 1.131 m³ (approx.)

Question 8. A spherical glass vessel has a cylindrical neck 8 cm long, and 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Solution:

Radius of cylindrical part \(r=\frac{2}{2}=1 \mathrm{~cm}\)

and height h = 8 cm

∴ The volume of the cylindrical part = nr2h

= π(1)²(8) = 8πcm³

Radius of spherical part, \(R=\frac{8.5}{2}=\frac{17}{4} \mathrm{~cm}\)

∴ The volume of the spherical part,

= \(\frac{4}{3} \pi R^3=\frac{4 \pi}{3} \times \frac{17}{4} \times \frac{17}{4} \times \frac{17}{4} \mathrm{~cm}^3\)

= \(\frac{4913}{48} \pi \mathrm{cm}^3\)

∴ Volume of cylinder = \(\left(8 \pi+\frac{4913 \pi}{48}\right) \mathrm{cm}^3\)

= \(\frac{384 \pi+4913 \pi}{48} \mathrm{~cm}^3[/latex

= [latex]\frac{5297 \pi}{48} \mathrm{~cm}^3=\frac{5297}{48} \times 3.14 \mathrm{~cm}^3\)

= 346.51 cm³

So the answer 345 cm³ of a child is not correct.

∴ Correct volume of cylinder = 346.51 cm³

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids Exercise 13.3

Question 1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Solution:

Given

A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm.

The radius of sphere R = 4.2 cm

The radius of cylinder r = 6 cm

Let the height of the cylinder = h

Now, the volume of the cylinder = volume of a sphere

⇒ \(\pi r^2 h=\frac{4}{3} \pi R^3\)

⇒ \(h=\frac{4 R^3}{3 r^2}=\frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6} \mathrm{~cm}\)

= 2.744 cm

∴ Height of cylinder = 27.44 cm

Question 2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution:

Given

Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere.

Let r₁ = 6 cm, r₂ = 8 cm and r₃ = 10 cm

Let the radius of a bigger solid sphere = R

The volume of a bigger solid volume

= sum of volumes of three given spheres

⇒ \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi r_1^3+\frac{4}{3} \pi r_2^3+\frac{4}{3} \pi r_3^3\)

⇒ \(R^3=r_1^3+r_2^3+r_3^3\)

= 63 + 83+ 10²

= 216 + 512 + 1000 = 1728 = 12²

⇒ R = 12 cm

∴ The radius of the new solid sphere = 12 cm

Question 3. A 20 m deep well with a diameter of 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Solution:

Given

A 20 m deep well with a diameter of 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m.

Radius of well, \(r=\frac{7}{2} m\)

and depth h = 20 m

Let the height of the platform be H metre.

∴ The volume of the platform = the volume of the well.

⇒ 22 x 14 x H = πr²h

⇒ \(22 \times 14 \times H=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20\)

⇒ \(H=\frac{22 \times 7 \times 7 \times 20}{7 \times 2 \times 2 \times 22 \times 14}=2.5 \mathrm{~m}\)

∴ Height of platform = 2.5 m

Question 4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution:

Given

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment.

Diameter of well 2r = 3 m

⇒ \(r=\frac{3}{2}=1.5 \mathrm{~m}\)

and depth h = 14 m

∴ The volume of earth taken out from the well = nr2h

= \(\frac{22}{7} \times 1.5 \times 1.5 \times 14=99 \mathrm{~m}^3\)

Now, the outer radius of the well, R = 1.5 + 4 = 5.5m

∴ Area of the ring of platform = π (R² – r²)

= \(\frac{22}{7}\left[(5.5)^2-(1.5)^2\right]\)

= \(\frac{22}{7} \times 7 \times 4=88 \mathrm{~m}^2\)

Let the height of the embankment = H

∴ 88 x H = 99

⇒ \(H=\frac{99}{88}=\frac{9}{8}=1.125 \mathrm{~m}\)

Height of embankment = 1.1 25 m

Question 5. A container shaped like a right circular cylinder having a diameter of 12 cm and a height of 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Answer:

Given

A container shaped like a right circular cylinder having a diameter of 12 cm and a height of 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top.

The radius of the cylindrical container

∴ \(r=\frac{12}{2} \mathrm{~cm}=6 \mathrm{~cm}\)

and height h= 15 cm

∴ Volume = πr²h = 71 x 6 x 6 x 15

= 540 π cm³

⇒ The total volume of ice cream = 540cm³

The radius of cone = radius of hemisphere = R

⇒ \(R=\frac{6}{2}=3 \mathrm{~cm}\)

Height of cone, H = 12 cm

The volume of ice cream in one cone + hemisphere

= \(\frac{1}{3} \pi R^2 H+\frac{2}{3} \pi R^3=\frac{1}{3} \pi R^2(H+2 R)\)

= \(\frac{1}{3} \pi \times 3 \times 3 \times(12+2 \times 3)=54 \pi\)

Now, number of cones = \(\frac{\text { Total volume of ice cream }}{\text { Volume of ice cream in one cone }}\)

= \(\frac{540 \pi}{54 \pi}=10\)

Question 6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?

Solution:

Let the number of silver coins = n

Radius of coin \(r=\frac{1.75}{2} \mathrm{~cm}=\frac{7}{8} \mathrm{~cm}\)

Height h = 2 mm = \(\frac{2}{10} \mathrm{~cm}=\frac{1}{5} \mathrm{~cm}\)

The volume of one coin = πr²h

= \(\frac{22}{7} \times \frac{7}{8} \times \frac{7}{8} \times \frac{1}{5}=\frac{77}{160} \mathrm{~cm}^3\)

∴ Volume of n coins = \(\frac{77 n}{160} \mathrm{~cm}^3\)

Volume of cuboid = 5.5 x 10 x 3.5 = 192.5 cm³

Now, the volume of n coins = volume of a cuboid

⇒ \(\frac{77 n}{160}=192.5 \Rightarrow n=\frac{192.5 \times 160}{77}\)

⇒ n = 400

∴ Number of silver coins = 400

Question 7. A cylinder bucket, 32 cm high and with radius of base of 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Solution:

Given

A cylinder bucket, 32 cm high and with radius of base of 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm

For cylindrical buckets,

Radius r = 18 cm

Height h = 32 cm

∴ The volume of sand = volume of the bucket

= πr²h = π x 18 x 18 x 32 cm³

= 10368π cm³

For conical heap,

Let Raidus = R

Height H = 24 cm

∴ Volume of conical heap = \(\frac{1}{3} \pi R^2 H=\frac{1}{3} \pi R^2 \times 24=8 \pi R^2\)

Now, the volume of conical heap = volume of sand

⇒ 8πR² = 10368π ⇒ R² = 1296

⇒ R = 36 cm

∴ l² = H² + R² = 24² + 36²

= 576 + 1296 = 1872

⇒ \(l=\sqrt{1872}=12 \sqrt{13} \mathrm{~cm}\)

Question 8. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Solution:

Given

Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h.

Speed of water in canal = 10 km/h

= \(\frac{10 \times 1000}{60} \mathrm{~m} / \mathrm{min}\)

= \(\frac{500}{3} \mathrm{~m} / \mathrm{min}\)

Width of canal = 6 m and depth = 1.5 m

Now, the canal will transfer the water equal to the volume of a cuboid of dimensions.

⇒ \(6 \mathrm{~m} \times 1.5 \mathrm{~m} \times \frac{500}{3} m\) in 1 minute.

∴ Volume of water transfer in 30 minutes

= \(30 \times 6 \times 1.5 \times \frac{500}{3}=45000 \mathrm{~m}^3\)

If the depth of the irrigating region = 8 cm

= \(\frac{8}{100} \mathrm{~m}\) then

area x depth = 45000

⇒ \({Area} \times \frac{8}{100}=45000\)

⇒ \(\text { Area }=\frac{45000 \times 100}{8}=562500 \mathrm{~m}^2\)

Therefore, the area of the region irrigated by the canal in 30 minutes = 562500 m²

Question 9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Solution:

Given

farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h

Diameter of tank 2r = 10 m

⇒ r = 5 m

and depth h = 2m

∴ Volume of tank = r2h = (5)2 x 2 = 50 m3

Again, diameter of pipe = 2R = 20 cm

⇒ \(R=10 \mathrm{~cm}=\frac{10}{100} \mathrm{~m}=\frac{1}{10} \mathrm{~m}\)

Speed of water in pipe = 3 km/h

= \(\frac{3 \times 1000}{60} \mathrm{~m} / \mathrm{min}\)

= 50m/min

Now the pipe will transfer the water into the tank in 1 minute equal to the volume of a cylinder of radius \(\frac{1}{10}\) m and length 50 m.

∴ Time taken to fill the tank = \(\frac{\text { volume of cylindrical tank }}{\text { volume of water transfer in tank in 1 minute }}\)

= \(\frac{50 \pi}{\pi \times\left(\frac{1}{10}\right)^2 \times 50}=100 \text { minutes }\)

∴ Time taken to fill the tank completely = 100 minutes

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids Exercise 13.4

Question 1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Solution:

Given

A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm.

The diameters of the frustum of the cone are 4 cm and 2 cm.

∴ Radius r₁ = 2 cm and r₂ = 1 cm

∴ Height of glass h = 14 cm

∴ The volume of a glass of the shape of a frustum of a cone

= \(\frac{1}{3} \pi h\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \times \frac{22}{7} \times 14\left[(2)^2+2 \times 1+(1)^2\right]\)

= \(\frac{44}{3}[4+2+1]\)

= \(\frac{44 \times 7}{3}=\frac{308}{3} \mathrm{~cm}^3\)

= \(=102 \frac{2}{3} \mathrm{~cm}^3\)

So, capacity of glass = = \(102 \frac{2}{3} \mathrm{~cm}^3\)

Question 2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Solution :

Given

The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm.

The slant height of the frustum of cone l = 4 cm.

Circumference of one end 2πr₁ = 18 cm

∴ πr₁ = 9 cm

Circumference of other ends 2πr₂ = 6 cm

∴ πr₂ = 3 cm

Curved surface area of frustum = π(r₁ + r₂) l

= (πr₁ + πr₂) l

= (9 + 3) x 4

= 48 cm²

Therefore, the curved surface area of the frustum of the cone = 48 cm².

Question 3. A fez, tire cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, its radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution:

Given

A fez, tire cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, its radius at the upper base is 4 cm and its slant height is 15 cm

The cap is in the form of a frustum of a cone whose slant height is l = 15 cm.

Radius r₁ = 10 cm and radius r₂= 4 cm

∴ Curved surface of cap = π(r₁ + r₂) l

= \(\frac{22}{7}(10+4) \times 15=660 \mathrm{~cm}^2\)

Area of the closed end of the cap

= \(\pi r_2^2=\frac{22}{7} \times(4)^2 \mathrm{~cm}^2\)

= \(\frac{352}{7} \mathrm{~cm}^2=50 \frac{2}{7} \mathrm{~cm}^2\)

∴ Total canvas used in cap = Curved surface of cap + area of closed-end

= \(\left(660+50 \frac{2}{7}\right) \mathrm{cm}^2\)

= \(710 \frac{2}{7} \mathrm{~cm}^2\)

Therefore, the area of material used for making cap

∴ \(710 \frac{2}{7} \mathrm{~cm}^2\)

Question 4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 1 6 cm with radii of its lower arid upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of the metal sheet used to make the container, if it costs ₹ 8 per 100 cm². (Take π = 3.14)

Solution:

The vessel is in the shape of a frustum of a cone whose height is h = 16 cm.

And radius of upper end r₁ = 20 cm and radius of lower end r₂ = 8 cm

Then, the volume of the vessel = volume of the frustum

= \(\frac{1}{3} \pi h\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \pi \times 16 \times\left[(20)^2+20 \times 8+(8)^2\right]\)

= \(\frac{16}{3} \pi \times 624 \mathrm{~cm}^3=3328 \pi \mathrm{cm}^3\)

= 3328 x 3.14 cm³ = 10449.92 cm³ [∴ π = 3.14]

The milk required to fill the vessel is 10449.92 cm³ or 10.450 litre.

Then, cost of milk at ₹ 20 per litre = 20 x ₹ 10.45 = ₹ 209

The sheet will be used to make the curved surface and base of the vessel.

Then, the area of the base of the vessel = nr₂²

= 3.14 x (8)² = 3.14 x 64 = 200.96 cm²

The slant height of the vessel

⇒ \(l=\sqrt{h^2+\left(r_1-r_2\right)^2}=\sqrt{(16)^2+(20-8)^2}\)

= \(\sqrt{256+144}=\sqrt{400}=20 \mathrm{~cm}\)

Then, curved surface of vessel = π (r₁ + r₂) l

= 3.14 (20 + 8) x 20 cm²

= 3.14 x 28 x 20 cm²

= 1758.4 cm²

∴ Area of sheet used in vessel

= (1758.4 + 200.96) cm²

= 1959.36 cm²

Cost of the sheet at the rate of ₹ 8 per 1 00 cm²

= \(₹ \frac{8}{100} \times ₹ 1959.36=₹ 156.7488\)

= ₹ 156.75

∴ Cost of milk = ₹ 209

and cost of sheet = ₹ 156.75

Question 5. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \({1}{16}\) cm, find the length of the wire.

Solution:

In the given, the diameter of the base of the cone is A’OA and the vertex is V. The angle of the vertex is A’VA = 60°, and then the semi-vertical angle of the cone is α = 30°.

Height of cone = 20 cm

Then, in right ΔOAV,

⇒ \(\tan \alpha=\frac{O A}{O V} \Rightarrow \tan 30^{\circ}=\frac{r_1}{20} \Rightarrow \frac{1}{\sqrt{3}}=\frac{r_1}{20}\)

⇒ \(r_1=\frac{20}{\sqrt{3}} \mathrm{~cm}\)

∵ ΔVO’B and ΔVOA are similar.

