NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Important Points
1. The numbers expressed as the product of the number with itself are called square numbers or perfect squares.
For example, 1 = 1×1 = 1²
4 = 2 ×2 = 2²
9 = 3 × 3 = 3²
16 = 4 × 4 = 4²
25 = 5 × 5 = 5²………………..
2. If a natural number m can be expressed as n², where n is also a natural number, then, m is called a square number.
For example, 100 is the square of 10, as
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100 – 10 × 10 – 10².
So, 100 is a square number.
121 is the square of 11, as
121 – 11 × 11 = n²
So, 121 is a square number.
Question 1. Consider the following numbers and their squares.
From the above table, can we enlist the square numbers between 1 and 100? Are there any natural square numbers up to 100 left out ? From the above table, can we enlist the square numbers between 1 and 100? Are there any natural square numbers up to 100 left out?
Solution:
From, the above table, we can enlist the square numbers between 1 and 100.
Such numbers are listed below:
4, 9, 16, 25, 36, 49, 64, and 81 Also, we see that no natural square numbers up to 100 are left out.
Question 2. Find the perfect square numbers between
- 30 and 40
- 50 and 60.
Solution:
1. We know that 5²= 5 × 5 = 25, 6²= 6 × 6 = 36 and 7² = 7 × 7 = 49.
So, the perfect square number between 30 and 40 is 36.
2. We know that 7² = 7 × 7 = 49, and 8² = 8 × 8 = 64.
So there is no perfect square number between 50 and 60.
Properties Of Square Numbers
- All square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place, i.e., all square numbers have the unit’s digit as 0, 1, 4, 5, 6, or 9 only.
- No square number ends with 2, 3, 7 or 8 at the unit’s place.
- Square numbers can have only an even number of zeros at the end.
- No square number ends with an odd number of zeros at the end.
- If a square number ends in 6 then the number whose square it is will end in 4 or 6.
Question 1. Can we say that if a number ends in 0, 1, 4, 5, 6, or 9, then it must be, a square number? Think about it,
Solution:
If a number ends with the digit 0, 1, 4, 5, 6, or 9 at its unit’s place, then it is not necessarily a square number.
For example:
The numbers 10, 21, 34, 325, 146, and 209 are not square numbers although they end with 0, 1, 4, 5, 6, and 9 respectively at their unit’s places.
Note 1: The non-perfect, square numbers end with the digit 2, 3, 7, or 8. i.e., if a number ends with the digit 2, 3, 7, or 8 at its unit place, then it is necessarily a non-perfect square number.
Examples: 32, 42, 43, 53, 63
Note 2: If a number ends with an odd number of zeros at the end, then it is necessarily a non-perfect square number.
Examples: 70, 690, 1430
Question 2. Can we say whether the following numbers are perfect squares? How do we know?
- 1057
- 23453
- 7928
- 222222
- 1069
- 2061.
Write five numbers which you can decide by looking at their unit’s digit that they are not square numbers.
Solution:
1. 1057
The number 1057 is not a perfect square because it ends with 7 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.
2. 23453
The number 23453 is not a perfect square because it ends with 3 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.
3. 7928
The number 7928 is not a perfect square because it ends with 8 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.
4. 222222
The number 222222 is not a perfect square because it ends with 2 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.
5. 1069
The number 1069 ends with 9 at the unit’s place. But we cannot say that 1069 is a perfect square number. It may or may not be a square number.
Also, 30 × 30 = 900
31 × 31 = 961
32 × 32 = 1024
33 × 33 = 1089
This shows that there is no natural number lying between 1024 and 1089 which is a square number.
As 1024 < 1069 < 1089, i.e., the number 1069 lies between 1024 and 1089, so 1069 is not a square number.
4. The number 2061 ends with 1 at the unit’s place. But we cannot say that 2061 is a perfect square number. It may or may not be a perfect square number.
Also, 45 × 45 = 2025
46 × 46 = 2116
This shows that there is no natural number lying between 2025 and 2116 which is a square number. Since 2025 < 2061 < 2116, i.e., the number 2061 lies between 2025 and 2116, therefore 2061 is not a square number.
The number 2061 ends with 1 at & the unit’s place. But we cannot say that 2061 is a j perfect square number. It may or may not be an O-perfect square number.
Also, 45 × 45 = 2025
46 ×46 = 2116
This shows that there is no natural number lying between 2025 and 2116 which is a square number. Since 2025 < 2061 < 2116, i.e., the number 2061 lies between 2025 and 2116, therefore 2061 is not a square number.
Question 3. Write five numbers that you can decide by looking at their unit’s digit that they are not square numbers.
Solution:
The five numbers that we can decide by looking at their one digit that they are not square numbers are 1032, 453, 5567, 13258, and 13293.
[Note that the non-perfect square numbers end with the digits 2, 3, 7, or 8 at the unit’s place]
Question 4. Write five numbers that you can not decide just by looking at their unit digit (or unit place) whether they are square numbers or not
Solution:
100000, 3246, 56556, 12331 and 799
Note that the numbers ending with the digits 0, 1, 4, 5, 6, or 9 at the unit’s place may or may not be perfect square numbers.
Question 5. Which of 123², 77², 82², and 161², 109²would end with the digit 1?
Solution:
We know that if a number has 1 or 1 is 9 in the unit’s place, then its square ends with the digit 1.
So, 161² and 109² would end with the digit 1
Question 6. Write the next two square numbers respectively. after 441 which ends in 1 and their corresponding numbers
Solution:
From Table 1 on page 91 ). We find that 441 = 21 x 21= 21s. So, the next two square numbers that end in are
29² (= 841) and 31² ( = 961) and their corresponding numbers are 29 and 31respectively.
[Note that if a number has 1 or 9 in the unit’s place, then its square ends with the digit 1.]
Question 7. Which of the following numbers has the digit 6 at the unit place ?
- 19²
- 26²
- 36²
- 34².
Solution:
Note that when a square number ends with digit 6, then the number whose square it is, will have either 4 or 6 at the unit’s place. So,
1. The digit at the unit’s place = 9
∴ 19² would not have 6 in place.
2. The digit at the unit’s place = 4
∴ 24² would have 6 at unit’s place.
3. The digit at the unit’s place = 6
∴ 26² would have 6 at the unit place
Question 8. Can you find more such rules by observing the numbers and their squares
Solution:
Yes; We can find more such rules by observing the numbers and their squares
These are as follows:
- A square number will end with digit 4 only and only when the number, of which it is the square, ends either with digit 2 or with digit 8.
- A square number will end with the digit 9 only and only when the number of which, it is square, ends either with digit 3 or with the digit 7.
- A square number will end with the h digit if only and only when the number, of which it is square, ends with digit f >.
- A square number will end with digit 1 only mid only when ( ho number, of which it is the Hqunro, ends will) digit 4 or 0,
- A square number will end with a digit Only and only when the number, of which it is the square, ends with the digit 0.
Question 9. What will be the one digit in the square of the following number?
- 1234
- 26387
- 52698
- 99880
- 21222
- 9106
Solution:
1. 1234
Ending digit = 4 and 4² = 4× 4 = 16
∴ The one digit in the square of the number 1234 will be 6.
2. 26387
Ending digit = 7 and 7² = 7 × 7 = 49
∴ The one digit in the square of the number 26387 will be 9.
3. 52698
Ending digit = 8 and 8² = 8×8 = 64
∴ The one digit in the square of the number 52698 will be 4.
4. 99880
Ending digit = 0 and 0² = 0×0 = 0
∴ The one digit in the square of the number 99880 will be 0.
5. 21222
Ending digit = 2 and 2² = 2×2 = 4
∴ The one’s digit in the square of the number 21222 will be 4
6. 9106
Ending digit = 6 and 6² = 6 × 6 = 36
∴ The one’s digit in the square of the number 9106 will be 6
Question 10. If a number contains 3 zeros at the end, how many zeros will its square have? What do you notice about the number of zeros at the end of the number and the number of zeros at the end of its square? Can we say that square numbers can only have an even number of zeros at the end?
Solution:
If a number contains 3 zeros at the end, then its square will have 3 × 2 = 6 zeros at the end
The number of zeros at the end of a number
=\(\frac{1}{2}\) ×The number of zeros at the end of its square.
Yes; we can say that square numbers can only have an even number of zeros at the end.
Question 11. With numbers and their squares. What can you say about the squares of even numbers and squares of odd numbers?
Solution:
- The squares of even numbers are always even numbers.
- The squares of odd numbers are always odd numbers.
Question 12. Square of which of the following numbers would be an odd number/an even number? Why?
- 727
- 158
- 269
- 1980
Solution:
1. 727
The square of 727 will be an odd number because the number 727 is odd
2. 158
The square of 158 will be an oven number because be number
∴ 158 is an even number
3. 269
The square of 269 will be an odd number because the number ‘
∴ 269 is an odd number
4. 1980
The squares of 1980 will be an even number because the number
∴ 1980 is an even number
Question 13. What will be the number of zeros in the square of the following numbers?
