# RD SHARMA Solutions for Class 10 Maths Chapter 8 - Circles

## Chapter 8 - Circles Exercise Ex. 8.2

A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC.

AP = AQ , BP = PR and CR = CQ (tangents from an external point)

Perimeter of ∆ABC = AB + BR + RC + CA

= AB + BP + CQ + CA

= AP + AQ

= 2AP

∆APO is a right-angled triangle. AO^{2} = AP^{2} + PO^{2}

13^{2} = AP^{2} + 5^{2}

AP^{2} = 144

AP = 12

∴ Perimeter of ∆ABC = 24 cm

In Fig., BC is a tangent to the circle with centre O. OE bisects AP. Prove that ∆AEO ~ ∆ABC

In ∆AOP,

OA = OP (radii) ∆AOP is an isosceles triangle. OE is a median.

In an isosceles triangle,the median drawn∴∠OEA = 90^{o}

In ∆AOE and ∆ABC,

∠ABC = ∠OEA = 90^{o}

∠A is common.

∆AEO ~ ∆ABC…(AA test)

In Fig., there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5 cm, find the length of PS.

PR = PQ…(tangents fromexternal points)

PQ = 5 cm

Also,

OQ is perpendicular to PS …(tangent is perpendicular to the radius)

Now, in a circle,a perpendicular drawn from the centre of a circle bisects the chord.

So, OQ bisects PS.

PQ = QS

QS = 5 cm

PS = 10 cm

In Fig., PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOR is a diameter. If ∠POR = 130˚ and S is a point on the circle, find ∠1 + ∠2.

In DPQR,

∠POR is an external angle.

So,

∠POR = ∠PQO + ∠OPQ

Now, PQ is tangent to the circle with radius OQ.

∠PQO = 90^{o}

130˚ = 90˚ + ∠OPQ

∠OPQ = 40^{o}

∠1 = 40^{o}

Now,

Minor arc RT subtends a 130˚ angle at the centre.

So, it will subtend a 65˚angle at any other point on the circle.∠RST = 65˚

∠2 = 65˚

∠1 + ∠2 =105^{˚}

In Fig., PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.

AP = PB = 12 cm, AC = CQ = 3 cm and QD = DB = 3 cm …(tangent from external point)

PA = 12 cm, PC + CA = 12

PC + 3 = 12

PC= 9 …(i)

Now,

PB = 12

PD + DB = 12

PD + 3 = 12

PD = 9 …(ii)

PC + PD = 18 cm…from (i) and (ii)

If from any point on the common chord of two intersecting circles, tangents be drawn to the circles, prove that they are equal.

Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find the radius of the inner circle.

A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.

Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

If AB, AC, PQ are tangents in Fig. 8.56 and AB = 5 cm, find the perimeter of ΔAPQ.

In Fig., PQ is tangent at a point R of the circle with centre O. If ∠TRQ = 30°, find m∠PRS.

In a right triangle ABC in which ∠B =90°, 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.

From an external point P, tangents PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn, which intersects PA and PB at C and D respectively. If PA = 14 cm, find the perimeter of Δ PCD.

In fig., ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incirde.

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

Two tangent segments PA and PB are drawn to a circle with centre O such that APB = 120^{o}. Prove that OP = 2 AP.

AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC =BD

In fig., a circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

In Fig., a circle is inscribed in a quadrilateral ABCD in which ∠B= 90°. If AD=23 cm, AB =29 cm and DS =5 cm, find the radius r of the circle.

In Fig., there are two concentric circles with centre O of radii 5 cm and 3 cm. From an external point P, tangents PA and PB are drawn to these circles. If AP = 12 cm, find the length of BP.

In Fig., AB is a chord of length 16 cm of a circle of radius 10 cm. The tangents at A and B intersect at a point P. Find the length of PA.

In Fig., PA and PB are tangents from an external point P to a circle with centre O. LN touches the circle at M. Prove that PL + LM = PN + MN.

In Fig., BDC is a tangent to the given circle at point D such that BD = 30 cm and CD = 7 cm. The other tangents BE and CF are drawn respectively from B and C to the Calculate (i) AF (ii) radius of the circle.

In the given figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQ. Find ∠RQS.

Since RS is drawn parallel to the tangent PQ,

∠SRQ = ∠PQR

Also, PQ = PR

⇒ ∠PQR = ∠PRQ

In ∆PQR,

∠PQR + ∠PRQ + ∠QPR = 180°

⇒∠PQR + ∠PQR + 30° = 180°

⇒2∠PQR = 150°

⇒∠PQR = 75°

⇒∠SRQ =∠PQR = 75° (alternate angles)

Also, ∠RSQ =∠RQP = 75° (the angle between a tangent and a chord through the point of contact is equal to an angle in the alternate segment.)

