## RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS

**Other Exercises**

- RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.1
- RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2
- RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS
- RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

**Answer each of the following questions either in one word or one sentence or as per requirement of the questions :**

**Question 1.**

In the figure, PA and PB are tangents to the circle drawn from an external point P. CD is a third tangent touching the circle at Q. If PB = 10 cm and CQ = 2 cm, what is the length PC ?

**Solution:**

In the figure, PA and PB are the tangents to the circle drawn from P

CD is the third tangent to the circle drawn at Q

PB = 10 cm, CQ = 2 cm

PA and PB are tangents to the circle

PA = PB = 10 cm

Similarly CQ and CA are tangents to the circle

CQ = CA = 2 cm

PC = PA – CA = 10 – 2 = 8 cm

**Question 2.**

What is the distance between two parallel tangents of a circle of radius 4 cm ?

**Solution:**

TT’ and SS’ are two tangents of a circle with centre O and radius 4 cm and TT’ || SS’

OP and OQ are joined

Now OP is the radius and TPT’ is the tangent

OP ⊥ TPT’

Similar OQ ⊥ SS’

But TT’ || SS’

POQ is the diameter

Which is 4 x 2 = 8 cm

Distance between the two parallel tangents is 8 cm

**Question 3.**

The length of tangent from a point A at a distance of 5 cm from the centre of the circle is 4 cm. What is the radius of the circle ?

**Solution:**

PA is a tangent to the circle from P at a distance of 5 cm from the centre O

PA = 4 cm

OA is joined and let OA = r

Now in right ∆OAP,

OP² = OA² + PA²

=> (5)² = r² + (4)²

=> 25 = r + 16

=> r² = 25 – 16 = 9 = (3)²

r = 3

Radius of the circle = 3 cm

**Question 4.**

Two tangents TP and TQ are drawn from an external point T to a circle with centre O as shown in the following figure. If they are inclined to each other at an angle of 100°, then what is the value of ∠POQ ?

**Solution:**

TP and TQ are the tangents from T to the circle with centre O and ∠PTQ = 100°

OT, OP and OQ are joined

OP and OQ are radius

OP ⊥ PT and OQ ⊥ QT

Now in quadrilateral OPTQ,

∠POQ + ∠OPT + ∠PTQ + ∠OQT = 360° (Sum of angles of a quadrilateral)

=> ∠POQ + 90° + 100° + 90° = 360°

=> ∠POQ + 280° = 360°

=> ∠POQ = 360° – 280° = 80°

Hence ∠POQ = 80°

**Question 5.**

What is the distance between two parallel tangents to a circle of radius 5 cm?

**Solution:**

In a circle, the radius is 5 cm and centre is O

TT’ and SS’ are two tangents at P and Q to the circle

Such that TT’ || SS’

Join OP and OQ

OP is radius and TPT’ is the tangent

OP ⊥ TT’

Similarly OQ ⊥ SS’

POQ is the diameter of the circle

Now length of PQ = OP + OQ = 5 + 5 = 10 cm

Hence distance between the two parallel tangents = 10 cm

**Question 6.**

In **Q. No. 1**, if PB = 10 cm, what is the perimeter of ∆PCD ?

**Solution:**

In the figure, PB = 10 cm, CQ = 2 cm

PA and PB are tangents to the give from P

PA = PB = 10 cm

Similarly, CA and CQ are the tangents

CA = CQ = 2 cm

and DB and DQ are the tangents

DB = DQ

Now, perimeter of ∆PCD

PC + PD + CQ + DQ

= PC + CQ + PD + DQ

= PC + CA + PD + DB {CQ = CA and DQ = DB}

= PA + PB = 10 + 10 = 20 cm

**Question 7.**

In the figure, CP and CQ are tangents to a circle with centre O. ARB is another tangent touching the circle at R. If CP = 11 cm and BC = 7 cm, then find the length of BR. **(C.B.S.E. 2009)**