- 60
- 400
Solution:
1. 60
Number of zeros at the end of (the number 60 = 1)
∴ The number of zeros in the square of (be number 60 will be 2 ×1 = 2.
2. 400
Number of zeros at (the end of the number 400 = 2
∴ The number of zeros in the square of the number 400 will be 2 × 2 = 4
Note that the number of zeros at the end of the square of a number = 2 × number of zeros at the end of that number
Some More Interesting Patterns
1. A triangular number is one whose dot patterns can be arranged as triangles. If we combine two consecutive triangular numbers, we get a square number.
2. There are 2n non-perfect square numbers between the squares of the numbers n and (n + 1) which is 1 less than the difference of two squares.
3. The sum of the first n odd natural numbers is n².
4. We can express the square of any odd number as the sum of two consecutive positive integers
Note: \(n^2=\frac{n^2-1}{2}+\frac{n^2+1}{2}\)
Example: \((11)^2=\frac{121-1}{2}+\frac{121+1}{2}\)
= 60 + 61
=121
5. If (n + 1) and (n – 1) are two consecutive even or odd natural numbers, then (n + 1) x (n – 1) = n² – 1
Question 1. How many numbers are there between 6² and 7²
Solution:
Let n = 6
Then, n + 1 = 6 + 1 = 7
∴ There are 2n = 2 × 6 = 12
Nonperfect square numbers between 62 and 72
Question 2. Check for n = 5, n = 6, n = 7 etc., and verify
Solution:
1. n = 5
.’. n + 1 = 6
There are 2n = 2 × 5 = 10 non perfect square numbers between 5² and 6²
These are 26, 27, 28, 29, 30, 31, 32, 33, 34 and 35.
2. n = 6
n + 1 = 7
There are 2n = 2 × 6 = 12 nonperfect square numbers between 62 and 72.
These are 37, 38, 39, 40, 41, 42, 43,44, 45, 46, 47 and 48
3. n = 7
n + 1 = 8
There are 2n = 2 × 7 = 14
Nonperfect square numbers between 7² and 8².
These are 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62 and 63
Question 3. How many natural numbers lie between 9² and 10² Between 11² and 12²
Solution:
1. Here, n = 9
2n = 2 × 9 = 18.
So, 18 natural numbers lie between 9² and 10².
2. Here, n = 11
2n = 2 × 11 = 22.
So, 22 natural numbers lie between 11² and 12²
Question 4. How many non square numbers lie between the following pairs of numbers?
- 100² and 101²
- 90² and 91²
- 1000² and 1001²
Solution:
1. Here, n = 100
∴ 2n = 2 × 100 = 200.
So, 200 non square numbers lie between the pair of numbers 100² and 101²
2. Here, n = 90
∴ 2n = 2×90 = 180.
So, 180 non square numbers lie between the pair of numbers 90² and 91².
3. Here, n = 1000
∴ 2n = 2 × 1000 = 2000.
So, 2000 non square numbers lie between the pair of numbers 1000² and 1001²
Question 5. 1 + 3 + 5 + 7 […] = 16 = 42 1+3 + 5 + 7+ 9[..]= 25 = 52 1+3 + 5+7 + 9 + 11 […] =36 = 62
Solution:
⇒ 1 + 3 + 5 + 7
[Sum of first four odd numbers]
= 16 = 4²
⇒ 1 +3 + 5 + 7 + 9
[Sum of first five odd numbers]
= 25
= 5²
1 + 3 + 5 + 7 + 9+11
[Sum of first six odd numbers]
= 36
= 36
Question 6. Find whether each of the following numbers is a perfect square or not.
- 121
- 55
- 81
- 49
- 69
Solution:
1. Successively subtract consecutive odd natural numbers starting with l. i.e .. 1. 3, 5, 7. 5), … from 121.
121 – 1 = 120
120 – 3 = 117
117 – 5 = 112
112 – 7 = 105
105 – 9 = 96
96 – 11 = 85
85 – 13 = 72
72 – 15 = 57
57 – 17 = 40
40 – 19 = 21
21 – 21 = 0
This means that
121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
i.e., we can express 121 as the sum of consecutive odd numbers starting with 1.
∴ 121 is a perfect square number. number.
2. Successively subtract consecutive odd natural numbers starting with 1, i.e., 1, 3, 5, 7, 9, … from 55.
55 – 1 = 54
54 – 3 = 51
51 – 5 = 46
46 – 7 = 39
39 – 9 = 30
30 – 11 = 19
19 – 13 = 6
6 – 15 = -9
This shows that we are not able to express 55 as the sum of consecutive odd numbers starting with 1
∴ 55 is not a perfect square number
3. Successively subtract consecutive odd natural numbers starting with 1, i.e., 1, 3, 5, 7, 9, … from 81
81 – 1 = 80
80 – 3 = 77
77 – 5 = 72
72 – 7 = 65
65 – 9 = 56
56 – 11 = 45
45 – 13 = 32
32 – 15 = 17
17 – 17 = 0
This means that
81 = 1 + 3 + 5 + 7 + 9+11 + 13 + 15 + 17.
i.e., we can express 81 as the sum of consecutive odd numbers starting with 1.
∴ 81 is a perfect square number
4. Successively subtract consecutive odd natural numbers starting with 1, i.e., 1, 3, 5, 7, 9, … from 49
49 – 1 = 48
48 – 3 = 45
45 – 5 = 40
40 – 7 = 33
33 – 9 = 24
24 – 11 = 13
13 – 13 = 0
This means that
9 = 1 + 3 + 5 + 7 + 9 + 11 + 13
We can express 49 as the sum of consecutive odd numbers starting with 1.
∴ 49 is a perfect square number.
5. Successively subtract consecutive odd natural numbers starting with 1, i.e. 3. 5. 7. 9. … from 69.
69 – 1 = 68
68 – 3 = 65
65 – 5 = 60
60 – 7 = 53
53 – 9 = 44
44 – 11= 33
33 – 13 = 20
20 – 15 = 5
5 – 17 = -12
This shows that we are not able to express 69 as the sum of consecutive odd numbers starting with 1
.
∴ 69 is not a perfect square number
Question 7. Express the following as the sum of two consecutive integers.
- 21²
- 13²
- 11²
- 19²
Solution:
1. Here, n = 21
∴ \(\frac{n^2-1}{2}=\frac{21^2-1}{2}=\frac{441-1}{2} \)
= \(\frac{440}{2}=220 \\\)
And \(\frac{n^2+1}{2}=\frac{21^2+1}{2}=\frac{441+1}{2}\)
= \(\frac{442}{2}=221 \)
∴ 21² = 220 + 221
= 441
2. Here n= 13
∴ \(\frac{n^2-1}{2}=\frac{13^2-1}{2}=\frac{169-1}{2} \)
=\(\frac{168}{2}=84 \)
And \(\frac{n^2+1}{2}=\frac{13^2+1}{2}=\frac{169+1}{2} \)
= \(\frac{170}{2}=85 \)
13² = 84 + 85
= 169
3. Here n= 11
∴ \(\frac{n^2-1}{2}=\frac{11^2-1}{2}=\frac{121-1}{2}\)
= \(\frac{120}{2}=60 \)
And \(\frac{n^2+1}{2}=\frac{11^2+1}{2}=\frac{121+1}{2} \)
= \(\frac{122}{2}=61 \)
11² = 60 + 61
= 121 .
4. Here n= 19
⇒ \(\frac{n^2-1}{2}=\frac{19^2-1}{2}=\frac{361-1}{2} \)
=\(\frac{360}{2}=180 \)
And\(\frac{n^2+1}{2}=\frac{19^2+1}{2}=\frac{361+1}{2} \)
= \(\frac{362}{2}=181\)
19² = 180 + 181
= 361 .
Question 8. Do you think the reverse is also true, i.e., is the sum of any two consecutive positive integers is perfect square ofa number? Give an example to support your answer
Solution:
No; The reverse is not always true.
For example :
1. Consider two consecutive positive integers 3 and 4. Their sum is 3 + 4 = 7 which is not a perfect square ofa number.
2. Consider two consecutive positive integers 8 and 9. Their sum is 8 + 9 = 17 which is not a perfect square ofa number
Question 9. Write the square, making use of the above pattern.
- 111111²
- 1111111²
Solution:
1. 11111²
11111²= 1234565432
2. 1111111²
1111111² = 1234567654321
Question 10. Can you find the square, of the following numbers using the above pattern?
- 6066667²
- 66666667²
Solution:
1. 6666667²
66666672 = 44444448888889
2. 66666667²
66666667² = 4444444488888889
NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Exercise 5.1
Question 1. What will be the unit digit of the squares of the following numbers?
- 81
- 799
- 1234
- 52698
- 12796
- 272
- 3853
- 26387
- 99880
- 55555
Solution:
1. Unit’s digit of 81 = 1
⇒ 1 × 1 = 1
∴ The unit digit of the square of the number 81 will be 1.