In ∆RSQ,

∠RSQ + ∠SRQ + ∠RQS = 180°

⇒75° + 75° + ∠RQS = 180°

⇒ ∠RQS =30°

From an external point P, tangents PA = PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB.

The common tangents AB and CD to two circles with centres O and O' intersect at E between their centres. Prove that the points O, E and O' are collinear.

In Fig, common tangents PQ and RS to two circles intersects at A. Prove that PQ = RS.

Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Since AC is the tangent to the circle with radius 9 cm, we have OB ⊥ AC.

Hence, by applying the Pythagoras Theorem, we have,

OA^{2} = OB^{2} + AB^{2}

⇒ 15^{2} = 9^{2} + AB^{2}

⇒ AB^{2} = 15^{2} - 9^{2}

⇒ AB^{2} = 225 - 81 = 144

∴ AB = 12 cm

We know that the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.

Here, OB is the perpendicular and AC is the length of the chord of the circle with radius 15 cm.

So,

AC = 2 × AB = 2 × 12 = 24 cm

Length of the chord of the larger circle which touches the smaller circle = 24 cm.

AB and CD are common tangents to two circles of equal radii. Prove that AB =CD.

A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of ∆ PQR is 336 cm^{2}, find the sides PQ and PR.

Let PA = PB = x

Tangents drawn from an external point are equal in length. QB = QT = 14 cm , RA = RT = 16 cm

PR = (x + 16) cm, PQ = (x + 14)cm,

QR = 30 cm

= x + 30

Area of ∆PQR

Area of ∆PQR = 336 cm^{2}

Side PR = (12 + 16) = 28 cm

Side PQ = (12 + 14) = 26 cm

In Fig., the tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA =110°, find ∠CBA.

AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB.

In the given figure, a ∆ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC, when area of ∆ ABC is 84 cm^{2}.

Let M and N be the points where AB and AC touch the circle respectively.

Tangents drawn from an external point to a circle are equal

⇒ AM=AN

BD=BM=8 cm and DC=NC=6 cm

In the given figure, AB is a diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ.

∠AOQ=58° (given)

In right ∆BAT,

∠ABT + ∠BAT + ∠ATB=180°

29° + 90° + ∠ATB=180°

∠ATB = 61°

that is, ∠ATQ = 61°

In Fig., OQ : PQ = 3 : 4 and perimeter of ΔPOQ = 60 cm. Determine PQ, QR and OP.

In fig., PO ⊥ QO . The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.

In fig., O is the centre of the circle and BCD is tangent to it at C. Prove that BAC + ACD = 90^{o}.

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

## Chapter 8 - Circles Exercise Ex. 8.1

Fill in the blanks:

(i) The common point of a tangent and the circle is called ....... .

(ii) A circle may have ....... parallel tangents.

(iii) A tangent to a circle intersects it in ....... point(s).

(iv) A line intesecting a circle in two points is called a ....... .

(v) The angle between tangent at a point on a circle and the radius through the point is ....... .

Fill in the blanks:

(i) The common point of a tangent and the circle is called point of contact .

(ii) A circle may have two parallel tangents.

(iii) A tangent to a circle intersects it in one point(s).

(iv) A line intesecting a circle in two points is called a secant .

(v) The angle between tangent at a point on a circle and the radius through the point is 90^{o} .

## Chapter 8 - Circles Exercise 8.48

A tangent PQ at a point P of a circle of radius 5 cm meets line through the centre O at a point Q such that OQ = 12 cm. Length PQ is

(a) 12 cm

(b) 13 cm

(c) 8.5 cm

(d)

radius = 5 cm

So, OP = 5 cm

OQ = 12 cm

So, the correct option is (d).

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

(a) 7 cm

(b) 12 cm

(c) 15 cm

(d) 24.5 cm

PQ is a tangent to the circle

So, OP^{2 }+ PQ^{2} = OQ^{2}

OP^{2 }= OQ^{2 }- PQ^{2}

= (25)^{2 }- (24)^{2}

= 49

OP = 7

So, the correct option is (a).

The length of tangent from point A at a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is

(a)

(b) 7 cm

(c) 5 cm

(d) 25 cm

Given OP = 3 cm

PA = 4 cm

Hence, OA^{2 }= OP^{2} + PA^{2}

OA^{2} = 3^{2 }+ 4^{2}

= 25

OA = 5 cm

So, the correct option is (c).

## Chapter 8 - Circles Exercise 8.49

Tangents from same point to circle have equal length.

