# Class 10 RD Sharma Solutions- Chapter 10 Circles – Exercise 10.2 | Set 2

### Question 11. In the figure, PQ is tangent at a point R of the circle with centre O. If âˆ TRQ = 30Â°, find m âˆ PRS

**Solution:**

âˆ TRQ = 30Â° and PQ is tangent at point R

To Find: âˆ PRS

Now âˆ SRT = 90Â° (angle in a semicircle)

Also, âˆ TRQ + âˆ SRT + âˆ PRS = 180Â° (Angles of a line)

=> 30Â° + 90Â° + âˆ PRS = 180Â°

=> âˆ PRS = 180Â° – 120Â°

=> âˆ PRS = 60Â°

### Question 12. If PA and PB are tangents from an outside point P, such that PA = 10 cm and âˆ APB = 60Â°. Find the length of chord AB.

**Solution:**

PA = 10 cm and âˆ APB = 60Â°

We also know that

PA = PB = 10 cm (Tangents drawn from a point outside the circle are equal)

=> âˆ PAB = âˆ PBA

Now, In âˆ†APB, we have:

âˆ APB + âˆ PAB + âˆ PBA = 180Â° (Angles of a triangle)

=> 60Â° + âˆ PAB + âˆ PAB = 180Â°

=> 2 âˆ PAB = 180Â° â€“ 60Â° = 120Â°

=> âˆ PAB = 60Â°

âˆ PBA = âˆ PAB = 60Â°

Hence. PA = PB = AB = 10 cm

Thus, length of chord AB = 10 cm

### Question 13. In a right triangle ABC in which âˆ B = 90Â°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC.

**Solution:**

Given: Let O be the centre of the circle with diameter AB.

Tangents to the circle at P meets BC at Q.

To Prove: PQ bisects BC, i.e., BQ = QC

Proof:

âˆ ABC = 90Â°

In âˆ†ABC, âˆ 1 + âˆ 5 = 90Â° [angle sum property, âˆ ABC = 90Â°]

âˆ 3 = âˆ 1 [angle between tangent and the chord equals angle made by the chord in alternate segment]

=> âˆ 3 + âˆ 5 = 90Â° (1)

And, âˆ APB = 90Â° [angle in semi-circle]

=> âˆ 3 + âˆ 4 = 90Â° (2) [âˆ APB + âˆ BPC = 180Â°, linear pair]

From Equation (1) and (2), we get

âˆ 3 + âˆ 5 = âˆ 3 + âˆ 4

âˆ 5 = âˆ 4

=> PQ = QC (3) [sides opposite to equal angles are equal]

Also, QP = QB (4) [tangents drawn from an internal point to a circle are equal]

From equation (3) and (4), we get:

=> QB = QC

Thus, PQ bisects BC is proved

### Question 14. From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of âˆ†PCD.

**Solution:**

PA = 14 cm

PA = PB = 14 cm (PA and PB are the tangents to the circle from P)

Also,

CA = CE (1) (CA and CE are the tangents from C)

DB = DE (2) (DB and DE are the tangents from D)

So, perimeter of âˆ†PCD:

= PC + PD + CD

=> PC + PD + CE + DE

=> PC + CE + PD + DE

=> PC + CA + PD = DB (From (1) and (2))

=> PA + PB

=> 14 cm + 14 cm

= 28 cm

### Question 15. In the figure, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle.

**Solution:**

âˆ B = 90Â°, BC = 6 cm, AB = 8 cm and r be the radius of incircle with centre O

Applying Pythagoras Theorem in right-angled âˆ†ABC:

ACÂ² = ABÂ² + BCÂ²

=> ACÂ² = (8)Â² + (6)Â² = 64 + 36 = 100

=> AC = 10 cm

=> AR + CR = 10 cm (1)

Now, AP = AR (AP and AR are the tangents to the circle from A)

Similarly, CR = CQ and BQ = BP

OP and OQ are radii of the circle = r (2)

OP âŠ¥ AB and OQ âŠ¥ BC (3) (angle between the radius to the point of contact of tangent is 90Â°)

and âˆ B = 90Â° (4)

From equation (2), (3) and (4):