2. Unit’s digit of 272 = 2
⇒ 2 × 2 = 4
∴ The unit digit of the square of the number 272 will be 4.
3. Unit’s digit of 799 = 9
⇒ 9 × 9 = 81
∴ The unit digit of the square of the number 799 will be 1.
4. Unit’s digit of 3853 = 3
⇒ 3 × 3 = 9
∴ The unit digit of the square of the number 3853 will be 9.
5. Unit’s digit of 1234 = 4
⇒ 4 × 4 = 16
∴ The unit digit of the square of the number 1234 will be 6.
6. Unit’s digit of 26387 = 7
⇒ 7 × 7 = 49
∴ The unit digit of the square of the number 26387 will be 9.
7. Unit’s digit of 52698 = 8
⇒ 8 × 8 = 64
∴ The unit digit of the square of the number 52698 will be 4.
8. Unit’s digit of 99880 = 0
⇒ 0 × 0 = 0
∴ The unit digit of the square of the number 99880 will be 0.
9. Unit’s digit of 12796 = 6
⇒ 6 × 6 = 36
∴ The unit digit of the square of the number 12796 will be 6.
10. Unit’s digit of 55555 = 5
⇒ 5 × 5 = 25
∴ The unit digit of the square of the number 55555 will be 5.
Question 2. The following numbers are not perfect squares. Give reasons.
- 1057
- 23453
- 798
- 22222
- 64000
- 89722
- 222000
- 505050
Solution:
1. 1057
The number 1057 is not a perfect square because it ends with 7 at the unit’s place whereas the square numbers end with 0. I, 4. 5, 6 or 9 only at unit’s place.
2. 23453
The number 23453 is not a perfect square because it ends with 3 at the unit’s place whereas the square numbers end with 0, 1. 4, 5, 6, or 9 only at the unit’s place.
3.7928
The number 7928 is not a perfect square because it ends with 8 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6, or only 9 at the unit’s place.
4. 222222
The number 222222 is not a perfect square because it ends with 2 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6 or 9 only at the unit’s place.
5. 64000
The number 64000 is not a square number because it has 3 (an odd number of) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even
6. 89722
The number 89722 is not a square number because it ends in 2 at the unit’s place whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.
7. 222000
The number 222000 is not a square number because it has 3 (an odd number) zeros at the end whereas the number of zeros at the end of a square number ending with zeros is always even.
8. 505050
The number 505050 is not a square number because it has 1 (an odd number) zeros i at the end whereas the number of zeros at the end of a square number ending with zeros is always even.
Question 3. The squares of which of the following would be odd numbers?
- 431
- 2826
- 7779
- 82004
Solution:
1. 431
431 is an odd number
∴ Its square will also be an odd number
2. 2826
2826 is an oven number
∴ Its square will not be an odd number.
3. 779
779 is an odd number
∴ Its square will be an odd number.
4. 82004
82004 is an even number
∴ Its square will not be an odd number.
[Note that the squares of odd numbers are odd numbers and the squares of even numbers are even numbers
Question 4. Observe the following pattern and find the missing digits :
- 112²= 121
- 101² = 10201
- 1001² = 1002001
- 100001² = 1…………2 ……………..1
- 10000001² =………………………………
Solution:
4. 100001² = 10000200001
5. 10000001² = 100000020000001
Question 5. Observe the following pattern and supply the missing numbers
- 11² = 121
- 101² = 10201
- 10101² = 102030201
- 1010101² = ………………………
- …………………………..² = 10203040504030201
Solution:
4. 1010101² = 1020304030201
5. 101010101² = 10203040504030201
Question 6. Using the given pattern, find the jJJ missing numbers
- 1² + 2²+ 2² = 3²
- 2² + 3² + 6² = 7²
- 3² + 4² + 12² = 13²
- 4² + 5² + ………………²= 21²
- 5²+ ………………..² + 30²= 31²
- 6² + 7² + …………….² = …………..²
Solution:
4. 4² + 5² + 20² = 21²
5. 5² + 6²+ 30² = 31²
6. 6² + 7² + 4²= 43²
Question 7. Without adding, find the sum :
- 1+3 + 5 + 7 + 9
- 1+3 + 5 + 7 + 9 + 11 + 13 + 15+ 17 + 1
- 1 + 3 + 5+ 7 + 9 + 11 +13 + 15 + 17 + 19 + 21 + 23
Solution:
1. 1 + 3 + 5 + 7 + 9
= Sum of first five odd natural numbers
= 5² = 25
2. 1 + 3 + 5+ 7 + 9 + 11 + 13 + 15+17+19
= Sum of first ten odd natural numbers
= 10² = 100
1 + 3 + 5 + 7 + 9+11+13 + 15+17+19 + 21 + 23
= Sum of first twelve odd natural
= 12² = 144.
[Note that the sum of the first n odd natural numbers is n².]
Question 8.
- Express 49 as the sum of 7 odd numbers.
- Express 121 as the sum of 11 odd numbers
Solution:
1. 49 = 7 × 7 = 7²
= 1 + 3 + 5 + 7 + 9+11 + 13
2. 121 = 11 × 11 = 11²
= 1 + 3 + 5 + 7 + 9+11 + 13 + 15 + 17 + 19 + 21
Question 9. How many numbers lie between squares of the following numbers?
- 12 and13
- 25 and 26
- 99 and 100.
Solution:
1. Here, n = 12 and n + 1 = 13
∴ 2n, = 2 × 12 = 24
So, 24 numbers lie between squares of the numbers 12 and 13.
2. Here, n = 25 and n + 1 = 26
∴ 2n = 2 × 25 = 50
So, 50 numbers lie between squares of the numbers 25 and 26.
3. Here, n = 99 and 11+ 1 = 100
∴ 2n = 2 x 99 = 198
So, 198 numbers he between squares of the numbers 99 and 100
.
[Note that 2n natural numbers lie between the squares of the numbers n and (n + 1
Finding The Square Of A Number
23s = (20 + 3)² = (20 + 3) (20 + 3)
= 20 (20 + 3) + 3 (20 + 3)
= 20² + 20 × 3 + 3 × 20 + 32
= 400 + 60 + 60 + 9 = 529
Other Patterns In Squares
Let 5 be a number with the unit digit 5.
Then, (a.5)² = a (a + 1) hundred + 25
Question 1. Now can you find the square of 95?
Solution:
(95)² = 9 (9 + 1) ×100 + 25
= 9 × 10 × 100 + 25
= 9000+ 25
= 9025
Question 2. Find the the. squares of Ike following numbers containing 5 in unit’s place:
- 15
- 95
- 105
- 205
Solution:
1. 15² = (1×2) hundred + 25
= 200 + 25 = 225
2. 952 = (9 ×10) hundred + 25
= 9000 + 25 = 9025
3. 1052 = (10 × 11) hundred + 25
= 11000 + 25 = 11025
4. 2052 = (20 × 21) hundred + 25
= 42000 + 25 = 42025
Pythagorean Triplets Points
1. If a, b, and c are three numbers such that any one of the following three relations holds :
- a² + b² = c²
- b² + c² = a²
- c² + a² = b²
Then the numbers a, b, and c are said to form a Pythagorean triplet.
For example : 3, 4, 5 is a Pythagorean triplet because
3² + 42 = 9 + 16 = 25 = 5² .
Can you find more such triplets?
Yes; we can find more such triplets.
For example: 8, 15, 17; 9, 12, 15; 12, 35, 37, etc.
2. For any natural number m > 1, (2m, m² – 1, m² + 1) forms a Pythagorean triplet.
Try to find some more Pythagorean triplets using this form.
Some more Pythagorean triplets using this form are as follows :
6, 8, 10; taking m = 3
8, 15, 17; taking m. = 4
10, 24, 26; taking m = 5
NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Exercise 5.2
Question 1. Find the square of the following numbers
- 32
- 35
- 86
- 93
- 71
- 46
Solution:
1. 32 = 30 + 2
⇒ 32 = 30 + 2
Therefore, 32² = (30 + 2)²
= (30 + 2) (30 + 2)
= 30(30 + 2) +2(30 + 2)
= 900 + 60 + 60 + 4
= 1024
2. 35 =30+5
⇒ 35 =30+5
Therefore, 35² = (30 + 5)²
= (30 + 5) (30 + 5)
= 30 (30 + 5) + 5 (30 + 5)
= 900+ 150+ 150 + 25
= 1225
3. 86 = 80 + 6
⇒ 86 = 80 + 6
Therefore, 86² = (80 + 6)²
= (80 + 6) (80 + 6)
= 80 (80 + 6) + 6 (80 + 6)
= 6400 + 480 + 480+36
= 7396
4. 93 = 90 + 3
⇒ 93 = 90 + 3
Therefore, 93² = (90 + 3)²
= (90 + 3) (90 + 3)
= 90 (90 + 3) + 3 (90 + 3)
= 8100 + 270+270+9
= 8649
5. 71 = 70+1
⇒ 71 = 70+1
Therefore, 71² = (70 + 1)²
= (70+ 1) (70+ 1)
= 70(70+ 1)+ 1(70+ 1)
= 4900+70 + 70+1
= 5041
6. 46 =40 + 6
⇒ 46 =40 + 6
Therefore, 46² = (40 + 6)²
= (40 + 6) (40 + 6)
= 40 (40 + 6) + 6 (40 + 6)
= 1600 + 240 + 240 + 36
= 2116
Question 2. Write a Pythagorean triplet whose one number is
- 6
- 14
- 16
- 18.