∴ \(\frac{V O^{\prime}}{V O}=\frac{O^{\prime} B}{O A} \Rightarrow \frac{10}{20}=\frac{r_2}{r_1} \Rightarrow \frac{r_2}{r_1}=\frac{1}{2}\)

⇒ \(r_1=2 r_2 \Rightarrow r_2=\frac{1}{2} r_1=\frac{1}{2} \times \frac{20}{\sqrt{3}}=\frac{10}{\sqrt{3}} \mathrm{~cm}\)

and height of frustum \(\dot{h}=\frac{1}{2}\) x height of cone

= 10 cm

Then, volume of frustum = \(\frac{1}{3} \pi h\left(r_1^2+r_1 r_2+r_2^2\right)\)

⇒ \(\frac{1}{3} \pi(10)\left[\left(\frac{20}{\sqrt{3}}\right)^2+\frac{20}{\sqrt{3}} \times \frac{10}{\sqrt{3}}+\left(\frac{10}{\sqrt{3}}\right)^2\right]\)

⇒ \(\frac{1}{3} \pi 10\left[\frac{400}{3}+\frac{200}{3}+\frac{100}{3}\right]\)

⇒ \(\frac{7000}{9} \pi \mathrm{cm}^3\)

Diameter of cylindrical wire = \(\frac{1}{16}\) cm

Radius of wire \(r=\frac{1}{32} \mathrm{~cm}\) cm

Let the length of the wire drawn be l cm.

Then, the volume of wire = πr²l

⇒ \(\pi \times \frac{1}{32} \times \frac{1}{32} \times l=\frac{\pi}{1024} l \mathrm{~cm}^3\)

∵ The wire is drawn from the frustum of the cone.

∴ The volume of wire = volume of the frustum

⇒ \(\frac{\pi}{1024} l=\frac{7000}{9} \pi\)

⇒ \(l=\frac{7000 \pi}{9} \times \frac{1024}{\pi} \mathrm{cm}^3\)

= \(\frac{70}{9} \times 1024 \mathrm{~m}\)

= \(\frac{71680}{9} \mathrm{~m}=7964.44 \mathrm{~m}\)

Therefore, the length of the wire = 7964.44 m

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume and Surface Area of Solids Exercise 13.5

Question 1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.

Solution:

Given

A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, to cover the curved surface of the cylinder.

The diameter of the cylinder = 10 cm and

height of cylinder = 12 cm

∴ Circumference of cylinder

= π x diameter = π x 10 = 10π cm

∴ Length of wire used in one round about a cylinder

= 10 cm

∵ The length of the cylinder is 12 cm or 120 mm. When one round of wire is wound on the cylinder then it covers 3mm length of the cylinder.

When two rounds of wire are wound on the cylinder then it covers the (2 x 3) mm length of the cylinder.

When three rounds of wire are wound on the cylinder then it covers (3×3) mm length of the cylinder.

When four rounds of wire are wound on the cylinder then it covers the (4×3) mm length of the cylinder.

The number of wounds of wire to cover the cylinder = \(\frac{120}{3} = 40\)

Length of required wire to wound 40 rounds on cylinder

= 40 x 10 π = 400 π cm

= 400 x 3.14 cm = 1256 cm(approx.)

= 12.56 m

So, the required length of wire = 12.56 m

The volume of wire = length x area of the wire

= \(1256 \times \pi \frac{d^2}{4} \quad\left[d=3 \mathrm{~mm}=\frac{3}{10} \mathrm{~cm}\right]\)

= \(1256 \times 3.14 \times \frac{9}{100 \times 4}\)

= \(\frac{314 \times 3.14 \times 9}{100}=88.74 \mathrm{~cm}^3\)

and mass of wire = 88.74 x 8.88 g

= 788.01 g = 0.788 kg

Question 2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose the value of π as found appropriate.)

Solution:

Given

A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse.

In right ΔABC, ∠B = 90°, AB = 4 cm,

BC = 3 cm

Then, area of \(\triangle A B C=\frac{3 \times 4}{2}=6 \mathrm{~cm}^2\)

Hypotenuse \(A C=\sqrt{A B^2+B C^2}\)

= \(\sqrt{(4)^2+(3)^2}=\sqrt{25}=5\)

BOB’ is perpendicular to AC, if BO = r, then area of \(\triangle A B C=\frac{A C \times B O}{2}=\frac{5}{2} B O=\frac{5}{2} r\)

Then, \(\triangle A B C=\frac{A C \times B O}{2}=\frac{5}{2} B O=\frac{5}{2} r\)

Then, \(\frac{5}{2} r=6\) (∵\(\frac{5}{2} r\)and 6 both are area of AABC)

∴ \(r=\frac{6 \times 2}{5}=2.4 \mathrm{~cm}\)

Now, the radius r = 2.4 cm of the double cone formed by rotating the right ΔABC.

Then, volume of double cone (two cones) = volume of cone (A, BB’) + volume of cone (C, BB’)

= \(\frac{1}{3} \pi r^2(A O)+\frac{1}{3} \pi r^2(O C)\)

= \(\frac{1}{3} \pi r^2(A O+O C)\)

= \(\frac{1}{3} \pi r^2(A C)\) [ AO + OC = AC]

= \(\frac{1}{3} \pi \times(2.4)^2 \times 5=9.6 \pi \mathrm{cm}^3\)

= 9.6 x 3.14 cm3 (π = 3.14)

= 30.144 cm³

and surface area of the double cone (both cones)

= curved surface of the cone (A, BB’) + curved surface of the cone (C, BB’)

= πr (AB) + πr (BC) = nr (AB +BC)

= 3.14 x 2.4 x (4 +3) =3.14 x 2.4 x 7

= 52.75 cm²

Therefore, the volume of the double cone = 30.144 cm³ and surface area = 52.75 .cm² (approximately).

Question 3. A cistern, internally measuring 150cm x 120cm x 110cm, has 129600 cm³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?

Solution:

Volume of cistern

= 150 x 120 x 110 cm³

= 1980000 cm³

The volume of water filled in cistern = 129600 cm³

Volume of each brick = 22.5 x 7.5 x 6.5 cm³

= 1096.875 cm³

Let on placing x bricks, the water rises upto the brim in the cistern.

Then, volume of x bricks = 1096.875 x cm³

Then, volume of absorbs water by bricks = \(1096.875 x \times \frac{1}{17}=\frac{1096.875 x}{17} \mathrm{~cm}^3\)

Then, the volume of remaining water in cistern = \(\left(129600-\frac{1096.875 x}{17}\right) \mathrm{cm}^3\)

Now, volume of x bricks + volume of water in cistern = volume of cistern

∴ \(1096.875 x+129600-\frac{1096.875 x}{17}\) = 1980000

or \(1096.875 x-\frac{1096.875 x}{17}\) = 1980000 – 129600

or \(1096.875 x\left(1-\frac{1}{17}\right)=1850400\)

or \(1096.875 x=\frac{1850400 \times 17}{16}\)

or \(x=\frac{1850400 \times 17}{16 \times 1096.875}\)

= 1792.4 = 1792 (approximately)

Therefore, the number of bricks placed in the cistern is 1792 (approximately).

Question 4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km², show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Solution:

The volume of each river

= 10721cm x 75 m x 3 m

= 1072 x 75 x 3 x 1000 m³

= 241200000 m³

∴ The volume of total water in three rivers

= 3 x 241200000 m³

∴ Total water of rivers = 723600000 m³

∴ Area of valley = 7280 km²

= 7280 x (1000)² m²

= 7280000000 m

∴ Volume of rainwater

= \(7280000000 \times \frac{10}{100} \mathrm{~m}^3\) (∵\(10 \mathrm{~cm}=\frac{10}{100} \mathrm{~m}\))

= 728000000 m³

These two volumes are not equal.

So, it is clear that the given data given in the question are incorrect.

Question 5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion is 8 cm and the diameter portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

Solution:

Height of cylindrical part h = 10 cm

Total height of funnel = 22 cm

∴ Height of frustum of cone (H) = 22 – 10 = 12 cm

Upper radius of frustum of cone = \(R_1=\frac{18}{2}=9 \mathrm{~cm}\)

Lower radius of frustum of cone = \(R_2=\frac{8}{2}=4 \mathrm{~cm}\)

The radius of cylindrical part r = 4 cm

The curved surface of the cylindrical part = 2 πrh

= 2π x 4 x 10 = 80 π cm²

The slant height of the frustum of a cone

⇒ \(l=\sqrt{H^2+\left(R_1-R_2\right)^2}\)

= \(\sqrt{(12)^2+(9-4)^2}=\sqrt{144+25}\)

= \(\sqrt{169}=13 \mathrm{~cm}\)

So, the total surface area of the funnel

∴ The curved surface of the frustum of a cone

= π (R₁+ R₂) l

= π (9 + 4) x 13 = 169 π cm²

∴ The curved surface area of the cylindrical part + curved surface area of the frustum of a cone

= 80 π + 169π = 249π cm²

= \(249 \times \frac{22}{7} \mathrm{~cm}^2\)

= \(\frac{5478}{7}=782 \frac{4}{7} \mathrm{~cm}^2\)

Therefore, area of tin sheet used in funnel = \(782 \frac{4}{7} \mathrm{~cm}^2\)

Question 6. Derive the formula for the curved surface area and total surface area of the frustum of a cone.

Solution:

Let for the (V, AB), V is the vertex, r₂ the base radius and l₂ the slant height. A cone (V, CD) is cut off from this cone from a point O’ below h₁ from the vertex V of this cone, parallel to the base whose’ radius is r₁ and slant height is l₁.

Draw the perpendicular DE from point D to the base.

In ΔVO’D and ΔDEB,

∠VO’D = ∠DEB (VO and DE both are perpendicular to the base)

∠VDO’ = ∠DBE (the bases of two cones are parallel to each other)

∴ ΔVO’D and ΔDEB are similar

\(\frac{V D}{B D}=\frac{O^{\prime} D}{E B}\)

or \(\frac{l_1}{l}=\frac{O^{\prime} D}{O B-O E}=\frac{O^{\prime} D}{O B-O^{\prime} D}\)

while BD = l = slant height of the frustum

⇒ \(\frac{l_1}{l}=\frac{r_1}{r_2-r_1} \Rightarrow l_1=\left(\frac{r_1}{r_2-r_1}\right) l\) → (1)

The curved surface area of a frustum

= curved surface area of cone (V, AB) – curved surface area of cone ( V, CD)

= πr₂l₂ -πr₁l₁ = πr₂(l₁ + BD) – πr₁l₁

= πr₂l₁ +πr₂ (BD) – πr₁l₁

= π(r₂-r₁) l₁ + πr2 l

= \(\pi\left(r_2-r_1\right)\left(\frac{r_1}{r_2-r_1}\right) l+\pi r_2 l\) [from eqn. (1)]

= π r₁l + πr₂l

So, curved surface area of frustum = π (r₁ + r₂) l

Hence Proved

And total surface area of a frustum

= curved surface + area of first end + area of second end

= π(r₁ + r₂) / + πr₁² + πr₂²

= π(r₁ + r₂)l + π (r_1² + r_2²)

Hence Proved.

Question 7. Derive the formula for the volume of the frustum of a cone.

Solution:

From the last question, for the cone (V, AB), height = h₂ and radius = r₂

∴ Volume of cone (V, AB) = \(\frac{1}{3} \pi r_2^2 h_2\)

and volume of cone (V, CD) = \(\frac{1}{3} \pi r_1^2 h_1\)

∴ Volume of frustum = volume of cone (V, AB) – volume of cone (V, CD)

∴ Volume of frustum (V)

= \(\frac{1}{3} \pi r_2^2 h_2-\frac{1}{3} \pi r_1^2 h_1\) → (1)

∴ h₂ = VO’= VO’ + O’O = h₁ + h

∴ Put h₂ = h₁ +h in eqn. (1),

Volume of frustum V = \(\frac{1}{3} \pi r_2^2\left(h_1+h\right)-\frac{1}{3} \pi r_1^2 h_1\)

Volume of frustum V = \(\frac{1}{3} \pi\left(r_2^2-r_1^2\right) h_1+\frac{1}{3} \pi r_2^2 h\) → (2)

In similar ΔVO’D and ΔDEB,

\(\frac{h_1}{h}=\frac{r_1}{r_2-r_1} \quad \Rightarrow \quad h_1=\left(\frac{r_1}{r_2-r_1}\right) h\)

Put \(h_1=\left(\frac{r_1}{r_2-r_1}\right) h\) in eqn. (2),

⇒ \(V=\frac{1}{3} \pi\left(r_2^2-r_1^2\right) \frac{\eta_1}{\left(r_2-r_1\right)} h+\frac{1}{3} \pi r_2^2 h\)

= \(\frac{1}{3} \pi\left(r_2+r_1\right) r_1 h+\frac{1}{3} \pi r_2^2 h\)

= \(\frac{1}{3} \pi\left(r_1^2+r_1 r_2\right) h+\frac{1}{3} \pi r_2^2 h\)

= \(\frac{1}{3} \pi\left(r_1^2+r_1 r_2+r_2^2\right) h\)

Therefore, the volume of the frustum of the cone

= \(\frac{1}{3} \pi\left(r_1^2+r_1 r_2+r_2^2\right) h\) Hence Proved.

NCERT Exemplar Solutions for Class 10 Maths Chapter 13  Volume And Surface Area Of Solids Multiple Choice Questions

Question 1. A surah is a combination of:

1. A sphere and a cylinder
2. A hemisphere and a cylinder
3. Two hemispheres
4. A cylinder and a cone

Answer: 1. A sphere and a cylinder

Question 2. A glass is generally of the shape of:

1. A cone
2. A frustum of a cone
3. A cylinder
4. A sphere

Answer: 2. A frustum of a cone

Question 3. A plummet is a combination of:

1. A cone and a cylinder
2. A hemisphere and a cone
3. A frustum of a cone and a cylinder
4. A sphere and a cylinder

Answer: 2. A hemisphere and a cone

Question 4. An iron piece in the shape of a cuboid of dimensions 49 cm x 33 cm x 24 cm is melted and recast into a solid sphere. The radius of the sphere is:

1. 21 cm
2. 23 cm
3. 25 cm
4. 19 cm

Answer: 1. 21 cm

Question 5. While converting a shape of a solid into another shape, the volume of the new shape:

1. Increases
2. Decreases
3. Remains same
4. Becomes twice.

Answer: 3. Remains same

Question 6. The ratio of the surface of two spheres is 16:9. The ratio of their volumes is:

1. 3:4
2. 64: 27
3. 27: 64
4. 4:3

Answer: 2. 64: 27

Question 7. The diameter of a sphere exactly inscribed in a right circular cylinder of radius r cm and height h cm (h > 2r) is:

1. r cm
2. 2r cm
3. h cm
4. 2h cm

Answer: 2. 2r cm

NCERT Exemplar Solutions for Class 10 Maths Chapter 12  Area Related To Circles

Area Related To Circles Introduction

We are familiar with the shape of a circle. The circumference of the wheels of cars, and coins appears as a circle.

