NCERT Exemplar For Class 10 Maths Chapter 11 Constructions

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.