Solution:
1. Let 2m. = 6
⇒ m = – = 3
m² – 1 = 3² – 1 = 9 – 1 = 8
And m² + 1 = 3² + 1 = 9 + 1 = 10
So, a Pythagorean triplet, whose one member is 6, is 6, 8, 10.
2. Let 2m = 14
⇒ m = \(\frac{14}{2}\) = 7
∴ m²- 1=7² -1 = 49 – 1= 48
and m² + 1 = 7² + 1 = 49 + 1 = 50
So, a Pythagorean triplet, whose one member is 14, is 14, 48, 50.
3. Let 2m = 16
⇒ m = \(\frac{16}{2}\)
∴ m²- 1= 8² – 1 = 64 – 1 = 63
m² + 1 = 8² + 1 = 64 + 1 = 65
So, a Pythagorean triplet, whose one number is 16, is 16, 63, 65.
4. Let 2m =18
⇒ m = \(\frac{18}{2}\)
∴ m²- 1= 9² – 1 = 81- 1 = 80
m² + 1 = 9² + 1 = 81 + 1 = 82
So, a Pythagorean triplet, whose one the number is 18, is 18, 80, 82.
Square Roots
The square root of a number ‘a is that number which when multiplied by itself gives that number ‘a as a product.
Thus, if b is the square root of a;
Then b × b = a or b² = a
Symbolically, we write
b = \(\sqrt{a}\)
Note: b = \(\sqrt{a}\) ⇔ b² = a i.e., is the square root iofaaisthesquare ofb
Finding Square Roots
To /meet a number whose square is known is known as finding the squarewot.
Finding the square root is the inverse (opposite) operation of squaring.
There are two integral square roots of a perfect square number.
For example : 4 = (2)² = (- 2)²
∴ \(\sqrt{4}\) = 2 and – 2 both. Here, we shall take up only the positive square rootof a natural number.
Thus, \(\sqrt{4}\) = 2 (not- 2)
The positive square root of a number is denoted by the symbol V.
For example,
3² = 9
⇒ \(\sqrt{9}\)
= 3
Question 1. 11² = 121. What is the square root
Solution:
The square root of 121 is 11.
Question 2. 14² = 196. What is the square root of
Solution:
The square root of 196 is 14
Question 3.
- (- 1)² =1. Is- 1 a square root of 1?
- (- 2)² = 4.Is – 2 a square root of 4?
- (- 9)²= 81. Is – 9a square root of 81?
Solution:
- Yes; – 1 is a square root of 1.
- Yes; – 2 is a square root of 4.
- Yes; -9 is a square root of 81.
Finding Square Root Through Repeated Subtraction
We subtract successive odd numbers starting from 1 from the given square number till we get zero. The number of times, we have to make subtractions, gives the square root of the given square number.
Question 1. By repeated subtraction of odd (iii) 0) numbers starting from 1, find whether the following numbers are. perfect squares or not? If the number is a perfect square then find its square root.
- 121
- 55
- 36
- 49
- 90.
Solution:
1. 121
- 121 – 1 = 120
- 120 – 3 = 117
- 117 – 5 = 112
- 112- 7 = 105
- 105 – 9 = 96
- 96 – 11 = 85
- 85 – 13 = 72
- 72 – 15 = 57
- 57 – 17 = 40
- 40 – 19 = 21
- 21 – 21 = 0
Since from 121 we subtracted successive odd numbers starting from 1 and obtained 0 at the 11th step,
∴ \(\sqrt{121}\) = 11.
2. 55
- 55 – 1 = 54
- 54 – 3 = 51
- 51 – 5 = 46
- 46 – 7 = 39
- 39 – 9 = 30
- 30 – 11 = 19
- 19 – 13 = 6
- 6 – 15 = -9
This shows that we are not able to get zero at any step while subtracting successive odd numbers starting from 1 to 55.
∴ 55 is not a perfect square
3.36
- 36 – 1 = 36
- 35 – 3 = 32
- 32 – 5 = 27
- 27 – 7 = 20
- 20 – 9 = 11
- 11 – 11 = 0
Since from 36 we subtracted successive odd numbers starting from 1 and obtained 0 at the 6th step, therefore, = 6.
4. 49
- 49 – 1 = 48
- 48 – 3 = 45
- 45 – 5 = 40
- 40 – 7 = 33
- 33 – 9 = 24
- 24 – 11 = 13
- 13 – 13 = 0
Since from 49 we subtracted successive odd numbers starting from 1 and obtained 0 at the 7th step,
∴ \(\sqrt{49}\) =7
- 90 – 1 = 89
- 89 – 3 = 86
- 86 – 5 = 81
- 81 – 7 = 74
- 74 – 9 = 65
- 65 – 11 = 54
- 54 – 13 = 41
- 41 – 15 = 26
- 26 – 17 = 9
- 9 – 19 = – 10
This shows that we are not able to get zero at any step while subtracting successive odd numbers starting from 1 to 90.
∴ 90 is not a perfect square.
Finding Square Root Through Prime Factorisation
We find the prime factors of the given perfect square and arrange in pairs. Then, we choose one factor from each pair and multiply it together. The product thus obtained gives the required square root.
Note: A square number has complete pairs of prime factors.
NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Exercise 5.3
Question 1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?
- 9801
- 99856
- 99856
- 657666025.
Solution:
1. 9801
∴ 1 × 1 = 1 and 9 × 9 = 81
The possible one’s digit of the square root ofthe number 9801 could be1 or 9.
2. 99856
∴ 4 × 4= 16 and 6 × 6 = 36
The possible one digit of the square root ofthe number 99856 could be 4 or 6.
3. 99856
∴ 1 × 1 = 1 and 9 × 9 = 81
The possible one digit of the square root of the number 998001 could be 1 or 9.
4. 657666025
∴ 5 × 5 = 25
The possible one digit of the square root of the number 657666025 could be 5
Question 2. Without doing any calculation, find the numbers which are surely not perfect squares.
- 153
- 408
- 257
- 441.
Solution:
1. 153
The number 153 is surely not a perfect square because it ends in 3 whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only.
2. 257
The number 257 is surely not a perfect square because it ends in 7 whereas the square numbers end with 0, 1, 4, 5, 6 or 9
3. 408
The number 408 is surely not a perfect square because it ends in 8 whereas the square numbers end with 0, 1, 4, 5, 6, or 9 only.
4. 441
The number 441 may be a perfect square as the square numbers end with 0, 1, 4, 5, 6, or 9 only.
Question 3. Find the square roots of 100 and 169 by the method of repeated subtraction.
Solution:
1.100
- 100 – 1 = 99
- 99 – 3 = 96
- 96 – 5 = 91
- 91 – 7 = 84
- 84 – 9 = 75
- 75 – 11 = 64
- 64 – 13 = 51
- 51 – 15 = 36
- 36 – 17 = 19
- 19 – 19 = 0
Since from 100, we subtracted successive odd numbers starting from 1 and obtained 0 at the 10th step, therefore, \(\sqrt{100}\)= 10.
2. 169
- 169 – 1 = 168
- 168-3=165
- 165-5=160
- 160 – 7 = 153
- 153-9 = 144
- 144-11=133
- 133 – 13 = 120
- 120 – 15 = 105
- 105 – 17 = 88
- 88 – 19= 69
- 69 – 21 = 48
- 48 – 23 = 25
- 25 – 25 = 0
Since from 169, we subtracted successive odd numbers starting from 1 and obtained 0 at I the 13th step, therefore,\(\sqrt{169}\)= 13.
Question 4. Find the square roots of the following numbers by the Prime Factorisation Method
- 729
- 400
- 1764
- 4096
- 7744
- 9604
- 5929
- 9216
- 529
- 8100
Solution:
1. The prime factorization of 729 is
729 = 3 × 3 × 3 × 3 × 3 × 3.
By pairing the prime factors, we get
729 = \(3 \times 3 \times 3 \times 3 \times 3 \times 3\)
So, \(\sqrt{729}=3 \times 3 \times 3\) = 27
2. The prime factorization of 400 is
400 = 2× 2× 2× 2× 5× 5.