A circle can be defined as “A figure with an outline, every point on this outline is at the same constant distance from a certain point inside the figure, which is called the centre”.

NCERT Exemplar Solutions for Class 10 Maths Chapter 12  Area Related To Circles Area And Circumference Of A Circle

Let r be the radius of the circle. The distance travelled once around a circle is its perimeter, usually called its circumference. Now

Area of circle = πr²

Circumference of circle = 2πr

Read and Learn More Class 10 Maths Solutions Exemplar

π(pi) is a fixed irrational number whose approximate value is \(3.1416 \text { as } \frac{22}{7} \text { or } \frac{355}{113}\) (sometimes)

Diameter: A chord of a circle passing through its centre is called the diameter of the circle.

Diameter = 2 x radius

NCERT Exemplar Solutions for Class 10 Maths Chapter 12  Area Related To Circles Formulae Of Circle

1. Area of circle = πr²
2. Circumference of a circle = 2πr

3. Area of semicircle = \(\frac{1}{2} \pi r^2\)

4. Perimeter of semicircle = (πr + 2r)

5. For a ring having outer radius = R and inner radius = r
Area of ring = π(R² – r²)

6. For rotation of the hands of a clock

1. Angle described by minute hand in 60 minutes = 360°
2. Angle described by hour hand in 12 hours = 360°

7. For rotating wheels

1. Distance moved by a wheel in 1 rotation = its circumference
2. Number of rotation in unit time = \(\frac{\text { distance moved in unit time }}{\text { circumference }}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 12  Area Related To Circles Solved Examples

Question 1. The circumference of a field is 220 m. Find

1. Its radius
2. Its area.

Solution:

Circumference of circle = 220 m

2πr = 220

⇒ \(r=\frac{220 \times 7}{2 \times 22}=35 \mathrm{~m}\)

Area of circle = \(\pi r^2=\frac{22}{7} \times 35 \times 35=3850 \mathrm{~m}^2\)

Question 2. Find the area of a circular park whose radius is 4.5 m.

Solution:

Area of circular park = r²

⇒ Area = \(\text { Area }=\frac{22}{7} \times 4.5 \times 4.5\)

Area = 63.63 m²

The area of a circular park = 63.63 m²

Question 3. The area of a circular plot is 346.5 m². Calculate the cost of fencing the plot at the rate of ₹ 6 per metre.

Solutions:

Area of plot = 346.5 m²

⇒ πr² = 346.5 m² ⇒ \(r^2=\frac{346.5 \times 7}{22}\)

⇒ r² = 110.25 ⇒ r = 10.5 m

Circumference of plot = \(2 \pi r=2 \times \frac{22}{7} \times 10.5=66 \mathrm{~m}\)

Cost of fencing = Circumference x Cost of fencing per metre

= ₹ 66 x 6 = ₹ 396

The cost of fencing the plot at the rate of ₹ 6 per metre = ₹ 396

Question 4. The diameter of a cycle wheel is 28 cm. How many revolutions will it make in moving 13.21cm?

Solution:

Given

The diameter of a cycle wheel is 28 cm.

Distance travelled by the wheel in one revolution = \(2 \pi r=\frac{22}{7} \times 28=88 \mathrm{~cm}\)

Total distance travelled by the wheel = 13.2 x 1000 x 100 cm

Number, of revolutions made by the wheel = \(\frac{\text { total distance }}{\text { circumference }}\)

= \(\frac{13.2 \times 1000 \times 100}{88}=15000 \text { revolutions }\)

15000 revolutions will it make in moving 13.21cm.

Question 5. The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius of the circle.

Solution:

Given

The circumference of a circle exceeds the diameter by 16.8 cm.

Let the radius of the circle be r.

Diameter = 2r

Circumference of circle = 2πr

Using the given information, we have

2πr = 2r + 16.8

⇒ \(2 \times \frac{22}{7} \times r=2 r+16.8\)

⇒ 44r = 14r + 16.8 x 7

⇒ 30r = 117.6

∴ \(r=\frac{117.6}{30}=3.92 \mathrm{~cm}\)

The radius of the circle =3.92 cm

Question 6. Find the area of a ring whose outer and inner radii are respectively 20cm and 1 5cm.

Solution:

Outer radius R = 20 cm

Inner radius r = 15 cm

∴ Area of ring = π(R² – r²)

⇒ Area of ring = \(\frac{22}{7}\left[(20)^2-(15)^2\right]=\frac{22}{7}(400-225)=\frac{22}{7} \times 175 \mathrm{~cm}^2\)

= 22 x 25 = 550 cm².

The area of a ring = 550 cm².

Question 7. A race track is in the form of a ring whose inner circumference is 352 m and the outer circumference is 396 m. Find the width of the track.

Solution:

Given

A race track is in the form of a ring whose inner circumference is 352 m and the outer circumference is 396 m.

Let R and r be the outer and inner radii of the circle.

The width of the track = (R – r) cm

Now, 2πr = 352 ⇒ \(2 \times \frac{22}{7} \times r=352\)

⇒ \(r=\frac{352 \times 7}{2 \times 22}=7 \times 8=56 \mathrm{~m}\)

Again, 2πR = 396 ⇒ \(2 \times \frac{22}{7} \times R=396\)

⇒ \(R=\frac{396 \times 7}{2 \times 22}=7 \times 9=63 \mathrm{~m}\)

∴ R = 63 m, r = 56 m

Width of the track = (R- r) m = (63 – 56) m = 7 m.

Question 8. Two circles touch internally. The sum of their areas is 116π cm² and the distance between their centres is 6 cm. Find the radii of the circles.

Solution:

Given

Two circles touch internally. The sum of their areas is 116π cm² and the distance between their centres is 6 cm.

Let two circles with centres O’ and O having radii R and respectively touch each other at P.

It is given that OO’ = 6

R – r = 6 ⇒ R = 6 + r → (1)

Also, πR² + πr² = 116π (given)

π(R² + r²) = 116π

⇒ R² + r²= 116 → (2)

∴ From equations (1) and (2), we get

(6 + r)² + r² = 116

⇒ 36 + r² + 12r + r²= 116 ⇒ 2r² + 12r – 80 = 0

⇒ r² + 6r – 40 = 0 = (r + 10) (r – 4) = 0

∴ r = -10 or r = 4

But the radius cannot be negative. So, we reject r = -10

∴ r = 4 cm

∴ R = 6 + 4 = 10cm [from(1)]

Hence, the radii of the two circles are 4 cm and 10 cm.

Question 9. The radius of a wheel of a bus is 45 cm. Determine its speed in kilometres per hour, when its wheel makes 315 revolutions per minute.

Solution:

Given

The radius of the wheel of the bus = 45 cm

∴ Circumference of the wheel = 2ar

= \(2 \times \frac{22}{7} \times 45=\frac{1980}{7} \mathrm{~cm}\)

∴ Distance covered by the wheel in one revolution = \(\frac{1980}{7} \mathrm{~cm}\)

Distance covered by the wheel in 315 revolutions = \(315 \times \frac{1980}{7}\)

= 45 x 1 980 = 89 100 cm

= \(\frac{89100}{1000 \times 100} \mathrm{~km}=\frac{891}{1000} \mathrm{~km}\)

∴ Distance covered in 60 minutes or 1 hr =\(\frac{891}{1000} \times 60=\frac{5346}{100}=53.46 \mathrm{~km}\)

Hence, speed of bus = 53.46 km/hr

Question 10. Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm are drawn in the adjoining figure. Find the area of the shaded region.

Solution:

Given

Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm are drawn in the adjoining figure.

Required area = (area of larger semicircle with radius 4.5 cm) – (area of two smaller semicircles with radius of each \(\frac{3}{2}\) cm and a circle with radius \(\frac{4.5}{2}\) cm + (area of smaller semicircle with radius \(\frac{3}{2}\) cm)

= \(\frac{1}{2} \pi(4.5)^2-\left[2 \times \frac{1}{2} \pi\left(\frac{3}{2}\right)^2+\pi\left(\frac{4.5}{2}\right)^2\right]+\left[\frac{1}{2} \pi\left(\frac{3}{2}\right)^2\right]\)

= \(\frac{1}{2} \pi(4.5)^2-\pi\left(\frac{3}{2}\right)^2-\pi\left(\frac{4.5}{2}\right)^2+\frac{1}{2} \pi\left(\frac{3}{2}\right)^2\)

= \(\frac{1}{4} \pi\left[2 \times(4.5)^2-9-(4.5)^2+\frac{9}{2}\right]=\frac{1}{4} \pi \times 4.5[2 \times 4.5-2-4.5+1]\)

= \(\frac{1}{4} \pi \times 4.5(3.5)=\frac{1}{4} \times \frac{22}{7} \times 4.5 \times 3.5=12.375 \mathrm{~cm}^2\).

The area of the shaded region =12.375 cm².

Question 11. In the adjoining figure, find the area of the shaded region. (Use π = 3.14)

Solution:

Diameter BD = \(\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10 \mathrm{~cm}\)

Radius = \(]frac {10}{2}\) = 5 cm

Area of circle = πr² = 3. 1 4 x 52 = 3. 14 x 25 = 78.50 cm²

Area of rectangle ABCD = 8 x 6 = 48 cm²

Hence, area of shaded region = 78.50- 48 = 30.5 cm²

Question 12. A park is of the shape of a circle of diameter 7 m. It is surrounded by a path of width 0.7 m. Find the expenditure of cementing the path, if its cost is ₹ 110 per sq. m.

Solution:

Given

A park is of the shape of a circle of diameter 7 m. It is surrounded by a path of width 0.7 m

The radius of the circular park = 3.5 m

There is a path of width 0.7 m.

So, the radius of the external circle R = 3.5 + 0.7

⇒ R = 4.2 cm

Area of path = πR² – πr² = n(R- r) (R + r)

= \(\frac{22}{7}(4.2-3.5)(4.2+3.5)\)

= \(\frac{22}{7} \times 0.7 \times 7.7=16.94 \mathrm{~m}^2\)

Now, the cost of cementing 1 m² of path = ₹ 110

Cost of cementing 16.94 m² of path = ₹ ( 110 x 16.94) = ₹ 1863.40

Question 13. Find the area of the region between the two concentric circles, if the length of a chord of the outer circle just touching the inner circle at a particular point on it is 10 cm. \(Take, \pi=\frac{22}{7}\)

Solution:

Let the chord AB touch the inner circle at C and let 0 be the centre of both the circles, then

OC = r, OA = R and AB = 10 cm

Now, since OC ⊥ AB (∵ radius through the point of contact is perpendicular to the tangent)

∴ C is the mid-point of AB (∵ ⊥ drawn from the centre to the chord, bisects the chord)

⇒ \(A C=\frac{1}{2} A B=\frac{1}{2} \times 10=5 \mathrm{~cm}\)

Now, in right ΔOCA,

OA² = OC² +AC² (by Pythagoras theorem)

R² – r² = 25 → (1)

The required area of the region between two concentric circles

= πR² – πr² = π(R²– r²) – 25π [from(1)]

= \(25 \times \frac{22}{7}=78.57 \mathrm{~cm}^2\)

Question 14. In the adjoining figure, CM = 5 cm, RB = 9 cm, CD ⊥ AB, O is the centre of the larger circle and K is the centre of the smaller circle. Find the area of the shaded region.

Solution:

Let the radius of the larger circle be R and the radius of the smaller circle is r.

Since, CM = 5 cm

∴ MO = R – 5

Also OB = R and RB = 9

∴ OR = R – 9

In ΔAOM,

∠1 + ∠2 = 90° → (1)

In ΔAMR,

∠2 + ∠3 = 90° (the angle in a semicircle is the right angle) → (2)

From eqs. (1) and (2), we get

∠1 + ∠2 = ∠2 + ∠3

⇒ ∠1 = ∠3

In Δs AMO and MOR

∠1 = ∠3 (just proved)

∠4 = ∠5 (each 90°)

∴ ΔAMO ~ ΔMRO (AA corollary)

∴ \(\frac{M O}{R O}=\frac{A O}{M O}\)

⇒ \(\frac{R-5}{R-9}=\frac{R}{R-5}\)

⇒ (R – 5)² = R (R-9) = R² + 25 – 10R = R² – 9R

⇒ R = 25 cm

Now, 2R – 9 = 2r ⇒ 2r = 41

⇒ r = 20.5 cm

Required area = πR² – πr² = π(R + r) (R – r)

= \(\frac{22}{7} \times 45.5 \times 4.5 = 643.5 \mathrm{~cm}^2\)

Question 15. A square of the largest area is cut out of a circle. What % of the area of the circle is lost as trimmings?

Solution:

Let the radius of the circle be r units.

∴ Area of circle = πr² sq. units

Let ABCD be the largest square.

∴ Length of diagonal = 2r = side √2

side = \(\frac{2 r}{\sqrt{2}}=\sqrt{2} r\)

∴ Area (square ABCD) = (side)² = (√2r)² = 2r²

∴ Area of circle lost by cutting out of a square of largest area = 2r²

∴ Required percentage of the area of circle lost = \(\frac{2 r^2}{\pi r^2} \times 100=\frac{200}{\pi} \%\).

Question 16. In a circular table covering of radius 32 cm, a design (shade) is formed leaving an equilateral triangle ABC in the middle as shown in the adjacent figure. Find the area of the shaded region.

Solution:

Since ABC is an equilateral triangle.

∴ ∠A = 60°

⇒ ∠BOC = 2 x ∠BAC = 2 x 60° = 120° (degree measure of an arc is twice the angle subtended by it in an alternate segment)

Draw OM ⊥ BC

So, we can prove

ΔOMB ≅ ΔOMC (R.H.S. congruency)

∴ ∠BOM = ∠COM = 60° (c.p.c.t.)