BPOQ is a square

=> BP = BQ = r

=> AR = AP = AB â€“ BD = 8 â€“ r (5)

and CR = CQ = BC â€“ BQ = 6 â€“ r (6)

From equation (1), (5) and (6), we get:

AR + CR = 10

=> 8 â€“ r + 6 â€“ r = 10 (from (5) and (6))

=> 14 â€“ 2r = 10

=> 2r = 14 â€“ 10 = 4

=> r = 2

Radius of the incircle = 2 cm

### Question 16. Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the endpoints of the arc.

**Solution:**

Given: Let mid-point of arc ACB be C and, DCE be the tangent to it

To prove: AB || DCE

Proof:

Arc AC = Arc BC

=> Chord AC = Chord BC

Now, In âˆ†ABC,

AC = BC

=> âˆ CAB = âˆ CBA (1) (equal sides corresponding to the equal angle)

Since, DCE is a tangent line.

âˆ ACD = âˆ CBA (angle in alternate segment are equal)

=> âˆ ACD = âˆ CAB (from Eq. (1))

=> âˆ ACD and âˆ CAB are alternate angles

Which is only possible only when AB || CDE

Hence, the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

### Question 17. From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle show that âˆ†APB is equilateral.

**Solution:**

Given : Two tangents PA and PB are drawn to the circle and OP is diameter

To prove: âˆ†APB is equilateral

Proof: OP = 2r (let r be the radius of the circle)

=> OQ + QP = 2r

=> OQ = QP = r (OQ is the radius)

Now in right âˆ†OAP,

OP is the hypotenuse and Q is the mid point of it

=> OA = AQ = OQ (mid-point of hypotenuse of a right triangle is equidistant from its vertices)

Thus, âˆ†OAQ is equilateral triangle

=> âˆ AOQ = 60Â°

Also, âˆ APO = 90Â° â€“ 60Â° = 30Â° (Sum of all the angles of triangle is 180Â° )

=> âˆ APB = 2 âˆ APO = 2 x 30Â° = 60Â° (1)

We also know that, PA = PB (Tangents from P to the circle)

=> âˆ PAB = âˆ PBA (2)

Now, in âˆ†APB:

âˆ PBA + PAB + âˆ APB = 180Â° (Sum of all angles)

=> 2âˆ PBA = 120Â° (from (1) and (2))

=> âˆ PAB = âˆ PBA = 60Â°

Hence, âˆ†APB is an equilateral triangle.

### Question 18. Two tangents segments PA and PB are drawn to a circle with centre O such that âˆ APB = 120Â°. Prove that OP = 2 AP.

Solution:

Given : Two tangents to the circle from a point P and âˆ APB = 120Â°

To prove : OP = 2 AP

Proof : In right âˆ†OAP,

âˆ OPA = (1/2)âˆ APB = 60Â°

=> âˆ AOP = 90Â° â€“ 60Â° = 30Â°

Let, Q be mid point of hypotenuse OP of âˆ†OAP

=> QO = QA = QP

=> âˆ OAQ = âˆ AOQ = 30Â°

=> âˆ PAQ = 90Â° â€“ 30Â° = 60Â°

So, âˆ†AQP is an equilateral triangle

=> QA = QP = AP (1)

Also, Q is mid point of OP

=> OP = 2 QP = 2 AP (from (1))

Hence, proved.

### Question 19. If âˆ†ABC is isosceles with AB = AC and C (0, r) is the incircle of the âˆ†ABC touching BC at L. Prove that L bisects BC.

**Solution:**

Given: âˆ†ABC isosceles with AB = AC and incircle with centre O and radius r touches the side BC of âˆ†ABC at L.

To prove : L is mid point of BC.

Proof : AM and AN are the tangents to the circle from A

=> AP = AQ

But AB = AC (given)

=> AB â€“ AQ = AC â€“ AP

=> BQ = CP (1)

Now BL and BQ are the tangents from B

=> BL = BQ (2)

Similarly, CL and CP are tangents

=> CL = CP (3)

Also, BQ = CP (from (1))

=> BL = CL (from (2) and (3))

Hence, proved that L is mid point of BC.

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