By pairing the prime factors we get
400 = \(2 \times 2 \times 2 \times 2 \times 5 \times 5\)
∴ \(\sqrt{400}=2 \times 2 \times 5\)
= 20
3. The prime factorisation 2 1764of 1764 is 2
1764 = 2 × 2 × 3×3 × 7 × 7
By pairing the prime factors, we get
1764 = \(2 \times 2 \times 3 \times 3 \times 7 \times 7\)
So, \(\sqrt{1764}=2 \times 3 \times 7\)
= 42
4. The prime factorization 2 512 of 4096 is
4096 = 2 × 2 × 2 × 2 × 2 × 2× 2 × 2× 2×2× 2× 2
By pairing the prime factors, we get
4096 = \(\underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2}\) × \(\underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2}\)
So, \(\sqrt{4096}=2 \times 2 \times 2 \times 2 \times 2 \times 2\)
= 64
5. The prime factorisation of 7744 is
7744 = 2 × 2× 2 × 2 × 2 × 2 ×11× 11.
By pairing the prime factors, we get
7744 = \(\underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2}\)
So,\(\sqrt{7744}=2 \times 2 \times 2 \times 11\)
6. The prime factorisation of 9604 is
9604 = 2 × 2 × 7 × 7 × 7 ×7.
By pairing the prime factors, we get
9604 = \(\underline{2 \times 2} \times \underline{7 \times 7} \times \underline{7 \times 7} \)
So, \(\sqrt{9604}=2 \times 7 \times 7=98\)
7. The prime factorization of 5929 is
5929 = 7 × 7 × 11 × 11
By pairing the prime factors, we get
5929 = \(\underline{7 \times 7} \times \underline{11 \times 11}\)
So, \(\sqrt{5929} =7 \times 11=77\)
8. The prime factorisation of 9216 is
9216 = 2 × 2 × 2 × 2 × 2× 2 × 2 × 2× 2 × 2 × 3 × 3
By pairing the primary factors get
9216 = \( \underline{2 \times 2} \times \underline{2 \times 2} \)
So, \(\times \underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2} \times \underline{3 \times 3}\)
⇒ \(\sqrt{9216}=2 \times 2 \times 2 \times 2 \times 2 \times 3\)
= 96
9. The prime factorization of 529 is
529 = 23 × 23.
By pairing the prime factors, we get
529 = \(\underline{23 \times 23} \)
So,\(\sqrt{529}\) = 23
10. The prime factorisation of 8100 is
8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5.
By pairing the prime factors, we get
8100 =\(\underline{2 \times 2} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \underline{5 \times 5}\)
So, \(\sqrt{8100}\) = 2×3 × 3 × 5
= 90.
Question 5. For each of the following numbers, find the smallest whole number by which it should be multiplied to get a perfect square number. Also, find the square root of the square number so obtained
- 252
- 180
- 10008
- 2028
- 1458
- 768
Solution:
1. The prime factorization of 252 is
252 = 2 × 2 × 3× 3× 7
By pairing the prime factors, we get
252= \(\underline{2 \times 2} \times \underline{3 \times 3} \times 7\)
As the prime factor 7 has no pair, 252 is not a perfect square.
If 7 gets a pair, then the number will be a perfect square. So, we multiply 252 by 7 to get
252 × 7 = \(\underline{2 \times 2} \times \underline{3 \times 3} \times \underline{7 \times 7}\)
Now each prime factor has a pair. Therefore, 252× 7 = 1764 is a perfect square. Thus the required smallest number is 7.
Also,\(\sqrt{1764}\) =2 × 3× 7
= 42
2. The prime factorisation of 180 is
180 = 2 × 2 × 3 × 3×5
By pairing the prime factors, we get
180 = \(\underline{2 \times 2} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \) x 5
As the prime factor 5 has no pair, 180 is not a perfect square.
If 5 gets a pair, then the number will be a perfect square. So, we multiply 180 by 5 to get
180 × 5= \(\underline{2 \times 2} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \underline{5 \times 5}\)
Now each prime factor has a pair. Therefore, 180 × 5 = 900 is a perfect square. Thus the required smallest number is 5
Also, \(\sqrt{900}\)
= 2 × 3× 5 = 30
3. The prime factorization of 1008 is
1008 = 2 × 2 × 2×2 × 3 × 3 × 7
By pairing the prime factors, we get
1008 = \(\underline{2 \times 2} \times \underline{2 \times 2} \times \underline{3 \times 3} \times 7\)
As the prime factor 7 has no pair, 1008 is not a perfect square.
If 7 gets a pair, then the number wil be a perfect square. So, we multiply 1008 by 7 to get
1008 × 7 = \(\underline{2 \times 2} \times \underline{2 \times 2} \times\underline{3 \times 3} \times \underline{3 \times 3} \times \underline{7 \times 7}\)
Now each prime factor has a pair, Therefore, 1008 x 7 = 7056 is a perfect square, Thus the required smallest number is 7
Also, \(\sqrt{7056}\)
= 2 × 2 × 3× 7 = 84
4. The prime factorization of 2028
2028 = 2 × 2 × 3 × 13 × 13.
By pairing the prime factors, we get
2028 = \(\underline{2 \times 2}\) × 3 × \(\underline{13 \times 13}\)
As the prime factor 3 has no pair, 2028 is not a perfect square
If 3 gels a pair, then (.ho number will be a perfect square. So, we multiply 2028 by 3 to got
2028 × 3 = \(\underline{2 \times 2} \times \underline{3 \times 3} \times \underline{13 \times 13}\)
Now each prime factor has a pair. Therefore, 2028 x 3 = 6084 is a perfect square. Tims the required smallest number is 3.
Also,\(\sqrt{6084}\)
= 2 × 3 × 13 = 78.
5. The prime factorisation of 1458 is
1458 =2 × 3 × 3× 3 × 3 × 3 × 3
By pairing the prime 3 9 factors, we get
1458 = 2 ×\(\underline{3 \times 3} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \underline{3 \times 3}\)
As the prime factor 2 has no pair,1458 is not a perfect square.
If 2 gets a pair, then the number will be a perfect square. So, we multiply 1458 by 2 to get
1458 × 2 = \(\underline{2 \times 2} \times\underline{3 \times 3} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \underline{3 \times 3}\)
Now each prime factor has a pair. Therefore, 1458 x 2 = 2916 is a perfect square. Thus the required smallest number is 2
Also, \(\sqrt{2916}\)= 2 × 3 × 3 × 3
= 54
6. The prime 768 factorisation of 768 is
768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3.
By pairing the prime factors, we get
786= \(\underline{2 \times 2} \times\underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2}\) x 3
As the prime factor 3 has no pair, 768 is not a perfect square.
If 3 gets a pair, then the number will be a perfect square. So, we multiply 768 by 3 to get
768 × 3 = \(\underline{2 \times 2} \times\underline{2 \times 2} \times \underline{2 \times 2} \times \underline{2 \times 2} \times \underline{3 \times 3}\)
Now (the inch prime factor has a pair. Therefore, 768 x 3 = 2304 to a perfect square. Thus the required smallest number is 3.
Also, \(\sqrt{2304}\) = 2 × 2 × 2 × 2 × 3
= 48
Question 6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root of the square number so obtained
- 252
- 2925
- 396
- 2645
- 2800
- 1620
Solution:
1. The prime factorisation of 252 is
252 = 2 × 2 × 3 × 3 × 7
By pairing the prime factors, we get
252 = \(\underline{2 \times 2} \times\underline{3 \times 3} \times\)7
We see that the prime factor 7 has no pair. So, if we divide 252 by 7, then we get
252 ÷ 7 = \(\underline{2 \times 2} \times\underline{3 \times 3}\)
Now each prime factor has a pair.
∴ 252 ÷ 7 = 36 is a perfect square.
Thus, the required smallest number is 7
Also, \(\sqrt{36}\) = 2 × 3
= 6
2. The prime factorisation of 2925 is
2925 = 3 × 3 ×5 × 5 × 13.
By pairing the prime factors, we get
2925 = \(\underline{3 \times 3} \times\underline{5 \times 5} \times\) 13
We see that the prime factor 13 has no pair.
So, if we divide 2925 by 13, then we get
2925 ÷ 13 =\(\underline{2 \times 2} \times\underline{5 \times 5}\)
Now each prime factor has a pair.
∴ 2925 ÷ 13 = 225 is a perfect square.
Thus, the required smallest number
Also, \(\sqrt{224}\)= 3 × 5
= 15
3. The prime factor ion of 306 is
306 = 2 × 2 × 3 × 3 × 11
By pairing the prime 2 factors, we get
396 = \(\underline{3 \times 3} \times\underline{5 \times 5}\) ×11
We see that the prime factor 11 has no pair. So, if we divide 396 by 11, then we get
= \(\underline{2 \times 2} \times\underline{3 \times 3}\)
Now each prime factor has a pair.
∴ 396 ÷ 11 = 36 is a perfect square.
Thus, the required smallest number is 11.
Also , \(\sqrt{2}{3}\)
= 6
4. The prime factorisation of 2645 is
2645 = 5 × 23 × 23.
By pairing the prime factors, we get
2645 = \(\underline{23 \times 23}\)
Now the only prime factor 23 has a pair.
Therefore, 2645 ÷ 5 = 529 is a perfect square.