In the right ΔOMB, we have

⇒ \(\sin 60^{\circ}=\frac{B M}{O B}\)

∴ \(\frac{\sqrt{3}}{2}=\frac{B M}{32} \quad = \quad B M=16 \sqrt{3} \mathrm{~cm}\)

∴ BC = 2 x 16√3 = 32√3 cm

∴ Area of shaded region = Area of circle- ar(AABC)

= \(\pi r^2-\frac{(\text { side })^2 \sqrt{3}}{4}=\pi \times(32)^2-(32 \sqrt{3})^2 \times \frac{\sqrt{3}}{4}\)

= \((32)^2\left(\frac{22}{7}-\frac{3 \sqrt{3}}{4}\right)=1024\left(\frac{22}{7}-\frac{3 \sqrt{3}}{4}\right) \mathrm{cm}^2\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 12  Area Of Sector And Segment Of A Circle

Sector: The shaded region (shown in the figure) OAXB is called a sector of the circle. Its boundary consists of arc AXB and two radii OA and OB. This sector has an angle θ, subtended at the centre of the circle by the arc AXB.

The region bounded by two radii of a circle and intercepted by them is called a sector of the circle.

When θ < 180°, arc AB is a minor arc and when θ > 180°, arc AB is a major arc.

Now for sector AOB with ∠AOB = θ°, then the length of minor arc AB \(2 \pi r \times \frac{\theta}{360^{\circ}} \text { i.e., } \frac{\pi r \theta}{180^{\circ}}\)

(Actually, 2πr is the distance covered in travelling the whole circumference in which an angle of 360 is formed at the centre. But for the length of the arc AXB, we do not rotate 360°, here we needed only 0 part of 360°)

∴ \(l=\frac{2 \pi r \theta}{360}\)

and the area of sector is \(\pi r^2 \times \frac{\theta}{360^{\circ}}\)

∴ \(A=\frac{\pi r^2 \theta}{360^{\circ}}\)

Relation between Length of Arc and Area of Sector

Area of sector \(A=\pi r^2 \times \frac{\theta}{360^{\circ}}\)

= \(\pi r \times \frac{\theta}{360^{\circ}} \times r=\frac{1}{2} \times 2 \pi r \times \frac{\theta}{360} \times r=\frac{1}{2} l r\)

Segment of a Circle: A segment of a circle is defined as the part of a circle bounded by a chord and the circumference. The segment containing the major arc is the major segment while the segment containing the minor arc is a minor segment.

Area of minor segment = Area of the sector- an area of ΔOAB.

= \(\left(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{1}{2} r^2 \sin \theta\right) \text { sq. unit }\)

Area of major segment = Area of circle- Area of minor segment

Perimeter of sector = \(\frac{\pi r \theta}{180^{\circ}}+2 r\)

Perimeter of minor segment = \(\frac{\pi r \theta}{180^{\circ}}\) + Length of chord AB

Area of semicircle = \(\frac{\pi r^2}{2}\)

Perimeter of semicircle = \(\pi r+2 r\)

Area of quarter circle = \(\frac{\pi r^2}{4}\)

Perimeter of quarter circle = \(\frac{\pi r}{2}+2 r\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 12  Area Related To Circles Solved Questions And Answers

Question 1. The perimeter of a semi-circular protractor is 32.4 cm. Calculate:

1. The radius of the protractor in cm,
2. the arc of the protractor in cm².

Solution:

Let the radius of the protractor be r cm.

Perimeter of semicircle protractor = (πr + 2r)cm

∴ r(π + 2) = 32.4

⇒ \(r\left(\frac{22}{7}+2\right)=32.4\)

⇒ \(r \times \frac{36}{7}=32.4 \quad ⇒ \quad r=\frac{32.4 \times 7}{36} ⇒ r=6.3 \mathrm{~cm}\)

Hence, the radius of the protractor = 6.3 cm

Area of semi-circular protractor = \(\frac{1}{2} \pi r^2\)

= \(\frac{1}{2} \times \frac{22}{7} \times 6.3 \times 6.3=62.37 \mathrm{~cm}^2\)

Hence, the area of the protractor = 62.37 cm²

Question 2. The minute hand of a clock is √2I cm long. Find the area described by the minute hand on the face of the clock between 6 a.m. and 6.05 a.m.

Solution:

In 60 minutes, the minute hand of a clock move through an angle of 360°.

∴ In 5 minutes hand will move through an angle = \(\frac{360^{\circ}}{60} \times 5=30^{\circ}\)

Now, r = √21 cm and θ = 30°

The area of sector described by the minute hand between 6 a.m. and 6.05 a.m.

= \(\frac{\pi r^2 \theta}{360^{\circ}}=\frac{22}{7} \times(\sqrt{21})^2 \times 30^{\circ} \times \frac{1}{360^{\circ}}=5.5 \mathrm{~cm}^2\)

Question 3. In the adjoining figure, calculate:

1. The length of minor arc ACB
2. Area of the shaded sector.

Solution:

Here, θ = 150°, r = 14 cm

1. Length of minor arc = \(\frac{\pi r \theta}{180^{\circ}}\)

= \(\frac{22}{7} \times 14 \times 150^{\circ} \times \frac{1}{360^{\circ}}\)

= 36.67 cm

2. Area of shaded sector = \(\frac{\pi r^2 \theta}{360^{\circ}}=\frac{22}{7} \times(14)^2 \times 150^{\circ} \times \frac{1}{360^{\circ}}=256.67 \mathrm{~cm}^2\)

Question 4. A chord AB of a circle of radius 10 cm makes a right angle at the centre of the circle. Find the area of the major and minor segments. (Use π = 3.14).

Solution:

Area of sector OAB = \(\frac{\pi r^2 \theta}{360^{\circ}}=3.14 \times 10^2 \times \frac{90^{\circ}}{360^{\circ}}=78.5 \mathrm{~cm}^2\)

Now, area of \(\triangle A O B=\frac{1}{2} \times(10)^2=50 \mathrm{~cm}^2\)

Area of the minor segment = (78.5 – 50) cm² = 28.5 cm²

Area of the major segment = Area of the circle – Area of the minor segment

= (3.14 x 10² – 28.5) cm² = (314 – 28.5) cm² = 285.5 cm²

Question 5. In the adjoining figure, the side of the square is 28 cm and the radius of each circle is half of the length of the side of the square where O and O’ are the centres of the circle. Find the area of the shaded region.

Answer:

Side of square = 28 cm

and the radius of each circle = 14 cm

Required area = area of square excluding the two sectors + area of two circles

= \((28)^2-2\left[\pi(14)^2 \times \frac{90^{\circ}}{360^{\circ}}\right]+2\left[\pi(14)^2\right]\)

= \((28)^2-2 \times \frac{22}{7} \times 14 \times 14 \times \frac{1}{4}+2 \times \frac{22}{7} \times 14 \times 14\)

= 784 – 308 + 1232 = 1708 cm²

Question 6. In the adjoining figure, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and the distance between AB and DC is 14 cm. If arcs of equal radii 7 cm taking A, B, C and D as centres, have been drawn, then find the area of the shaded region.

Solution:

Area of the shaded region

= area of trapezium – area of four sectors

= \(\begin{aligned}
\frac{1}{2} \times 14(32+18)-\left[\pi(7)^2 \times \frac{\angle A}{360^{\circ}}+\right. & \pi(7)^2 \times \frac{\angle B}{360^{\circ}} \left.+\pi(7)^2 \times \frac{\angle C}{360^{\circ}}+\pi(7)^2 \times \frac{\angle D}{360^{\circ}}\right]
\end{aligned}\)

= \(=7 \times 50-\frac{\pi \times(7)^2}{360^{\circ}}(\angle A+\angle B+\angle C+\angle D)\)

= \(=350-\frac{22}{7} \times \frac{7 \times 7}{360^{\circ}} \times 360^{\circ}\) (∵ sum of all the four angles of a quad. is 360°)

= 350 – 154 = 196 cm²

Hence, the area of the shaded region is 196 cm².

Question 7. In the adjoining figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of areas of the lawn and the flower beds.

Solution:

We know that the diagonals of a square bisect each other perpendicularly.

∴ ∠DOC = 90°

Also diagonal BD = side √2 = 56√2m

∴ \(O D=\frac{1}{2} \times B D=28 \sqrt{2} \mathrm{~m}\)

∴ Area of \(\triangle D O C=\frac{1}{2} \times 28 \sqrt{2} \times 28 \sqrt{2}=(28)^2 m^2\)

Area of sector ODCO = \(\pi(O D)^2 \times \frac{\theta}{360^{\circ}}\)

= \(\pi(28 \sqrt{2})^2 \times \frac{90^{\circ}}{360^{\circ}}=\frac{11 \times 28 \times 28}{7} \mathrm{~m}^2\)

= 1232 m²

∴ Area of 1 flower bed = Area of sector- Area of A

= 1232 -(28)² = 448 m²

∴ Area of 2 flowers beds = 2 x 448 = 896 m²

Area of square lawn = (56)² = 3136 m²

∴ Required area = 896 + 3136 = 4032 m²

Question 8. In the adjoining, ABCD is a square of side 10 cm and two A semicircles with side of the square as diameter. A quarter circle is also seen with a side of the square as a radius. Find the area of the square region excluding the shaded part (Deepak).(Takeπ = 3.14)

Solution:

First of all, we will find the area of 1 and 2 i.e., Batti of Deepak, draw OM ⊥ AB and ON ⊥ BC.

ar(1) = ar(sector NBKO) – ar(ABNO)

= \(\pi(5)^2 \times \frac{90^{\circ}}{360^{\circ}}-\frac{1}{2} \times 5 \times 5\)

ar (2) = ar (1)

⇒ \(\operatorname{ar}(\mathrm{1}+\mathrm{2})=2\left(\frac{25 \pi}{4}-\frac{25}{2}\right)=\frac{25}{2}(\pi-2)\)

∴ \(\frac{25}{2}(3.14-2)=14.25 \mathrm{~cm}^2\)

Now, area of deepak = ar(quadrant APCB) – [ar(semicircle 4 + 1 + 2) + ar(semicircle 5 + 1 + 2) – 2 x ar (1 + 2)]

= \(\pi(10)^2 \times \frac{90}{360}-\left[2 \times \frac{\pi(5)^2}{2}-2 \times 14.25\right]\)

25π – (25π – 28.5) = 28.5 cm²

Area of unshaded portion = Area of the square- Area of deepak

= (10)²– 28.5 = 100 – 28.5 = 71.5 cm²

The area of the square region excluding the shaded part = 71.5 cm²

Question 9. If the hypotenuse of an isosceles right triangle is 7√2 cm, find the area of the circle inscribed in it.

Solution:

Let AB = BC = x cm

∴ In right ΔABC, by Pythagoras theorem

x² + y² = (7√2)²

⇒ 2x² = 98 = x² = 49 ⇒ x = 7 cm

∴ ar(ABC) = ar(AOB)+ ar(BOC) + ar(COA)

⇒ \(\frac{1}{2} \times x \times x=\frac{1}{2} \times x \times r+\frac{1}{2} \times x \times r+\frac{1}{2} \times 7 \sqrt{2} \times r\)

⇒\(r=\frac{7}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{7(2-\sqrt{2})}{2}\)

⇒ \(7 \times 7=7 \times r+7 \times r+7 \sqrt{2} \times r \Rightarrow r=\frac{7}{2+\sqrt{2}}\)

Area of circle = πr²

= \(\frac{22}{7} \times \frac{49(2-\sqrt{2})^2}{4}=\frac{77}{2} \times(4+2-4 \sqrt{2})\)

= 77(3 – 2√2) cm²

The area of the circle inscribed in it = 77(3 – 2√2) cm²

Question 10. In the adjoining figure, two concentric circles with centre O have radii of 21 cm and 42 cm. If ∠AOB = 60°, find the area of the shaded region. (\(\text { Use } \pi=\frac{22}{7}\))

Solution:

Required area = area of a larger circle with a radius of 42 cm – an area of a smaller circle with a radius of 21 cm – (area of unshaded portion CDBA)

= π(42)² – π(21 )² – [(area of the larger sector with radius 42 cm) – (area of the smaller sector with radius 21 cm)]

= \(\pi(42)^2-\pi(21)^2-\left[\pi(42)^2 \times \frac{60}{360}-\pi(21)^2 \times \frac{60}{360}\right]\)

= \(\pi(42)^2-\pi(21)^2-\frac{1}{6} \pi(42)^2+\frac{1}{6} \pi(21)^2=\frac{5}{6} \pi(42)^2-\frac{5}{6} \pi(21)^2\)

= \(\frac{5}{6} \pi \times(21)^2[4-1]=\frac{5}{6} \times \frac{22}{7} \times 21 \times 21 \times 3=3465 \mathrm{~cm}^2\)

The area of the shaded region =3465 cm²

Question 11. In the adjoining figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.

Solution:

Since O is the centre of the circle.

∴ BC is the diameter of the circle.

∴ ∠CAB = 90° (angle in a semicircle is a right angle)

Now in right ΔACB, by Pythagoras Theorem,

BC² = AC² + AB² = (24)² + (7)² = 576 + 49 = 625

∴ BC = = 25 m

∴ Diameter of circle = 25 m

∴ Radius of circle = \(\frac{25}{2}\) = 12.5 m

∴ OC = OD = 12.5 m (each radii)

∴ Area of shaded region = area of circle – ar (ΔABC) – ar (sector COD)

= \(\pi(12.5)^2-\frac{1}{2} \times A C \times A B-\pi(O C)^2 \times \frac{90^{\circ}}{360^{\circ}}\)

= \(\frac{22}{7} \times 12.5 \times 12.5-\frac{1}{2} \times 24 \times 7-\frac{22}{7} \times 12.5 \times 12.5 \times \frac{1}{4}\)

= 491.07 – 84 – 122.768 = 284.302 = 284.30 m² (approx.)

The area of the shaded region = 284.30 m² (approx.)

NCERT Exemplar Solutions for Class 10 Maths Chapter 12  Area Related To Circles Exercise 12.1

Question 1. The radii of the two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.

Solution:

Here, r₁ = 19 cm and r₂ = 9cm

Let the radius of the new circle = R cm

Given that,

Circumference of new circle = sum of the circumference of given two circles

2πR = 2πr₁ + 2πr₂

R = r₁ + r₂ = 19 + 9 = 28 cm.

The radius of the circle = 28 cm.

Question 2. The radii of the two circles are 8 cm and 6 cm respectively. Find the radius of the circle having an area equal to the sum of the areas of the two circles.

