# Ncert Solutions For Class 10 Maths Ex 11.2

## Ncert Solutions For Class 10 Maths Chapter 11 Ex 11.2

7) Draw a circle with 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:

Procedure for construction:

1. Draw a line segment of length AB = 10 cm. Bisect AB by constructing a perpendicular bisector of AB. Let M be the mid-point of AB.
2. With M as centre and AM as radius, draw a circle. Let it intersect the given circle at the points P and Q.
3. Join PB and QB. Thus, PB and QB are the required two tangents.

Justification: Join AP. Here ∠APB is an angle in the semi-circle. Therefore, ∠APB = 90°. Since AP is a radius of a circle, PB has to be a tangent to a circle. Similarly, QB is also a tangent to a circle.

In a Right ∆APB, AB2 = AP2 + PB2 (By using Pythagoras Theorem)
PB2 = AB2 – AP2 = 102 — 62 = 100 – 36 = 64

PB = 8 cm.

8) 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:

Procedure for construction:

1. Draw a line segment of length OA = 4 cm. With O as centre and OA as radius, draw a circle.
2. With O as centre draw a concentric circle of radius 6 cm(0B).
3. Let C be any point on the circle of radius 6 cm, join OC.
4. Bisect OC such that M is the mid point of OC.
5. With M as centre and OM as radius, draw a circle. Let it intersect the given circle of radius 4 cm at the points P and Q.
6. Join CP and CQ. Thus, CP and CQ are the required two tangents.

Justification:

Join OP. Here ∠OPC is an angle in the semi-circle. Therefore, ∠OPC = 90°. Since OP is a radius of a circle, CP has to be a tangent to a circle. Similarly, CQ is also a tangent to a circle.

In ∆COP, ∠P = 90°

CO2=CP2+OP2$CO^{2} = CP^{2} + OP^{2}$ CP2=CO2OP2$CP^{2} = CO^{2} – OP^{2}$

=6242$6^{2} – 4^{2}$

CP=25cm$CP = 2\sqrt{5}cm$

9) Draw a circle with radius 3 cm. On one of its extended diameter, take two points P and Q each at a distance of 7 cm from its centre. From two points P and Q, draw tangents to the circle.

Solution:

Given:

Two points P and Q on the diameter of a circle with radius 3 cm OP = OQ = 7 cm.

Aim:

To construct the tangents to the circle from the given points P and Q.

Procedure for construction:

1. Draw a circle with radius 3 cm with centreO.
2. Extend its diameter both the sides and cut OP = OQ = 7 cm.
3. Bisect OP and OQ.Let mid-points of OP and OQ be M and N.
4. With M as centre and OM as radius, draw a circle. Let it intersect (0, 3) at two points A and B. Again taking N as centre ON as radius draw a circle to intersect circle(0, 3) at points C and D.
5. Join PA, PB, QC and QD. These are the required tangents from P and Q to circle (0, 3).

10) Draw a pair of tangents to a circle which is of radius 5 cm, such that they are inclined to each other at an angle of 60°.

Solution:

To determine: To draw tangents at the ends of two radius which are inclined to each other at 120°

Procedure for  construction :

1. Keeping O as centre, draw a circle of radius 5 cm.
2. Take a point Q on the circle and join it to O.
3. From OQ, Draw∠QOR = 120°.
4. Take an external point P.
5. Join PR and PQ perpendicular to OR and OQ respectively intersecting at P.

The required tangents are RP and QP.