**Solution:**

Given : In the figure, CP and CQ are tangents to a circle with centre O

ARB is a third tangent to the circle at R

CP = 11 cm, BC = 7 cm

To find : The length of BR

BQ and BR are tangents to the circle drawn from B

BQ = BR ….(i)

Similarly CQ = CP

=> BC + BQ = CP = 11 (CP = 11 cm and BC = 7 cm)

=> 7 + BQ = 11

=> BQ = 11 – 7

BQ = 4 cm

But BQ = BR

BR = 4 cm

**Question 8.**

In the figure, ∆ABC is circumscribing a circle. Find the length of BC. **(C.B.S.E. 2009)**

**Solution:**

∆ABC is circumscribing a circle which touches it at P, Q and R

AC = 11 cm, AR = 4 cm, BR = 3 cm

Now we have to find BC

AR and AQ are tangents to the circle from A

AQ = AR = 4 cm

Then CQ = AC – AQ = 11 – 4 = 7 cm

Similarly,

CP and CQ are tangents from C

CP = CQ = 7 cm

and BP and BR are tangents from B

BP = BR = 3 cm

Now BC = BP + CP = 3 + 7 = 10 cm

**Question 9.**

In the figure, CP and CQ are tangents from an external point C to a circle with centre O. AB is another tangent which touches the circle at R. If CP = 11 cm and BR = 4 cm, find the length of BC. **[CBSE 2010]**

**Solution:**

CP and CQ are the tangents to the circle from C.

AB is another tangent to the same circle which touches at R and meets the first two tangents at A and B. O is the centre of the circle.

OC is joined

CP = 11 cm, BR = 4 cm

CP and CQ are tangents to the circle

CP = CQ = 11 cm

Similarly from B, CR and BQ are the tangents

BQ = BR = 4 cm

Now BC = CQ – BQ = 11 – 4 = 7 cm

**Question 10.**

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

**Solution:**

Two concentric circles with centre O, have radii 5 cm and 3 cm

AB is a chord which touches the smaller circle at P

OP is joined which is radius of smaller circle

P is mid-point of AB

OP = 3 cm and OA = 5 cm

Now in right ∆OAP

OA² = OP² + AP²

(5)² = (3)² + AP²

=> 25 = 9 + AP²

=> AP² = 25 – 9 = 16 = (4)²

AP = 4 cm

AB = 2 AP = 2 x 4 cm = 8 cm

**Question 11.**

In the given figure, PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure of ∠OAB. **[CBSE 2015]**

**Solution:**

In the given figure,

PA and PB are tangents to the circle from P

PA = PB

∠APB = 50°, OA is joined

To find ∠OAB

In ∆PAB

PA = PB

**Question 12.**

In the figure, PQ is a chord of a circle and PT is the tangent at P such that ∠QPT = 60°. Then, find ∠PRQ. **[NCERT Exemplar]**

**Solution:**

∠OPQ = ∠OQP = 30°, i.e., ∠POQ = 120°

Also, ∠PRQ = reflex ∠POQ

**Question 13.**

In the figure, PQL and PRM are tangents to the circle with centre O at the points Q and R respectively and S is a point on the circle such that ∠SQL = 50° and ∠SRM = 60°. Then, find ∠QSR. **[NCERT Exemplar]**

**Solution:**

Here ∠OSQ = ∠OQS = 90° – 50° = 40°

and ∠RSO = ∠SRO = 90° – 60° = 30°.

Therefore, ∠QSR = 40° + 30° = 70°

**Question 14.**

In the figure, BOA is a diameter of a circle and the tangent at a point P meets BA produced at T. If ∠PBO = 30°, then find ∠PTA. **[NCERT Exemplar]**

**Solution:**

As ∠BPA = 90°,

∠PAB = ∠OPA = 60°

Also OP ⊥ PT.

Therefore, ∠APT = 30°

and ∠PTA = 60° – 30° = 30°

Hope given RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS are helpful to complete your math homework.

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