Solution:

Here, r₁ = 8 cm and r₂ = 6 cm

Let the radius of the new circle = R cm

Given that,

Area of new circle = sum of areas of given circles

⇒ πR² = πr² + πr²

⇒ R² = r²₁ + r²₂ = 82 + 62

= 64 + 36 = 100

⇒ R = 10 cm

The radius of the circle = 10 cm

Question 3. The depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing the Gold score is 21cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Solution:

The diameter of the Gold circle = 21 cm

∴ Radius of Gold circle = 10.5 cm

Now, the area of the Gold Circle

= \(\frac{22}{7} \times 10.5 \times 10.5 \mathrm{~cm}^2\)

= 346.5 cm²

Given, the width of each band = 10.5 cm

∴ Radius of Red circle = (10.5 + 10.5) = 21cm

Now, area of Red ring = [(external radius)² – (internal radius)²]

= \(\frac{22}{7}\left[(21)^2-(10.5)^2\right]\)

= \(\frac{22}{7} \times(441-110.25)\)

= \(\frac{22}{7} \times 330.75\)

= 1039.5 cm²

Again, radius of Blue circle = (21 + 10.5)cm = 31.5 cm

∴ Area of Blue ring

= \(\frac{22}{7}\left(31.5^2-21^2\right)\)

= \(\frac{22}{7} \times(992.25-441)\)

= \(\frac{22}{7} \times 551.25\)

= 1732.5 cm²

Again, the radius of the Black Circle

= (31.5 + 10.5)cm

= 42cm

∴ Area of Black circle

= \(\frac{22}{7} \times\left(42^2-31.5^2\right)\)

= \(\frac{22}{7} \times(1764-992.25)\)

= \(\frac{22}{7} \times 771.75=2425.5 \mathrm{~cm}^2\)

Again, radius of white circle = (42 + 10.5) cm

∴ Area of White circle

= \(=\frac{22}{7} \times\left(52.5^2-42^2\right)\)

= \(\frac{22}{7} \times(2756.25-1764)\)

= \(\frac{22}{7} \times 992.25=3118.5 \mathrm{~cm}^2\)

Question 4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Solution:

The diameter of the wheel of a car, 2r= 80 cm

⇒ r = 40 cm

∴ Distance covered in one revolution

= circumference

= \(2 \pi r=2 \times \frac{22}{7} \times 40=\frac{1760}{7} \mathrm{~cm}\)

Now, speed of car = 66 km/hr

= \(\frac{66 \times 1000 \times 100}{60} \mathrm{~cm} / \mathrm{min}\)

∴ Distance covered in 10 minutes

= \(\frac{66 \times 1000 \times 100}{60} \times 10 \mathrm{~cm}\)

= 11 x 1000 x 100 cm

Now, the number of revolutions made by the wheel

= \(\frac{\text { Total distance covered }}{\text { Distance covered by wheel in 1 revolution }}\)

= \(\frac{11 \times 1000 \times 100}{1760 / 7}\)

= \(=\frac{11 \times 1000 \times 100 \times 7}{1760}=4375 \text { Ans }\)

Question 5. Tick the correct answer in the following and justify your choice: if the perimeter and the area of a circle are numerically equal, then the radius of the circle is:

1. 2 units
2. units
3. 4 units
4. 7 units

Solution:

1. 2 units

Let r be the radius of the circle.

Given that, in numerical form, area of circle = perimeter of circle

⇒ πr² = 2πr ⇒ r = 2 units

NCERT Exemplar Solutions for Class 10 Maths Chapter 12  Area Related To Circles Exercise 12.2

Question 1. Find the area of a sector of a circle with a radius of 6 cm if the angle of the sector is 60°.

Solution:

Here, the radius of the circle, r = 6 cm

The angle of the sector, θ= 60°

∴ Area of sector = \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 6 \times 6 \mathrm{~cm}^2\)

= \(\frac{132}{7} \mathrm{~cm}^2 \text { or } 18.86 \mathrm{~cm}^2\)

The area of a sector of a circle is 18.86 cm²

Question 2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution:

Circumference of the circle, 2πr = 22

⇒ \(2 \times \frac{22}{7} \times r=22 \quad \Rightarrow \quad r=\frac{7}{2} \mathrm{~cm}\)

Now, the area of the quadrant of a circle

⇒ \(\frac{1}{4} \pi r^2=\frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)

∴ \(\frac{77}{8} \mathrm{~cm}^2\)

The area of a quadrant of a circle = \(\frac{77}{8} \mathrm{~cm}^2\)

Question 3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution:

Length of minute hand of clock = 14 cm

∴ The radius of the circle, r = 14 cm

∵ The angle subtended by minute hand in 60 min = 360°

∴ Angle subtended by minute hand in 1 minute = \(\frac{360^{\circ}}{60^{\circ}}=6^{\circ}\)

∴ The angle subtended by minute hand in 5 minutes = 30°

∴ From the formula,

area of sector of circle = \(\frac{\theta \pi r^2}{360^{\circ}}\)

= \(30^{\circ} \times \frac{22 \times(14)^2}{7 \times 360^{\circ}}\)

= \(\frac{22 \times 14 \times 2}{12}\)

= \(\frac{616}{12}=\frac{154}{3} \mathrm{~cm}^2\)

The area swept by the minute hand in 5 minutes =\(\frac{154}{3} \mathrm{~cm}^2\)

Question 4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:

1. Minor segment
2. Major sector

Solution:

Given, the radius of the circle, AO = 10 cm.

The perpendicular is drawn from the centre of the circle to the chord of the circle which bisects this chord.

AD = DC

And ∠AOD = ∠COD

= 45°

∴ ∠AOC = ∠AOD + ∠COD

= 45° + 45° = 90°

In the right ΔAOD,

⇒ \(\sin 45^{\circ}=\frac{A D}{A O} \quad \Rightarrow \quad \frac{1}{\sqrt{2}}=\frac{A D}{10}\)

AD = 5√2 cm

and \(\cos 45^{\circ}=\frac{O D}{A O} \Rightarrow \frac{1}{\sqrt{2}}=\frac{O D}{10}\)

⇒ OD = 5√2 cm

Now, AC = 2AD

2 x 5√2 = 10√2 cm

Now, the area of ΔAOC

= \(\frac{1}{2} A C \times O D\)

= \(\frac{1}{2} \times 10 \sqrt{2} \times 5 \sqrt{2}=50 \mathrm{~cm}^2\)

Now the area of the sector

= \(\frac{\theta \pi r^2}{360^{\circ}}=\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(10)^2\)

= \(\frac{314}{4}=78.5 \mathrm{~cm}^2\)

1. Area of minor segment AEC

= area of sector OAEC – area of AOC

= 78.5 – 50 = 28.5 cm²

2. Area of major sector OAFGCO

= area of circle- area of sector OAEC

= πr²– 78.5 = 3.14 x (10)² – 78.5

314 – 78.5 = 235.5cm²

Question 5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

1. The length of the arc
2. The area of the sector formed by the arc
3. The area of the segment is formed by the corresponding chord.

Solution:

Here, the radius of circle r = 21 cm

The angle subtended by are at the centre, θ = 60°

1. Length of arc

= \(l=\frac{\theta}{360^{\circ}} \times 2 \pi r\)

= \(\frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21=22 \mathrm{~cm}\)

2. Area of sector formed by the arc

= \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 21 \times 21\)

= 231 cm²

3. Area of segment formed by the corresponding chord

= \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{1}{2} r^2 \sin \theta\)

= \(231-\frac{1}{2} \times 21 \times 21 \times \sin 60^{\circ}\)

= \(\left(231-\frac{441 \sqrt{3}}{4}\right) \mathrm{cm}^2\)

Question 6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.

(Use π = 3.14 and √3 = 1.73)

Solution:

Given

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre.

Here, the radius of the circle, r = 15 cm

The angle subtended by a chord at the centre, θ = 60°

∴ Area of the corresponding minor segment

= \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{1}{2} r^2 \sin \theta\)

= \(r^2\left(\frac{\pi \theta}{360^{\circ}}-\frac{1}{2} \sin \theta\right)\)

= \(15 \times 15\left(\frac{3.14 \times 60^{\circ}}{360^{\circ}}-\frac{1}{2} \sin 60^{\circ}\right)\)

= \(225\left(\frac{3.14}{6}-\frac{1}{2} \times \frac{\sqrt{3}}{2}\right)\)

= \(225\left(\frac{3.14}{6}-\frac{1.73}{4}\right)=20.4375 \mathrm{~cm}^2\)

Now, the area of the circle = πr²

= 3.14 x 15 x 15 = 706.5 cm²

∴ Area of major segment = area of circle – an area of the minor segment

= (706.5 – 20.4375) cm²

= 686.0625 cm².

The areas of the corresponding minor and major segments of the circle = 686.0625 cm².

Question 7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.
(Use π = 3.14 and √3 = 1.73)

Solution:

Given

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre.

Here, the radius of the circle, r = 12 cm

The angle subtended by the chord at the centre, θ = 120°

Area of the corresponding minor segment

= \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{1}{2} r^2 \sin \theta\)

= \(r^2\left(\frac{\pi \theta}{360^{\circ}}-\frac{1}{2} \sin \theta\right)\)

= \(12 \times 12 \times\left(\frac{3.14 \times 120^{\circ}}{360^{\circ}}-\frac{1}{2} \times \sin 120^{\circ}\right)\)

= \(144\left(\frac{3.14}{3}-\frac{1}{2} \times \frac{\sqrt{3}}{2}\right)\)

= \(144\left(\frac{3.14}{3}-\frac{1.73}{4}\right)\)

= 88. 44 cm²

The area of the corresponding segment of the circle = 88. 44 cm²

Question 8. A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m long rope. Find:

1. The area of that part of the field in which the horse can graze.
2. The increase in the grazing area, if the rope were 10 m long instead of 5m. (Use π = 3.14)

Solution:

Given

A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m long rope.

Side of square field = 15 m

The radius of the circle, r = 5m

The angle formed by chord, θ = 90°

1. Area of that part of the field where the horse can graze the grass

= \(\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times 5^2\)

= 19.625 m²

Area of that part of the field where the horse can graze the grass = 19.625 m²

2. If the length of the rope is 10 m then the increase in the area of grazing.

= \(\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times\left(10^2-5^2\right)\)

= \(\frac{1}{4} \times 3.14 \times 75\)

= 58.875 m².

If the length of the rope is 10 m then the increase in the area of grazing = 58.875 m².

Question 9. A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown. Find:

1. The total length of the silver wire required.
2. The area of each sector of the brooch.

Solution:

Given

A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown.

1. Diameter of circular brooch,

2r = 35 mm

= \(r=\frac{35}{2} \mathrm{~mm}\)

Length of required silver wire = 2πr

= \(2 \times \frac{22}{7} \times \frac{35}{2} \mathrm{~mm}=110 \mathrm{~mm}\)

Now, the length of 5 diameters = 5 x 2r

= \(5 \times 2 \times \frac{35}{2}=175 \mathrm{~mm}\)

Length of total wire = (110 + 175) mm

= 285 mm

The total length of the silver wire required = 285 mm

2. For each sector,

angle, \(\theta=\frac{360^{\circ}}{10^{\circ}}=36^{\circ}\)

∴ Area = \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(\frac{36^{\circ}}{360^{\circ}} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2}\)

= \(\frac{385}{4} \mathrm{~cm}^2\)

The area of each sector of the brooch = \(\frac{385}{4} \mathrm{~cm}^2\)

Question 10. An umbrella has 8 ribs which are equally spaced. Assuming an umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:

Given

An umbrella has 8 ribs which are equally spaced. Assuming an umbrella to be a flat circle of radius 45 cm

The radius of the umbrella,

r = length of ribs of umbrella = 45 cm

Here, number of sectors = 8

∴ Angle of sector = \(\theta=\frac{360^{\circ}}{8}=45^{\circ}\)

Now, the area between two ribs = area of the sector

= \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(\frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 45 \times 45 \mathrm{~cm}^2\)

= \(\frac{22275}{28} \mathrm{~cm}^2\)

The area between the two consecutive ribs of the umbrella = \(\frac{22275}{28} \mathrm{~cm}^2\)

Question 11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Solution:

Given, the length of the wiper blade

r = 25 cm = r(say)

The angle formed by this blade, θ =115°

∴ Area cleaned by a blade = area of sector formed by blade

= \(\frac{\theta \pi r^2}{360^{\circ}}=115^{\circ} \times \frac{22}{7 \times 360^{\circ}} \times(25)^2\)

= \(\frac{23 \times 22}{7 \times 72} \times 625=\frac{23 \times 11 \times 625}{7 \times 36}\)

= \(\frac{158125}{252} \mathrm{~cm}^2\)

∴ Total area cleaned by two blades

= 2 x area cleaned by a blade

= \(\frac{2 \times 158125}{252}=\frac{158125}{126} \mathrm{~cm}^2\)

The total area cleaned at each sweep of the blades =\(\frac{158125}{126} \mathrm{~cm}^2\)

Question 12. To warn ships of underwater rocks, a lighthouse spreads a red-coloured light over a sector of angle 80° to a distance of 16.51cm. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Solution:

Given, the angle of the sector, θ = 80°

And distance or radius, r = 16.5 km

Area of sector

= \(\frac{\theta \pi r^2}{360^{\circ}}=\frac{80^{\circ} \times 3.14 \times(16.5)^2}{360^{\circ}}\)

= \(\frac{2 \times 3.14 \times 272.25}{9}=\frac{1709.73}{9}=189.97 \mathrm{~km}^2\) which is the area of the sea over which the ships are warned.