Thus, the required smallest number is 5
Also, \(\sqrt{529}\) =23
5. The prime factorisation of 2800 is
2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7.
By pairing the prime factors, we get
2800 = \(\underline{2 \times 2} \times\underline{2 \times 2} \times \times\underline{5\times 5} \times\)7
We see that the prime factor 7 has no pair. So, if we divide 2800 by 7, then we get
2800 ÷ 7 \(\underline{2 \times 2} \times\underline{2 \times 2} \times \times\underline{5\times 5}\)
Now each factor has a pair. Therefore, 2800 + 7 = 400 is a perfect square. Thus, the required smallest number is 7.
Also, \(\sqrt{400}\)
= 2 × 2 ×5 = 20
6. The prime factorisation of 1620 is
1620 = 2 × 2 × 3×3 × 3 × 3 × 5
By pairing the prime factors, we get
1620 ÷ 5 = \(\underline{2 \times 2} \times\underline{3 \times 3} \times \times\underline{3\times 3}\)
Now each factor has a pair. Therefore, 1620 ÷ 5 = 324 is a perfect square. Thus, the required smallest number is 5.
Hence, \(\sqrt{324}\)= 2 × 3 × 3
= 18.
Question 7. The students of Class 8 of a school donated ₹ 2401 in all, for Prime Minister’s National ReliefFund, Each student donated as many rupees as the number of students in the class. Find the number of students in the class
Solution:
Let the number of students in the
Then rupees donated by each student = ₹ x
Rupees denoted by x students
= ₹ x × x
= ₹ x²
∴ The students of class 8th of a school donated ₹ 2401 for the Prime Minister’s National ReliefFund
⇒ \(x^2= 2401 \)
= \(\sqrt{2401}\)
The prime factorisation of 2401 is
x = \(\sqrt{2401} \)
= \(\sqrt{7 \times 7} \times 7 \times 7\)
x = 7 × 7 = 49
Hence, the number of students in the class is 49
Question 8. 2025 plants arc to be planted in a Now prime factorization of 2 180 pardons in such a nay that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row
Solution:
Let the number of rows 3 2025 be x. 3 675
Then, the number of plants in each row = x.
Number of plants to be planted
= x × x
=x²
But 2025 plants are to be planted in the garden.
⇒ \(x^2=2025 \Rightarrow x=\sqrt{2025}\)
The prime factorization of 2025 is
2025= \(\underline{3 \times 3} \times\underline{3 \times 3} \times \times\underline{5\times 5}\)
x = \(\sqrt{2025}\)
x = \(\sqrt{3 \times 3} \times 3 \times 3 \times 5 \times 5\)
r = 3 × 3 × 5
x = 45
Hence, the number of rows is 45 and the number of plants in each row is 45
Question 9. Find the smallest square number that is divisible by each of the numbers 4, 9, and l0
Solution:
The least number divisible by each one of 4, 9, and 10 is their L.C.M
The L.C.M. of 4, 9 and 10 is
2 × 2 × 3 × 3 × 5 = 180
Now prime factorisation of 180 is
180 = \(\underline{2 \times 2} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \underline{5 \times 5}\)
The prime factor 5 is not in pairs. Therefore 180 is not a perfect square.
To get a perfect square, each factor of 180 must be paired. So, we need to make a pair of 5
Therefore, 180 should be multiplied by 5.
Hence, the required smallest square
number is 180 ×5 = 900.
Question 10. Find the smallest square number that is divisible by each of the numbers 8, 15, and 20.
Solution:
The least number divisible by each one of 8, 15, and 20 is their L.C.M.
The L.C.M. of 8, 15 and
2 × 2 × 2 × 3 × 5 = 120
Now prime factorisation of 120 is
120 = \(\underline{2 \times 2}\) × 2 × 3 × 5
The prime factors 2, 3, and 5 are not in pairs.
∴ 120 is not a perfect square.
To get a perfect square, each factor of 120 must be paired. So, we need to make pairs of 2, 3 and 5. Therefore 120 should be multiplied by 2 × 3 × 5; ie. 30.
Hence, the required smallest square number is 120 × 30 = 3600
Finding Square Root By Division Method
Steps
- Obtain the number whose square root is to be found. Place a bar for every pair of digits starting from the digit at one’s place. If the number of digits in it is odd, then the left-most single digit too will have a bar. Each pair and the left-most single digit (if any) is called a period.
- Think of the largest, number whose square is less than or equal to the number under the extreme left bar. Take this number as the divisor and the quotient with the number under the extreme left bar as the dividend. Divide and get the remainder.
- Bring down the number under the next bar to the right of the remainder. This becomes the new dividend.
- Double the divisor and enter it with a blank on its right. Guess the largest possible digit to fill the blank which will also become the new digit in the quotient, such that when the new divisor is multiplied by the new quotient, the product is less than or equal to the new dividend obtained in step 3.
- Continue this process till the remainder is 0 and no digits are left in the given number.
- The quotient thus obtained is the required square root of the given number.
Question 1. Can we say that if the perfect square is of n-digits, then its square root will have n/2 digits if n is even or \(\frac{(n+1)}{2}\)
Solution:
Yes. The following examples will establish the truth ofthe above fact.
1. Consider the perfect square 729. Here, n = 3 (which is odd)
∴ Number of digits in the square root
=\(\frac{n+1}{2}=\frac{3+1}{2}=2\)
= 2
Also =\(\sqrt{729}\)
= 27 (which has 2 digits)
2. Consider the perfect square 2025.
Here, n = 4 (which is even)
∴ Number of digits in the square root
= \(\frac{n}{2}=\frac{4}{2}\)
= 2
Also\(\frac{n}{2}=\frac{4}{2}\) =45(which has 2 digits)
Question 2. Without calculating square roots, find. the number of digits in the square root of the following numbers.
- 25600
- 100000000
- 36864
Solution:
1. By placing bars, we get \(\overline{2}\) \(\overline{56}\) \(\overline{00}\) Since there are 3 bars, the square root will be of 3 digits
Aliter:
Here, n = 5 (which is odd)
Number of digits in the square root
= \(\frac{n+1}{2}\)
= \(\frac{5+1}{2}\)
= \(\frac{6}{2}\)
= 3
2. By placing bars, we get \(\overline{3}\) \(\overline{68}\)\(\overline{64}\)
⇒ \(\overline{1} \overline{00} \overline{00} \overline{00} \overline{00}\)
Since there are 5 bars, the square root will be 5 digits.
Aliter:
Here, n = 9 (which is odd)
Number of digits in the square root
=\(\frac{n+1}{2}=\frac{9+1}{2}\)
= \(\frac{10}{2}=5\)
3. By placing bars, we get \(\)
Since there are 3 bars, the square root will be of 3 digits.
Aliter:
Here, n = 5 (which is odd)
Number of digits in the square root
=\(\frac{n+1}{2}=\frac{5+1}{2}\)
= \(\frac{6}{2}\)
= 3
Square Roots Of Decimals
Put bars on the integral part of the number in the usual manner. Place bars in the decimal part on every pair of digits beginning with the first decimal place. Proceed as usual to find the square root.
NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Exercise 5.4
Question 1. Find the square root of each of the following numbers by the Division method:
- 2304
- 4489
- 3481
- 529
- 3249
- 1369
- 5776
- 7921
- 576
- 1024
- 3136
- 900
Solution:
1. 2304
∴ \(\sqrt{2304}\)
= 48
2. 4489
∴ \(\sqrt{4489}\)
= 67
3. 3481
∴\(\sqrt{3481}\)
= 59
4. 529
∴ \(\sqrt{529}\)
= 23
5.3249
∴ \(\sqrt{3249}\)
= 57
6. 1369
∴ \(\sqrt{1369}\)
= 37
7. 5776
∴ \(\sqrt{5776}\)
= 76
8. 7921
∴ \(\sqrt{7921}\)
= 89
9. 576
∴ \(\sqrt{7921}\)
= 89
10. 1024
∴ \(\sqrt{1024}\)
= 32
11. 3136
∴ \(\sqrt{3136}\)
= 56
12. 900
∴ \(\sqrt{900}\)
= 30
Question 2. Find the number of digits in the square root of each of the following numbers (without any calculation)
- 64
- 144
- 4489
- 27225
- 390625
Solution:
1. Number (n) of digits in 64
= 2 which is even.
∴ Number of digits in the square root of 64 = \(\frac{n}{2}=\frac{2}{2}\)
= 1
2. Number (n) of digits in 144
= 3 which is odd.
∴ Number of digits in the square root of
144 = \(\frac{n+1}{2}=\frac{3+1}{2}\)
= \(\frac{4}{2}\)
= 2
3. Number (n) of digits in 4489
= 4 which is even.
∴Number of digits in the square root of
4489 = \(\frac{n}{2}=\frac{4}{2}\)
= 2
4. Number (n) of digits in 27225
= 5 which is odd.
∴ Number of digits in the square root of
27225 = \(\frac{n}{2}=\frac{6}{2}\)
= 3
= \(\frac{6}{2}\)
5. Number (n) of digits in 390625
= 6 which is even.