Question 13. A round table cover has six equal designs as shown. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm². (Use √3 = 1.7)

Solution:

Given

A round table cover has six equal designs as shown. If the radius of the cover is 28 cm

The radius of the table cover, r = 28 cm

Number of designs formed on table cover = 6

∴ The angle formed by each chord at the centre

= \(\frac{360^{\circ}}{6}=60^{\circ}\)

Now, the area of segment

= \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{1}{2} r^2 \sin \theta\)

= \(r^2\left(\frac{\pi}{360^{\circ}}-\frac{1}{2} \sin \theta\right)\)

⇒ Area of 6 segments

= \(6 r^2\left(\frac{\pi}{360^{\circ}}-\frac{1}{2} \sin \theta\right)\)

= \(6 \times 28 \times 28\left(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7}-\frac{1}{2}-\sin 60^{\circ}\right)\)

= \(6 \times 28 \times 28\left(\frac{11}{21}-\frac{1}{2} \times \frac{\sqrt{3}}{2}\right)\)

= \(=6 \times 28 \times 28\left(\frac{11}{21}-\frac{1.73}{4}\right)=464.8 \mathrm{~cm}^2\)

Now, the total cost of making the design

= ₹ 0.35 x 464.8 = ₹ 162.68

The total cost of making the design = ₹ 162.68

Question 14. The area of a sector of angle p (in degrees) of a circle with radius R is

1. \(\frac{p}{180} \times 2 \pi R\)
2. \(\frac{p}{180} \times \pi R^2\)
3. \(\frac{p}{360} \times 2 \pi R\)
4. \(\frac{p}{720} \times 2 \pi R^2\)

Solution:

4. \(\frac{p}{720} \times 2 \pi R^2\)

Area of sector = \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(\frac{p}{360^{\circ}} \times \pi R^2=\frac{p}{720} \times 2 \pi R^2\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 12  Area Related To Circles Exercise 12.3

Question 1. Find the area of the shaded region in the figure, if PQ= 24 cm, PR = 7 cm and O is the centre of the circle.

Solution:

Here, PQ = 24 cm and PR = 7 cm

∵ The angle in a semicircle is a right angle.

∴ ∠QPR = 90°

In ΔPQR,

QR² = PQ² + PR² = 24² + 7² = 625

QR = 25 cm

If r is the radius of the circle, then

2r = 25 cm

= \(r=\frac{25}{2} \mathrm{~cm}\)

Area of shaded region = area of semicircle – area of ΔPQR

= \(\frac{1}{2} \pi r^2-\frac{1}{2} \times P Q \times P R\)

= \(\frac{1}{2} \times \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2}-\frac{1}{2} \times 24 \times 7\)

= \(\frac{6875}{28}-84=\frac{4523}{28} \mathrm{~cm}^2\)

The area of the shaded region =\(\frac{4523}{28} \mathrm{~cm}^2\)

Question 2. Find the area of the shaded region, if the radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Solution:

Given

The radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Here, OA = 14 cm and OB = 7 cm

The angle of the sector, θ = 40°

∴ Area of the shaded portion

= \(\frac{\theta}{360^{\circ}} \times \pi \times\left(O B-O A^2\right)\)

= \(\frac{40^{\circ}}{360^{\circ}} \times \frac{22}{7} \times\left(14^2-7^2\right)\)

= \(\frac{1}{9} \times \frac{22}{7} \times 147=51.33 \mathrm{~cm}^2\)

The area of the shaded region =51.33 cm².

Question 3. Find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Solution:

Given

ABCD is a square of side 14 cm and APD and BPC are semicircles.

Side of square = 14

Area of square = 14 x 14 cm² = 196 cm²

Diameter of each semicircle 2r = side of square = 14 cm

⇒ r = 7 cm

Area of one semicircle = \(\frac{1}{2} \pi r^2\)

Area of two semicircles = \(2 \times \frac{1}{2} \pi r^2=\pi r^2\)

= \(\frac{22}{7} \times 7 \times 7=154 \mathrm{~cm}^2\)

Area of shaded portion = area of the square – the sum of areas of two semicircles

= (196 – 154) cm² = 42 cm²

The area of the shaded region = 42 cm²

Question 4. Find the area of the shaded region given, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as the centre.

Solution:

Given

A circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as the centre.

Each side of the equilateral triangle = 12 cm

and each angle = 60°

Now, the area of an equilateral triangle

= \(\frac{\sqrt{3}}{4} \times(\text { side })^2=\frac{\sqrt{3}}{4} \times 12 \times 12 \mathrm{~cm}^2\)

= 36√3 cm²

The radius of the circle, r = 6 cm

Angle of major sector, 0 = 360° – 60° = 300°

∴ Area of major sector

= \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(\frac{300^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 6 \times 6=\frac{660}{7} \mathrm{~cm}^2\)

Now, the area of the shaded region = area of the equilateral triangle + area of a major sector.

= \(\left(36 \sqrt{3}+\frac{660}{7}\right) \mathrm{cm}^2\)

The area of the shaded region = \(\left(36 \sqrt{3}+\frac{660}{7}\right) \mathrm{cm}^2\)

Question 5. From each A corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown. Find the area of the remaining portion of the square.

Solution:

Given

From each A corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown.

Side of square = 4 cm

∴ Area of square = (side)² = 4² = 16 cm²

The radius of a quadrant of a circle, r = 1 cm

∴ Area of four quadrants

= \(4 \times \frac{1}{4} \pi r^2\)

= \(\frac{22}{7} \times 1 \times 1=\frac{22}{7} \mathrm{~cm}^2\)

For the circle inside the square diameter, 2R = 2 cm ⇒ R = 1 cm

Area of this circle = \(\pi R^2=\frac{22}{7} \times 1 \times 1=\frac{22}{7} \mathrm{~cm}^2\)

Now, area of shaded portion = \(16-\left(\frac{22}{7}+\frac{22}{7}\right)=\frac{68}{7} \mathrm{~cm}^2\).

Question 6. In a circular table covered of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown. Find the area of the design.

Solution:

Given

In a circular table covered of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown.

ΔABC is an equilateral triangle.

∴ ∠A = ∠B = ∠C = 60°

AD is perpendicular to BC.

∴ OA, OB, and OC are the radii of the circle.

Given that,

OA = OB = OC = 32 cm

∠BOC = 2

∠BAC = 2 x 60° = 120°

∴ \(\angle B O D=\frac{1}{2} \times 120^{\circ}=60^{\circ}\)

In ΔBOD

⇒ \(\sin 60^{\circ}=\frac{B D}{O B} ⇒ \frac{\sqrt{3}}{2}=\frac{B D}{32}\)

BD = 16√3 cm

BC = 2 BD = 32√3 cm

Area of ΔABC = \(\frac{\sqrt{3}}{4} \times B C^2\)

= \(\frac{\sqrt{3}}{4} \times(32 \sqrt{3})^2=768 \sqrt{3} \mathrm{~cm}^2\)

Area of circle = π(OB)²

= \(\frac{22}{7} \times 32 \times 32=\frac{22528}{7} \mathrm{~cm}^2\)

Now, the area of the shaded portion

= area of a circle – an area of ΔABC

= \(\left(\frac{22528}{7}-768 \sqrt{3}\right) \mathrm{cm}^2\).

The area of the shaded portion = \(\left(\frac{22528}{7}-768 \sqrt{3}\right) \mathrm{cm}^2\).

Question 7. In the ABCD is a square of side 14 cm. With centres A. B. C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Solution:

Side of square ABCD = 14 cm

∴ Area of square = (14)² =196 cm²

Radius of first quadrant \(r=\frac{14}{2}=7 \mathrm{~cm}\)

∴ The sum of area’s four quadrants

= \(4 \times \frac{1}{4} \pi r^2=\pi r^2\)

= \(\frac{22}{7} \times 7 \times 7=154 \mathrm{~cm}^2\)

Now, the area of the shaded portion = area of the square – the sum of areas of four quadrants

= 196 = 154 = 42 cm²

The area of the shaded region = 42 cm²

Question 8. depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:

1. The distance around the track along its inner edge.
2. The area of the track.

Solution:

Hence OB O’C = \(\frac{60}{2}=30 \mathrm{~m}\)

AB = CD = 10m

∴ OA = O’D = (30 + 10) m = 40 m

1. Distance covered in one round along the inner edges of the path.

= BC + EH + 2 x circumference of semicircle

= \(106+106+2 \times \frac{1}{2} \times 2 \pi \times 30\)

= \(212+2 \times \frac{22}{7} \times 30=\frac{2804}{7} \mathrm{~m}\)

2. Inner radius r = OB = 30 m

Outer radius R = OA = 40 m

Area of path = 2 x area of ABCD + 2 x area of semicircular rings

= \(=2 \times 106 \times 10+\frac{1}{2} \times \pi\left(R^2-r^2\right)\)

= \(2120+\frac{22}{7} \times\left(40^2-30^2\right)\)

= 2120 + 2200 = 4320 m²

Area of path is 4320 m²

Question 9. In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution:

Given

In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm

The diameter of the smaller circle

OD = OA = 7 cm

∴ Radius r = \(\frac{7}{2}=3.5 \mathrm{~cm}\)

∴ Area of smaller circle = nr2

= \(\frac{22}{7} \times 3.5 \times 3.5 \mathrm{~cm}^2=38.5 \mathrm{~cm}^2\)

Area of semicircle OAQCPBO

= \(\frac{1}{2} \pi(O A)^2\)

= \(\frac{1}{2} \times \frac{22}{7} \times 7 \times 7=77 \mathrm{~cm}^2\)

Area of ΔABC = = \(\frac{1}{2} \times A B \times O C\)

= \(\frac{1}{2} \times 14 \times 7=49 \mathrm{~cm}^2\)

Now the area of the shaded portion = area of the smaller circle + area of OAQCPBO – an area of ABC

= 38.5 + 77 –  49 = 66. cm²

The area of the shaded region = 66. cm²

Question 10. The area of an equilateral triangle 11 ABC is 17320.5 cm². With each vertex of the triangle as the centre, a circle is drawn with a radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use π= 3.24 and √3 = 1.73205).

Solution:

Given

The area of an equilateral triangle 11 ABC is 17320.5 cm². With each vertex of the triangle as the centre, a circle is drawn with a radius equal to half the length of the side of the triangle.

Let the radius of each circle = r

Side of equilateral triangle = r + r = 2r

Now, area of equilateral triangle = \(\)

Given that, √3r² = 17320.5 ⇒ 1.73205r² = 17320.5

⇒ r² = 10000 ⇒ r = 100 cm

Each angle of an equilateral triangle, θ = 60° Now, the area of sectors of three circles

= \(3 \times \frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(3 \times \frac{60^8}{360^{\circ}} \times 3.14 \times(100)^2\)

= 15700 cm²

Area of shaded region = area of equilateral ABD = area of three sectors

= (17320.5 – 15700) cm²

= 1620.5 cm²

The area of the shaded region = 1620.5 cm².

Question 11. On a square handkerchief, nine circular designs each of a radius of 7 cm are made. Find the area of the remaining portion of the handkerchief.

Solution:

Given, the radius of each circle, r = 7 cm

Diameter of circle, d = 14 cm = 42 cm (∵ diameter = 2 x radius)

Three horizontal circles touch each other.

Length of square = 3 x 14 cm = 42 cm

Now area of a circle = πr² = (7)²

= \(\frac{22}{7} \times(7)^2=154 \mathrm{~cm}^2\)

∴ Area of 9 circles = 9 x 154 = 1386 cm²

Now, area of square ABCD = (side)²

= (42)² = 1764 cm²

∴ Area of the remaining part of the handkerchief

= 1764 – 1386 = 378 cm²

The area of the remaining portion of the handkerchief = 378 cm²

Question 12. The OACB is a quadrant of a circle with centre O and a radius of 3.5 cm. If OD = 2 cm. find the area of the

1. Quadrant OACB
2. Shaded region.

Solution:

Given

The OACB is a quadrant of a circle with centre O and a radius of 3.5 cm. If OD = 2 cm

Radius of quadrant AOB. r= 3.5 cm

1. Area of quadrant OACB

= \(\frac{1}{4} \pi^2\)

= \(\frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5=\frac{77}{8} \mathrm{~cm}^2\)

2. Area of ΔOBD

= \(\frac{1}{2} \times O B \times O D\)

= \(\frac{1}{2} \times 3.5 \times 2=3.5 \mathrm{~cm}^2=\frac{7}{2} \mathrm{~cm}^2\)

Area of the shaded portion

= area of quadrant OACB – area of OBD

= \(\left(\frac{77}{8}-\frac{7}{2}\right) \mathrm{cm}^2=\frac{49}{8} \mathrm{~cm}^2\)

Area of the shaded portion =\(\frac{49}{8} \mathrm{~cm}^2\)

Question 13. In the square O.ABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Solution:

Given

In the square O.ABC is inscribed in a quadrant OPBQ. If OA = 20 cm

Each side of square OABC = 20 cm

∴ Area of OABC = (20)² = 400 cm²

Area of OABC = (20)² = 400 cm²

In ΔOAB,

OB² = OA² ~ AB² = 20² + 20² = 800

⇒ OB = 20√2cm

which is the radius of quadrant OPBQ.

Area of quadrant OPBQ

= \(\frac{1}{4} \pi(O B)^2=\frac{1}{4} \times 3.14 \times(20 \sqrt{2})^2\)

= 628 cm²

Now, the area of the shaded portion

= area of quadrant OPBD – area of square OABC

= (628 – 400) cm2 = 228 cm²

The area of the shaded region = 228 cm²

Question 14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O . If ∠AOB = 30°, find the area of the shaded region.

Solution:

Given

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O . If ∠AOB = 30°

Here Q = 30°

Area of the shaded portion

= Area of sector OAB – area of sector OCD

= \(\frac{\theta}{360^{\circ}} \times \pi(O B)^2-\frac{\theta}{360^{\circ}} \times \pi(O C)^2\)

= \(\frac{\theta}{360^{\circ}} \times \pi\left[(O B)^2-(O C)^2\right]\)

= \(\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times\left(21^2-7^2\right)\)

= \(\frac{1}{12} \times \frac{22}{7} \times 392 \mathrm{~cm}^2=\frac{308}{3} \mathrm{~cm}^2\)

Area of the shaded portion \(\frac{308}{3} \mathrm{~cm}^2\)

Question 15. The ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Solution:

Given

The ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter.

For the quadrant ABC,

Radius r = 1 4 cm

∴ Area of quadrant ABC

= \(\frac{1}{4} \pi r^2\)

= \(\frac{1}{4} \times \frac{22}{7} \times 14 \times 14=154 \mathrm{~cm}^2\)

Area of \(\triangle A B C=\frac{1}{2} \times A B \times A C\)

= \(\frac{1}{2} \times 14 \times 14=98 \mathrm{~cm}^2\)

In right ABC,

BC² = AC² + AB² = 14² + 14² = 392

BC = 14√2 cm

∴ For the semicircle formed on BC

Radius \(R=\frac{B C}{2}=\frac{14 \sqrt{2}}{2}=7 \sqrt{2} \mathrm{~cm}\)

and area of semicircle

= \(\frac{1}{2} \pi R^2\)

= \(\frac{1}{2} \times \frac{22}{7} \times(7 \sqrt{2})^2=154 \mathrm{~cm}^2\)

Now, the area of the shaded portion

= area of ΔABC + area of semicircle – area of quadrant ABC

= (98 + 154 – 154) cm² = 98 cm²

The area of the shaded region = 98 cm²

Question 16. Calculate the area of the designed region in the figure common between the two quadrants of circles of radius 8 cm each.

Solution:

The radius of each of quadrants ABMD and BNDC, r = 8 cm

∴ Area of two quadrants

= \(2 \times \frac{1}{4} \pi r^2\)

= \(2 \times \frac{1}{4} \times \frac{22}{7} \times 8 \times 8\)

= \(\frac{704}{7} \mathrm{~cm}^2\)

and area of square ABCD = AB² = 8² = 64 cm²

Now, the area of the shaded portion

= area of two quadrants – area of square ABCD

= \(\left(\frac{704}{7}-64\right) \mathrm{cm}^2\)

= \(\frac{256}{7} \mathrm{~cm}^2\)

The area of the designed region = \(\frac{256}{7} \mathrm{~cm}^2\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 12  Area Related To Circles Multiple Choice Questions And Answers

Question 1. If the sum of the circumference of two circles of radii R₁ and R₂is equal to the circumference of a circle of radius R, then:

1. R₁ + R₂ = R
2. R₁ + R₂ > R
3. R₁ + R₂ < R
4. None of these

Answer: 1. R₁ + R₂ = R

Question 2. If the sum of the areas of two circles of radii and R₁ and R₂ is equal to the area of a circle of radius R, then:

1. R₁ + R₂ = R
2. R1 + R2 < R
3. R_1² + R_2² = R²
4. R_1² + R_2² < R²

Answer: 3. R_1² + R_2² = R²

Question 3. If the area of a circle is 154 cm², then its circumference is:

1. 11 cm
2. 22 cm
3. 44 cm
4. 66 cm

Answer: 3. 44 cm

Question 4. The area of the largest triangle inscribed in a semicircle of radius r is:

1. r²
2. \(\frac{1}{2} r^2\)
3. 2r²
4. r²√2

Answer: 1. r²

Question 5. The area of the largest square inscribed in a circle of radius 8 cm is:

1. 256 cm²
2. 64 cm²
3. 128 cm²
4. 32 cm²

Answer: 3. 128 cm²

Question 6. The area largest circle inscribed in a square of side 4 cm is:

1. 16π cm²
2. 8π cm²
3. 6π cm²
4. 4π cm²

Answer: 2. 8ir cm²

Question 7. If the circumference of a circle and the perimeter of a square are equal, then the ratio of their areas is:

1. 22:7
2. 14:11
3. 7:22
4. 11:14

Answer: 2. 14:11

Question 8. In the adjoining figure, the perimeter of sector OAB is:

1. \(\frac{64}{3} \mathrm{~cm}\)
2. 26 cm
3. \(\frac{64}{5} \mathrm{~cm}\)
4. 19 cm

Answer: 1. \(\frac{64}{3} \mathrm{~cm}\)

Question 9. The length of the minute hand of a clock is 14 cm. The area swept by hand in one minute will be:

1. 10.26 cm²
2. 10.50 cm²
3. 10.75 cm²
4. 11.0 cm²

Answer: 1. 10.26 cm²

NCERT Exemplar For Class 10 Maths Chapter 11 Constructions Division Of A Line Segment

To divide a line segment (internally) in a given ratio m: n

Working Rule: (Internal division)

Draw a line segment AB of a given length.

Draw a ray AX making an acute angle XAB with AB.

Mark (m + n) points A₁, A₂, A₃, …, A_m+n on AX such that AA₁ = A₁A₂ = A₂A₃ = … = A_m+n-1 A_m+n.

Join A_m+n B.

Through A_m, draw A_mY || A_m+n B (if m: n) meeting AB at Y. So, Y divides AB internally in the ratio m: n.

Read and Learn More Class 10 Maths Solutions Exemplar

Through A_n, draw A_n1Z || A_m+n B (if n: m) meeting AB at Z. So, Z divides AB internally in the ratio n: m.

In Short:

Sum (m +n) endpoint

first (m) Parallel (Y)  (say)

NCERT Exemplar For Class 10 Maths Chapter 11 Constructions Solved Problems

Question 1. Determine a point which divides a line segment 7 cm long, internally in the ratio 2:3.

Solution:

Steps of Construction:

1. Draw a line segment AB = 7 cm by using a ruler.
2. Draw any ray malting an acute ZBAC with AB.
3. Along AC, mark off (2 + 3) = 5 points A₁, A₂, A₃, A₄ and A₅ such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
4. Join BA₅
5. Through A₂ draw a line A₂P parallel to A₅B by making an angle equal to \(\angle A A_5 B\) at A₂ intersecting AB at A point P.

The point P so obtained is the required point.

Justification: In ΔA₅B,

A₂P || A₅B (Construction)

∴ \(\frac{A A_2}{A_2 A_5}=\frac{A P}{P B}\) (by B.P theorem)

⇒ \(\frac{2}{3}=\frac{A P}{P B}\) (Construction)

⇒ AP : PB = 2: 3

i.e., P divides AB internally in the ratio 2 : 3.

Alternate Method:

Draw the line segment AB = 7 cm.

Draw any ray AC making an acute angle ∠BAC with AB.

Draw a ray BD parallel to Ac by making ∠ABC equal to angle ∠BAC.

Mark off 2 points A₁ and A₂ on AC and 3 points B₁, B₂, B₃ on AD such that AA₁ = A₁A₂ = BB₁ = B₁B₂ = B₂B₃

Join B₃A₂, suppose it intersects AB at point P. Then, P is the required point.

To Divide a Line Segment (Externally) in a Given Ratio m: n

Working Rule: (External Division)

Question 2. Determine a point which divides a line segment 6 cm long externally in the ratio 5 : 3.

Solution:

Steps of Construction:

1. Draw a line segment AB = 6 cm.
2. Draw any ray making an acute ∠BAX with AB.
3. Along AX, mark off (larger among the ratios) 5 points A₁, A₂, A₃, A₄, A₅ such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
4. Join the point A₂ (5-3) with end point B.
5. Draw a line parallel to A₂B from A₅ (larger among the ratios) which meets AB produced at P.

The point P, so obtained is the required point such that AP: BP = 5 : 3.

Justification: In AA₅P,

Since A₂B || A₅P, (Construction)

∴ \(\frac{A P}{B P}=\frac{A A_5}{A_2 A_5}\) (by B.P theorem)

⇒ \(\frac{A P}{B P}=\frac{5}{3}\) (Construction)

Question 3. Determine a point which divides a line segment 6 cm long externally in the ratio 3:5.

Solution:

Steps of Construction:

1. Draw a line segment AB = 6 cm.
2. Draw any ray making an acute ∠ABX with AB.
3. Along BX, mark off (larger among the ratios) 5 points B₁, B₂, B₃, B₄ and B₅ such that BB₁ =B₁B₂ = B₂B₃ — B₃B₄ = B₄B₅.
4. Join the point B₂ (5 – 3) with endpoint A
5. Draw a line parallel to B₂A from B₅ (larger among the ratios) which meets BA produced at P.

The point P so obtained is the required point such that AP: BP = 3:5.

Justification: In ΔPBB₅,

Since, B₂A || B₅P (Construction)

∴ \(\frac{B_2 B_5}{B B_5}=\frac{A P}{B P}\) (by B.P theorem)

⇒ \(\frac{3}{5}=\frac{A P}{B P}\) (Construction)

i.e., P divides AB externally in the ratio of 3:5

NCERT Exemplar For Class 10 Maths Chapter 11 Constructions To Construct A Triangle Similar To A Given Triangle

Question 1. Construct a triangle similar to a given triangle ABC such that each of its sides is \(\frac{2}{3} \mathrm{rd}\) of the corresponding sides of the triangle ABC. It is given that AB = 4 cm, BC = 5 cm and AC = 6 cm.

Solutions:

Given

It is given that AB = 4 cm, BC = 5 cm and AC = 6 cm.

Steps of construction:

1. Take BC = 5cm and Construct ΔABC with BA = 4cm and CA = 6cm.
2. Divide BC into three equal parts.
3. Let C be a point on BC such that \(B C^{\prime}=\frac{2}{3} B C\)
4. Draw A’C parallel to AC through C” intersecting BA at A’. ΔA’BC” is the required triangle.

Question 2. Construct an isosceles triangle whose base is 6 cm and altitude 4 cm. Then construct another triangle whose sides are \(\frac{3}{4}\) times the corresponding sides of the first triangle.

Solution:

Steps to Construction:

1. Draw a line segment BC = 6 cm.
2. Draw a perpendicular bisector of BC.
3. From mid-point D of BC on perpendicular bisector mark DA = 4 cm, join AB and AC.
4. Below BC make an acute angle ∠CBZ.
5. Along BZ mark off four points B₁, B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
6. Join B₄C.
7. From B₃ draw B₃E parallel to B₄C meeting BC at E.
8. From E draw EF || CA meeting BA at F. Then, ΔFBE is the required triangle.

Question 3. Construct a quadrilateral ABCD with AB = 3 cm, AD = 2.7 cm, DB = 3.6 cm, ∠B =110° and BC = 4.2 cm. Construct another quadrilateral A’BC’D’ similar to quadrilateral ABCD so that diagonal BD’ = 4.8 cm.

Solutions:

Steps to construction:

1. Draw a line segment BC = 4.2 cm
2. At B, construct angle YBC = 110°
3. With centre B and a radius equal to 3 cm, draw an arc-cutting BT at A.
4. With centre A and a radius equal to 2.7 cm, draw an arc.
5. With centre B and radius equal to 3.6 cm, draw another arc cutting the previous arc at D.
6. Join AD, CD and BD. Then, ABCD is the required quadrilateral.
7. Produce BD to D’ such that BD’ = 4.8 cm.
8. From D’, draw a line parallel to DA which cuts BY at A’.
9. From D’, draw a line parallel to DC which cuts BC produced at D’.

Then, A’BC’D’ is the required quadrilateral similar to ABCD.

Question 4. Construct a cyclic quadrilateral ABCD in which AB = 4.2 cm, BC = 5.5 cm, CA = 4.6 cm and AD = 3 cm. Also, construct a quadrilateral similar to ABCD whose sides are 1 .5 times the corresponding sides of ABCD.

Solution:

Steps to construction:

1. Draw a line segment AB = 4.2 cm.
2. With centre A and a radius equal to 4.6 cm, draw an arc.
3. With centre B and radius equal to 5.5 cm, draw another arc cutting the previous arc at C.
4. Join AC and BC.
5. Draw the perpendicular bisectors of any two sides say AB and BC respectively of ΔABC. Let them intersect each other at O.
6. Taking O as the centre and radius as OA or OB or OC, draw a circle. This is the circumcircle of ΔABC.
7. With centre A and radius equal to 3 cm, cut an arc on the opposite side of B, to cut the circle at D.
8. Join AD and CD. Then, ABCD is the required cyclic quadrilateral.
9. Produce Ac to C’ such that \(A C^{\prime}=1.5 \times A C \text { i.e., }\left(1+\frac{1}{2}\right) A C \Rightarrow A C+\frac{1}{2} \times 4.6\) i.e., 2.3cm more.
10. From C’, draw a line parallel to CD which meets AD produced at D’.
11. From C’, draw a line parallel to CB which meets AB produced at B’. Then, AB’C’D’ is the required quadrilateral similar to cyclic quadrilateral ABCD.

NCERT Exemplar For Class 10 Maths Chapter 11 Constructions Of Tangents To A Circle

Question 1. Take a point O on the plane of the paper. With O as the centre draw a circle of radius 4 cm. Take a point on this circle and draw a tangent at P.

Solution:

Steps of Construction:

1. Take a point O on the plane of the paper and draw a circle of a given radius of 4 cm.
2. Take a point P on the circle and join OP.
3. Construct ∠OPT = 90°.
4. Produce TP to T’ to obtain the required tangent TPT’.

Question 2. Draw a circle of radius 3 cm. Take a point P on it. Without using the centre of the circle, draw a tangent to the circle at point P.

Solution:

Steps of construction:

1. Draw any chord PQ through the given point P on the circle.
2. Take a point R on the circle and join P and Q to a point R.
3. Construct ∠QPY = ∠PRQ and on the opposite side of the chord PQ.
4. Produce YP to X to get YPX as the required tangent.

Question 3. Draw a circle of radius 2.5 cm. Take a point at a distance of 5 cm from the centre of the circle. From point P, draw two tangents to the circle.

Solution:

Steps of Construction:

1. Take a point O in the plane of the paper and draw a circle of radius 2.5 cm.
2. Mark a point P at a distance of 5.0 cm from the centre O and, join OP.
3. Draw the right bisector of OP, intersecting OP at Q.
4. Taking Q as a centre and OQ = PQ as the radius, draw a circle to intersect the given circle at T and T.
5. Join PT and PT’ to get the required tangents.

Question 4. Draw a pair of tangents to a circle of radius 5 cm inclined to each other at an angle of 60°.

Solution:

Steps of Construction:

1. Take a point O on the plane of the paper and draw a circle with centre O and radius OA = 5 cm.
2. At O construct radii OA and OB such that ∠AOB equals 120° i.e., supplement of the angle between the tangents.
3. Draw perpendiculars to OA and OB at A and B respectively suppose these perpendiculars intersect at P. Then, PA and PB are required tangents.

Question 5. Draw a circle of radius 4 cm. Take a point P outside the circle. Without using the centre of the circle, draw two tangents to the circle from point P.

Solutions:

Steps of Construction:

1. Draw a line segment of 4 cm.
2. Take a point P outside the circle and draw a second PAB, intersecting the circle at A and B.
3. Produce AP to C such that AP = CP.
4. Draw a semi-circle with CB as the diameter.
5. Draw PD ⊥ CB, intersecting the semi-circle at D.
6. Widi P as centre and PD as radius draw arcs to intersect the given circle at T and T’.
7. Join PT and PT’. Then, PT and PT’ are the required tangents.

NCERT Exemplar For Class 10 Maths Chapter 11 Constructions Exercise 11.1

Question 1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.