∴ Number of digits in the square root of
= \(\frac{n+1}{2}=\frac{5+1}{2}\) = \(\frac{6}{2}\)
= 3
Question 3. Find the square root of the following decimal number
- 2.56
- 7.29
- 51.84
- 42.25
- 31.36
Solution:
1. 2.56
Hence, \(\sqrt{2.56}=1.6\)
2. 7.29
Hence, \(\sqrt{7.29}=2.7\)
3. 51.84
Hence, \(\sqrt{51.84}=7.2\)
4. 42.25
Hence, \(\sqrt{42.25}=6.5\)
5. 31.36
Hence, \(\sqrt{3136}=5.6\)
Question 4. Find the least number that must be subtracted from each of the following numbers to get a perfect square. Also, the square root of the perfect square so obtained
- 402
- 1989
- 3250
- 825
- 4000
Solution:
1. We have 402
This shows that 20² is less than 40² by 2. This means, that if we subtract the remainder of 2 from the number, we get a perfect square. So, the required least number is 2.
∴The required perfect square is 402-2 = 40
Hence,\(\sqrt{400}\)=20.
2. We have 1989
This shows that 44² is less than 1989 by 53. This means that if we subtract the remainder of 53 from the number, we get a perfect square. So, the required least number is 53
∴The required perfect square is 1989 – 53 = 1936.
Hence, \(\sqrt{1936}\) = 44
3. We have,3250
This shows that 57² is less than 3250 by
This means if we subtract the remainder (1) from the number, we get a perfect square. So, the required least number is 1.
∴ The required perfect square is
3250-1 = 3249
Hence, \(\sqrt{43249}\)= 57
4. We have, 825
This shows that 28² is less than 825 by 41. This means if we subtract the remainder (41) from the number, we get a perfect square. So, the required least number is 41.
∴ The required perfect square is 825-41 = 784
Hence,\(\sqrt{784}\) = 28
5. We have, 4000
This shows that 63² is less than 4000 by 31
This means if we subtract the remainder (31) from the number, we get a perfect square. So, the required least number is 31.
∴ The required perfect square is 4000 – 31 = 3969
Hence, \(\sqrt{3969}\) = 63
Question 5. Find the least number that must be added to each of the following numbers to get a perfect square. Also, find the square root of the perfect square to obtain
- 525
- 1750
- 252
- 1825
- 6412
Solution:
1. We have 525
This shows that 22² < 525.
The next perfect square is 23² = 529.
Hence, the least number to be added is
23² – 525 = 529 – 525 = 4
∴ The perfect square so obtained is 525 + 4 = 529
Hence,\(\sqrt{529}\) = 23
2. We have 1750
This shows that 41² < 1750.
The next perfect square is 42² = 1764.
Hence, the least number to be added is
42² – 1750 = 1764 – 1750
= 14
∴ The perfect square so obtained is 1750 + 14 = 1764
Hence \(\sqrt{1764}\)
= 42
3. We have, 252
This shows that 15² < 252.
The next perfect square is 16² = 256.
Hence, the least number to be added is
16² – 252 = 256 – 252 = 4.
∴ The perfect square so obtained is 252 + 4 = 256
Hence \(\sqrt{256}\)
= 16
4. We have, 6412
This shows that 42² < 1825
The next perfect square is 43² = 1849.
Hence, the least number to be added is
43² – 1825 = 1849 – 1825
= 24.
Therefore, the perfect square so obtained is 1825 + 24 = 1849
Hence, \(\sqrt{1849}\)
= 43
5. We have 1825
This shows that 80² < 6412.
The next perfect square is 81² = 6561.
Hence, the least number to be added is
81² – 6412 = 6561 – 6412 = 149.
Therefore, the perfect square so obtained is 6412+ 149 = 6561
Hence,\(\sqrt{6561}\)
= 81
Question 6. Find the length of the side of a square whose area is 441 m.2.
Solution:
Area of the square = 441 m
Length of the side of the square
⇒ \(\sqrt{441}\)m
∴ \(\sqrt{441}\) = 21
Hence, the length ofthe side ofthe square is 21 m
Question 7. In a right triangle ABC, ∠B – 30°.
- If AB = 6 cm, BC = 8 cm, find AC
- If AC =13 cm, BC = 5 cm, find AB.
Solution:
1. In the right triangle ABC,
∠B = 90°
By Pythagoras Theorem
AC² = AB² + BC²
AC² = 6² + 8²
AC² = 36 + 64
AC² = 100
AC = \(\sqrt{100}\)
⇒ \(\sqrt{100}\) = 10
Hence, AC is equal to 10 cm.
In the right triangle ABC,
∠B = 90°
By Pythagoras Theorem,
AC² = AB² + BC²
13² = AB² + 5²
169 = AB² + 25
AB² = 169-25
AB² = 144
AB = \(\sqrt{144}\)
Therefore,= \(\sqrt{144}\) 12.
Hence, AB is equal to 12 cm
Question 8. A gardener has 1000 plants. He wants to plant these in such a way that the number of rows and the number of columns remain the same. Find the minimum number of plants he needs for this
Solution:
Let the number of rows be x. Then the number of columns is x. Let us find out the square root of1000 by the division method
This shows that 31² < 1000
32² = 1024.
Hence, the minimum number of plants needs more for this = 1024 – 1000 = 24
So, number of rows
= number of columns = 32
Question 9. There are 500 children in a school. For a. P. T. drill they have to stand so that the number of rows is equal to several columns. How many children would be left out in this arrangement
Solution:
Let us find out the square root of the 500 division method.
We get the remainder of 16. It shows that 222 is less than 500 by 16.
This means that 16 children would be left out of this arrangement
NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Multiple-Choice Questions
Question 1. The perfect, square number out of 2, 3, 4, and 5 is
- 2
- 3
- 4
- 5
Solution: 3. 4
4 = 2 × 2 = 2²
Question 2. A perfect square number between 30 and 40 is
- 36
- 32
- 33
- 39
Solution: 1. 36
36 = 6 × 6 = 6²
Question 3. Between 50 and 60, the perfect square number is
- 56
- 55
- 54
- None
Solution: 4. None
None of 51,52,……59 is a perfect square
Question 4. Which ofthe following is a perfect square number?
- 1067
- 7828
- 4333
- 625
Solution: 4. 625
Perfect square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.
Question 5. Which ofthe following is a perfect square number?
- 2222
- 32543
- 888
- 10000
Solution: 4. 10000
Perfect square numbers end with 0, 1, 4, 5, 6, or 9 only at the unit’s place.
6. Which of 132², 87², 72² and 209² would end with digit 1?
- 132²
- 87²
- 72²
- 209²
Solution: 4. 209²
If a number has 1 or 9 in the unit’s place, then its square ends in 1
Question 7. Which of 105², 216², 333² and 111² would end with the digit 1
- 105²
- 216²
- 333²
- 111²
Solution: 4. 111²
If a number has 1 or 9 in the unit’s place, then its square ends in 1
Question 8. Which of 17², 34², 25², and 49² would have 6 at unit place?
- 17²
- 34²
- 25²
- 49²
Solution: 2. 34²
If a number has 4 or 6 in the unit’s place, then its square ends in 6.
Question 9. Which of 21², 33², 47², and 36² would have 6 at unit place?
- 21²
- 33²
- 47²
- 36²
Solution: 4. 36²
If a number has 4 or 6 in the unit’s place, then its square ends in 6.
Question 10. What will be the number of zeros in the square of the number 100
- 2
- 4
- 6
- 8
Solution: 2. 4
Number of zeros at the end of the number 100 = 2
Number of zeros at the end of the square ofthe number 100 = 2 ×2 = 4
Question 11. What will be the number of zeros in the square of the number 50?
- 1
- 2
- 3
- 4
Solution: 2. 2
Number of zeros at the end ofthe number 50 = 1
Number of zeros at the end of the square of the number 50 = 2×1 = 2
Question 12. What will be the number of zeros in the square ofthe number 9000?
- 2
- 3
- 4
- 6
Solution: 4. 6
Number of zeros at the end ofthe number 9000 = 3
Number of zeros at the end of the square ofthe number 9000 = 2 × 3 = 6
Question 13. The square of which of the following numbers will be even?
- 11
- 111
- 1111
- 112
Solution: 4. 112
∴ 112 is even 3.
∴ Its square will be even
Question 14. The square of which of the following numbers will be odd?
- 10
- 100
- 1000
- 99
Solution: 4. 99
∴ 99 is odd
∴ Its square will be odd.
Question 15. The square of which of the following numbers will be even?
- 21
- 27
- 35
- 50
Solution: 4. 50
∴ 50 is even.
∴ Its square will be even.
Question 16. The square of which of the following numbers will be odd?
- 42
- 54
- 66
- 81
Solution: 4. 81
∴ 81 is odd
∴ Its square will be odd
Question 17. How many natural numbers is he between 8² and 9²
- 16
- 17
- 18
- 19
Solution: 1. 16
2 × 8 = 16
Question 18. How many natural authors lie between 12² and 13²?
- 20
- 22
- 24
- 6
Solution: 3. 24
2× 12 = 24
Question 19. How many nonsquare numbers lie between the pair of numbers 80² and
- 162
- 160
- 161
- 154
Solution: 2. 160
2 × 80 = 160
Question 20. How many nonsquare numbers lie between the pair of numbers 36² and 37²?
- 36
- 37
- 74
- 72
Solution: 4. 72
2 × 36 = 72
Question 21. How many nonsquare numbers lie between the pair of numbers 500² and 501²
- 1000
- 999
- 1001
- 1002
Solution: 1. 1000
2 × 500=1000
Question 22. Express the square number 5² as the sum of two consecutive integers.
- 12+13
- 10+ 15
- 9+16
- 20 + 5
Solution: 1. 12+13
⇒ \(\frac{5^2-1}{2}=12, \quad \frac{5^2+1}{2}=13\)
Question 23. Express 9² as the sum of two consecutive integers.
- 40 + 41
- 50+31
- 36 + 45
- 72 + 9
Solution: 1. 40 + 41
⇒ \(\frac{9^2-1}{2}=40, \quad \frac{9^2+1}{2}=41\)
Question 24. Express 7² as the sum of two consecutive integers.
- 40 + 9
- 24+25
- 36+13
- 32+17
Solution: 2. 24+25
⇒ \(\frac{7^2-1}{2}=24, \quad \frac{7^2+1}{2}=25 .\)
Question 25. The unit digit in the square ofthe number 132 is
- 1
- 2
- 3
- 4
Solution: 4. 4
2 × 2 = 4
Question 26. The unit digit in the square ofthe number 1000 is
- 1
- 0
- 2
- None of these
Solution: 2. 0
0 × 0 = 0
Question 27. The unit digit in the* square of the number 1111 is
- 1
- 2
- 3
- 4.
Solution: 1. 1
1 × 1 = 1
Question 28. The unit digit in the square of the number 1333 is
- 3
- 6
- 9
- 1
Solution: 2. 6
3 × 3 = 9
Question 29. The unit digit in the square of the number 2644 is
- 4
- 6
- 8
- 2
Solution: 2. 6
4 × 4 = 1\(\underline{6}\)
Question 30. The unit digit in the square of the number 125 is
- 1
- 2
- 5
- 6
Solution: 3. 5
5 × 5 = 2 \(\underline{5}\)
Question 31. The unit digit in the square of the number 166 is
- 2
- 4
- 6
- 8.
Solution: 3. 6
6 × 6 = 3\(\underline{6}\)
Question 32. The unit digit in the square of the number 27 is
- 7
- 6
- 5
- 9.
Solution: 4. 9
7 × 7 =4 \(\underline{9}\)
Question 33. The unit digit in the square of number 78 is
- 1
- 2
- 3
- 4
Solution: 3. 3
8 × 8= 6\(\underline{4}\)
Question 34. The unit digit in the square of number 209 is
- 1
- 2
- 0
- 3
Solution: 1. 1
9× 9 = 8\(\underline{1}\)
Question 35. If 10² = 100, then the square root of 100 is
- 1
- 10
- 100
- 1000.
Solution: 2. 10
⇒ \(\sqrt{100}\) = 10
Question 36. If 25² = 625, (hen the square root of 625 is
- 5
- 26
- 125
- 625.
Solution: 2. 26
⇒ \(\sqrt{625}\) = 25
Question 37. What could be the possible one digit of the square root of 625?
- 2
- 3
- 4
- 5
Solution: 4. 5
5 × 5 = 25
Question 38. What could be the possible one digit of the square root of 121?
- 1, 9
- 3, 4
- 6, 7
- 7, 8.
Solution: 1. 1, 9
1 × 1 = 1
9 × 9 = \(1 \underline{81}\)
Question 39. What could be the possible one digit of the square root of 361?
- 1, 9
- 3, 4
- 6, 7
- 7, 8.
Solution: 1. 1, 9
1 × 1 = 1
9 × 9 = \(8 \underline{1}\)
Question 40. What could be the possible one digit of the square root of 576?
- 1, 9
- 5, 7
- 1, 8
- 2, 9.
Solution: 1. 1, 9
4 × 4 = \(1 \underline{6}\)
6 × 6 = \(1 \underline{36}\)
Question 41. What could be the possible one digit of the square root of 676?
- 4, 6
- 5, 7
- 1, 8
- 2, 9.
Solution: 1. 4, 6
4 × 4 = \(1 \underline{6}\)
6 × 6 = \(1 \underline{36}\)
Question 42. The smallest number by which 32 should be multiplied to get a perfect square is
- 2
- 3
- 4
- 8
Solution: 1. 2
32 × 2 = 64 = 8²
Question 43. The smallest number by which 48 should be multiplied to get a perfect square is
- 2
- 3
- 4
- 5
Solution: 2. 3
48 × 3 = 144 = 12²
Question 44. The smallest number by which 45 should be multiplied to get a perfect square is
- 2
- 3
- 5
- 7
Solution: 3. 5
45 × 5 = 225 = 15²
Question 45. The smallest number by which 54 should be multiplied so that we get a perfect square is
- 2
- 3
- 4
- 6
Solution: 1. 2
54 × 6 = 324 = 18²
Question 46. The smallest number by which 28 should be multiplied to get a perfect square is
- 2
- 4
- 3
- 7
Solution: 4. 7
28 × 7 = 196 = 14²
Question 47. The smallest number by which 1000 should be multiplied to get a perfect square is
- 5
- 10
- 4
- 8.
Solution: 2. 10
1000 × 10 = 10000 = 100²
Question 48. The smallest number by which 128 should be divided to get a perfect square is
- 2
- 3
- 4
- 8.
Solution: 1. 2
128 ÷ 2 = 64 = 8²
Question 49. The smallest number by which 48 should be divided to get a perfect square is
- 2
- 3
- 4
- 6.
Solution: 2. 3
48 ÷ 3 = 16 = 4²
Question 50. The smallest number by which 125 should be divided so as to get a perfect square is
- 3
- 5
- 25
- 125
Solution: 2. 5
125 ÷ 5 = 25 = 5²
Question 51. The smallest number by which 150 should be divided so as to get a perfect square is
- 4
- 2
- 5
- 6
Solution: 4. 6
150 ÷ 6 = 25 = 5²
Question 52. The smallest number by which 112 should be divided so as to get a perfect square is
- 6
- 4
- 3
- 7
Solution: 4. 7
112 ÷ 7 = 16 = 4²
Question 53. The smallest number by which 1000 should be divided so as to get a perfect square is
- 5
- 10
- 100
- 1000.
Solution: 2. 10
1000 ÷ 10 = 100 = 10²
Question 54. The smallest 3-digit, perfect square is
- 999
- 100
- 961
- 125
Solution: 2. 100
100 = 10²
Question 55. The number of digits in the square root of 62500 is
- 1
- 2
- 3
- 4
Solution: 3. 3
n = 5, \(\frac{n+1}{2}=3\)
Question 56. The number of digits in the square root of 441 is
- 1
- 2
- 3
- 4
Solution: 2. 2
n = 3, \(\frac{n+1}{2}=2\)
Question 57. The number of digits in the square root of 100 is
- 1
- 2
- 3
- 4
Solution: 2. 2
n = 3, \(\frac{n+1}{2}=\) = 2
Question 58. Find the length of the side of a square whose area is 100 cm2.
- 5 cm
- 10 cm
- 100 cm
- 4 cm
Solution: 2. 10 cm
⇒ \(\sqrt{100}=10\)
Question 59. The students of class VIII of a school donated? 10000 in all, for Prime Minister’s National Relief Fund. Each student donated as many rupees as the number of students in the class. The number of students in the class is
- 10
- 100
- 1000
- 10000
Solution: 2. 100
⇒ \(\sqrt{10000}=100\)
NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots True or False
1. That sum of two perfect squares is a perfect square – False
2. The product of two perfect squares is a perfect square – True
3. 1000 is a perfect square – False
4. All numbers of a Pythagorean triplet are even – False
5. There is only one square number between 20 and 30 – True
NCERT Solutions For Class 8 Maths Chapter 5 Squares And Square Roots Fill In The Blanks
1. 11² = 60 +→ 61
2. 112 + 113 =→ (15)²
3. There are non-square numbers between 42 and 52. → 8
4. The square of 1.1 is → 1.21
5. For every natural number m (> 1), 2m, m² – 1 and m² + 1 form a -Pythagorea
1. Write the greatest two-digit square number → 81
2. The smallest number of a Pythagorean triplet is 3. Find its other two numbers → 4,5
3. Which letter best represents the location of 716 on a number line → E
4. The area of a square board is 144 square units. Find the length ofthe side ofthe square board →12 units.
5. Find the value of \(\sqrt{17+\sqrt{64}}\) → 5