Solution:

Steps of construction :

1. Draw a line segment AB = 7.6 cm.
2. Draw a ray AX which forms an acute angle from AB.
3. Cut (8 + 5) = 13 equal marks on ray AX and mark them X₁, X₂, X₃, X₄, …, X₁₃.
4. Join X₁₃ to B.
5. Draw X₅C || X₁₃ B from X₅ which meets AB at C.

So, point C divides the line segment AB in the ratio 5:8.

On measuring two line segments, we get AC = 4.7 cm, BC = 2.9 cm

Verification: In ΔABX₁₃ and ΔACX₅, CX₅ || BX₁₃

∴ \(\frac{A C}{C B}=\frac{A X_5}{X_5 X_{13}}=\frac{5}{8}\)

⇒ AC: AB = 5:8

Question 2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.

Solutions:

Steps of construction:

1. Construct a ΔABC such that BC = 6 cm, AC = 5 cm and AB = 4 cm.
2. Draw a ray BX such that ∠CBX is at an acute angle.
3. Mark three points X₁, X₂ and X₃ on BX such that BX₁ = X₁X₂ = X₂X₃.
4. Join X₃ and C.
5. Draw a line parallel to line X₃C from X₂ which intersects BC at C’.
6. Draw a line parallel to line CA from C which meets BA at A’.

So, ΔA’B C’ is the required triangle.

Verification: By construction

X₃C || X₂C’  ⇒ \(\frac{B X_2}{X_2 X_3}=\frac{B C^{\prime}}{C^{\prime} C}\)

but \(\frac{B X_2}{X_2 X_3}=\frac{1}{2} \quad ⇒ \quad \frac{B C^{\prime}}{C^{\prime} C}=\frac{2}{1}\)

⇒ \(\frac{C^{\prime} C}{B C^{\prime}}=\frac{1}{2}\)

Adding 1 on both sides,

⇒ \(\frac{C^{\prime} C}{B C^{\prime}}+1=\frac{1}{2}+1\)

⇒ \(\frac{C^{\prime} C+B C^{\prime}}{B C^{\prime}}=\frac{1+2}{2} = \frac{B C}{B C^{\prime}}=\frac{3}{2}\)

Now, in ΔBC’A’ and ΔBCA,

CA || C’A’

from A.A. similarity, ΔBC’A’ ∼ ΔBCA

⇒ \(\frac{A^{\prime} B}{A B}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{B C^{\prime}}{B C} \quad\left[\text { each }=\frac{2}{3}\right]\)

Question 3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.

Solution:

Steps of construction:

1. Draw a line segment BC = 5 cm
2. Draw two arcs with centres B and C of radii 7 cm and 6 cm respectively which intersect each other at A.
3. Join BA and CA. ΔABC is the required triangle.
4. Draw a ray BX from B downwards, making an acute angle ∠CBX.
5. Mark seven points B₁, B₂, B₃, B₄, B₅, B₆ and B₇ on B₈ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇
6. Join B5C and draw B₇M || B₅C from B₇, which intersects the produced BC at M.
7. Draw MN || CA from point M which intersects the produced BA at N.

Now ΔNBM is the required triangle whose sides are \(\frac{7}{5}\) of the sides of ΔABC.

Justification:

By construction,

B₇M || B₅C

∴ \(\frac{B C}{C M}=\frac{5}{2}\)

Now, \(\frac{B M}{B C}=\frac{B C+C M}{B C}\)

= \(1+\frac{C M}{B C}=1+\frac{2}{5}=\frac{7}{5}\)

∴ \(\frac{B M}{B C}=\frac{7}{5}\)

and, MN || CA

∴ ΔABC ∼ ΔNBM

and \(\frac{N B}{A B}=\frac{B M}{B C}=\frac{M N}{C A}=\frac{7}{5}\)

Question 4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(1 \frac{1}{2}\) times the corresponding sides of the isosceles triangle.

Solutions:

Steps of construction:

1. Draw the line segment BC = 8cm.
2. Draw the perpendicular bisector OQ of BC which intersects BC at P.
3. Take PA = 4 cm along PO.
4. Join BA and CA. Now ΔABC is the required isosceles triangle.
5. Draw a ray BX from B making acute angle ∠CBX.
6. Mark three points B₁, B₂ and B₃ on BX such that BB₁ = B₁B₂ = B₂B₃
7. Join B₂C and draw B₃N || B₂C from B₃ which intersects the Produced BC at N.
8. Draw NM || CA from point N which intersects the produced BA at M.

Then ΔMBN is the required rectangle.

Justification:

∵ B₃ || B₂C (by construction)

∴ \(\frac{B C}{C N}=\frac{2}{1}\)

Now \(\frac{B N}{B C}=\frac{B C+C N}{B C}=1+\frac{C N}{B C}=1+\frac{1}{2}=1 \frac{1}{2}\)

Question 5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ABC = 60°. Then construct a triangle whose sides are \(\frac{4}{3}\) of the corresponding sides of the triangle ABC.

Solution:

Steps of construction:

1. Construct a triangle ABC such that BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
2. Draw a ray \(\overrightarrow{B X}\) such that CBX is an acute angle.
3. Mark four points X₁, X₂, X₃, and X₄ on BX such that BX₁ = X₁X₂ = X₂X₃ = X₃X₄.
4. Join X₄C.
5. Draw X₃C’ || X₄C which intersects BC at C.
6. Draw a line from C, parallel to CA which intersects BC at A’.

So, ΔA’BC’ is the required triangle.

Verification: By construction

X₄C || X₃C’ [from B.P.T.]

∴ \(\frac{B X_3}{B X_4}=\frac{B C^{\prime}}{B C} \text { but } \frac{B X_3}{B X_4}=\frac{3}{4}\) (by construction)

⇒ \(\frac{B C^{\prime}}{B C}=\frac{3}{4}\) → (1)

Now, CA || C’A’ (by construction)

ΔBC’A’ BCA [from A.A. similarity]

⇒ \(\frac{A^{\prime} B}{A B}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{B C^{\prime}}{B C}=\frac{3}{4}\) [from (1)].

Question 6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ΔABC.

Solution:

Steps of construction:

1. Construct a ΔABC such that BC = 7 cm, ∠B = 45° and ∠A = 105°.
2. Draw a ray BX such that ∠CBX is at an acute angle.
3. Mark four points X₁, X₂, X₃ and X₄ on BX such that:
   Bx₁ = X₁X₂ = X₂X₃ = X₃X₄.
4. Draw a line from X4 parallel to X₃C which intersects BC produced at C’.
5. Draw a line from C parallel to CA, that intersects BA produced at A’.
6. Thus, ΔA’BC’ is the required triangle.

Verification: By construction,

C’A’ || CA [from A.A. similarity]

ΔABC ∼ ΔA’BC’

⇒ \(\frac{A^{\prime} B}{A B}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{B C^{\prime}}{B C}\) → (1)

Again, by construction

X₄C’ || X₃C

∴ BX₄C’ BX₃C

⇒ \(\frac{B C^{\prime}}{B C}=\frac{B X_4}{B X_3}\)

but \(\frac{B X_4}{B X_3}=\frac{4}{3} \Rightarrow \frac{B C^{\prime}}{B C}=\frac{4}{3}\) → (2)

from (1) and (2),

∴ \(\frac{A^{\prime} B}{A B}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{B C^{\prime}}{B C}=\frac{4}{3}\).

Question 7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.

Solution:

Steps of construction:

1. Draw a line segment BC = 4 cm.
2. Draw a line segment AB = 3 cm from B which makes a 90° angle from BC.]
3. Join AC. ΔABC is the given right-angled triangle.
4. Draw an acute angle ∠CBY from B downwards.
5. Mark 5 points B₁, B₂, B₃, B₄ and B₅ on BY such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.
6. Join B₃C.
7. Draw B₅C’ || B₃ C from B₅, which meets the produced BC at C.
8. Draw C’A’ || CA from C’ which meets the produced BA at A’

So, ΔA’BC’ is the required triangle.

Justification:

By construction, B₅C’ || B₃C

∴ \(\frac{B C}{C C^{\prime}}=\frac{3}{2}\)

Now, \(\frac{B C^{\prime}}{B C}=\frac{B C+C C^{\prime}}{B C}=1+\frac{C C^{\prime}}{B C}\)

= \(1+\frac{2}{3}=\frac{5}{3}\)

and, C’A’ = CA

∴ ΔABC ∼ ΔA’BC’

and \(\frac{A^{\prime} B}{A B}=\frac{B C^{\prime}}{B C}=\frac{A^{\prime} C^{\prime}}{C A}=\frac{5}{3}\)

NCERT Exemplar For Class 10 Maths Chapter 11 Constructions Exercise 11.2

Question 1. Question 1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Solution:

Steps of construction:

1. Mark a point O.
2. Draw a circle with centre O and a radius of 6 cm.
3. Mark a point P at a distance of 10 cm from the centre.
4. Join O and P
5. Bisects OP at point M.
6. With the centre at point M, draw a circle with a radius MO or MP which intersects the given circle at A and B.
7. Join PA and PB. So, PA and PB are two required tangents. On measuring PA = PB = 9.6 cm.

Verification: Join OA and OB. Since OP is a diameter.

∠OAP = 90º; ∠OBP = 90º [angle in semicircle]

Again OA and OB are the radii of a circle.

⇒ PA and PB are tangents to the circle.

Question 2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.

Solution:

Steps of construction:

1. Draw two circles of radii 4 cm and 6 cm with a centre O.
2. Mark a point P on a larger circle.
3. Join O and P.
4. Find the point M of the perpendicular bisector of.OP.
5. With centre M and radius OM or PM draw a circle which intersects the smaller circle at A and B.
6. Join A and P.

So, PA is the required tangent. On measuring PA = 4.5 cm.

Verification: Join O and A.

∠PAO = 90° [angle in semcircle]

PA ⊥ OA

∵ OA is the radius of the small circle.

∴ PA is a tangent of the smaller circle

Question 3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameters each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

Solution:

Given : P and Q are two points on a diameter of the circle of radius 3 cm.

and OP = OQ = 7cm

We have to construct the tangents to the circle from P and Q.

Steps of construction:

1. Draw a circle of radius 3 cm with centre O.
2. Produce its diameter on both sides and take two points P and Q on it such that OP = OQ = 7 cm.
3. Bisect OP and OQ. Let E and F be the mid-points of OP and OQ respectively.
4. Draw a circle with centre E and radius OE, which intersects the given circle (0, 3) at M and N. Again draw a circle with centre F and radius OF which intersects the given circle at P’ and Q’.
5. Join PM, PN, QP’ and QQ’. These are the required tangents from P and Q to the circle (0, 3).

Justification:

Join OM and ON. ∠OMP lies in the semicircle, so ∠OMP = 90°. OM is the radius of the circle, so MP is the tangent to the circle. Similarly PN, QP’ and OQ’ are also the tangents to the circle.

Question 4. Draw a pair of tangents to a circle of radius 5 cm inclined to each other at an angle of 60°.

Solution:

Steps of construction:

1. Construct a circle with centre O and radius = 5 cm.
2. Draw ∠AOB = 120°.
3. Draw a perpendicular on OA from point A.
4. Draw a perpendicular on OB from B.
5. Both perpendiculars intersect each other at point C.

So, CA and CB are the required tangents to the circle, inclined at a 60° angle.

Verification:

In quadrilateral OACB, from angle sum property.

⇒ 120° + 90° + 90° + ∠ACB = 360°

⇒ 300° + ∠ACB = 360°

⇒ ∠ACB = 360°- 300° = 60°

Question 5. Draw a line segment AB of length S cm. Taking A as the centre, draw a circle of radius 4 cm and taking B as the centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Solution:

Steps of construction:

1. Draw a line segment AB = 8 cm.
2. Draw a circle of radius 4 cm taking A as the centre.
3. Draw another circle of radius 3 cm talcing B as the centre.
4. Draw the perpendicular bisector of AB and find the mid-point M of AB.
5. Draw the circle with centre M and radius MA or MB which intersects the circle with centre A at P and Q and the circle with centre F at R and S.
6. Join BP and FQ. So, BP and FQ are the required tangents on a circle with centre A from B.
7. Now join RA and SA.

So, RA and SA are the tangents on a circle with centre B from A.

Verification: Join A and P.

∠APB = 90° => BP ⊥ AP

but AP is the radius of a circle with centre A.

⇒ AP is a tangent of the circle with centre A.

Similarly, BQ is also a tangent of the circle with centre A. Similarly AR and AS are the tangents of the circle with centre B.

Question 6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and B = 90° BD is the perpendicular from b on AC. The circle through B, C, and D is drawn. Construct the tangents from A to this circle.

Solution:

Given

Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and B = 90° BD is the perpendicular from b on AC. The circle through B, C, and D is drawn.

Steps of construction:

1. Draw line segments AB = 6 cm and BC= 8 cm perpendicular to each other. Join AC. Now ΔABC is a right-angled triangle.
2. Taking the mid-point F of BC as the centre and radius of 4 cm, draw a circle which passes through points B, C and D.
3. Join AF and bisects A it. Let O be the midpoint of AF.
4. The circle drawn with f centre O and radius OA intersects the given circle at B and M.
5. Join AB and AM, which are the required tangents.

Justification:

Join FM and FB. Now ∠AMF lies in a semicircle,

so ∠AMF = 90° ⇒ FM ⊥ AM

∵ FM is the radius of the circle, so AM is the tangent to the circle and F is the centre.

Similarly, AB is also the tangent to the circle with centre F.

Question 7. Draw a circle with the help of a bangle Take a point outside the circle. Construct the pair of tangents from this point to the circle.

Solution:

Steps of construction:

1. Draw a circle using a bangle.
2. Draw two chords AP and MT. The perpendicular bisectors of AP and MT intersect each other at O, which is the centre of the circle.
3. Take a point R outside the circle. Join OR and bisect it.
4. Let Q be the mid-point of OR. The circle drawn with centre Q and radius OQ intersects the given circle at S and N.
5. Join RS and RN, so RS and RN are the required tangents from R.

Justification:

Join OS and ON. ∠OSR lies in a semicircle, so

∠OSR = 90° ⇒ OS || SR

∵ OS is the radius of a circle with a centre of O, so SR is the centre of that circle whose centre is O. Similarly, RN is also the tangent to the circle with a centre of O.