Hence Bb = Ba

bC = Cc

Ac = Aa

Let Ba = x then Bb = x

bc = 6 - x and Aa = 8 - x

and Cc = 6 - x and Ac = 8 - x

So AC = AC + cC

= 6 - x + 8 - x

AC = 14 - 2x ......(1)

Also AC^{2 }= AB^{2 }+ BC^{2}

= 8^{2} + 6^{2 }

= 100

AC = 10 .....(2)

from (1) & (2)

14 - 2x = 10

4 = 2x

x = 2 also aB = Ob = radius = 2 cm

So, the correct option is (b).

If four sides of a quadrilateral ABCD are tangential to a circle, then

(a) AC + AD = BD + CD

(b) AB + CD = BC + AD

(c) AB + CD = AC + BC

(d) AC + AD = BC + DB

Tangents from same point are of equal length.

AP = AS, PB = BQ

QC = CR, RD = DS

AB = AP + PB .....(1)

BC = BQ + QC ......(2)

CD = CR + RD .....(3)

AD = AS + DS .....(4)

Adding (1) & (3)

AB + CD = AP + BP + CR + RD

= AS + BQ + CQ + DS

= (AS + DS) + (BQ + CQ)

from (2) & (4)

AB + CD = AD + BC

So, the correct option is (b).

The length of the tangent drawn from a point 8 cm away from the centre of a circle of radius 6 cm is

(a)

(b)

(c) 10 cm

(d) 5 cm

Given OQ = 8 cm

OP = 6 cm

OP^{2 }+ PQ^{2} = OQ^{2}

6^{2 }+ PQ^{2} = 8^{2}

PQ^{2 }= 64 - 36

= 28

PQ =

So, the correct option is (b).

AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm, then AB is equal to

(a) 4 cm

(b) 6 cm

(c) 8 cm

(d) 12 cm

DA and DC are tangents to circle from same point

so, DA = DC ......(1)

similarly DB = DC ......(2)

(1) + (2)

2DC = DA + DB

2DC = AB

AB = 2 × 4

= 8 cm

So, the correct option is (c).

In Figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. Then,

(a) AD = AB + BC + CA

(b) 2AD = AB + BC + CA

(c) 3AD = AB + BC + CA

(d) 4AD = AB + BC + CA

AD = AE .......(1)

CD = CF ......(2)

BF = BE .....(3)

from (1)

2AD = 2AE

= AE + AD

= (AB + BE) + (AC + CD)

= AB + BF + AC + CF

= AB + AC + BC

So, the correct option is (b).

In figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR =

(a) 8 cm

(b) 3 cm

(c) 2.5 cm

(d) 5 cm

## Chapter 8 - Circles Exercise 8.50

AP and PQ are tangents drawn from a point A to a circle with centre O and radius 9 cm. If OA = 15 cm, then AP + AQ =

(a) 12 cm

(b) 18 cm

(c) 24 cm

(d) 36 cm

AP = PQ ....(1)

and OA^{2 }= OP^{2 } + PA^{2}

(15)^{2 }= (9)^{2 }+ AP^{2}

AP^{2 }= 225 - 81

= 144

AP = 12

AP + AQ = 2AP

= 24 cm

So, the correct option is (c).

At one end of a diameter PQ of a circle of radius 5 cm, tangent XPY is drawn to the circle. The length of chord AB parallel to XY and at a distance of 8 cm from P is

(a) 5 cm

(b) 6 cm

(c) 7 cm

(d) 8 cm

## Chapter 8 - Circles Exercise 8.51

In the adjacent figure, if AB = 12 cm, BC = 8 cm and AC = 10 cm, then AD =

(a) 5 cm

(b) 4 cm

(c) 6 cm

(d) 7 cm

AB = 12 cm

BC = 8 cm

AC = 10 cm

Let AD = x

AF = x

BD = 12 - x

and BE = BD = 12 - x

CE = BC - BE

= 8 - (12 - x)

= x - 4

and CE = CF = x - 4

AC = AF + FC

= x + x - 4

AC = 2x - 4

Given, AC = 10 cm

so 2x - 4 = 10

2x = 14

x = 7 cm

AD = 7 cm

So, the correct option is (d).

In figure, if AP = PB, then

(a) AC = AB

(b) AC = BC

(c) AQ = QC

(d) AB = BC

AP = BP given

and AP = AQ

also BP = BR

from this, we conclude that

AQ = BR .....(1)

We know CR = CQ .....(2)

from (1) & (2)

AQ + CR = BR + CR

AQ + CQ = BR + CR

AC = BC

So, the correct option is (b).

In figure, if AP = 10 cm, then BP =

AP = 10 cm

AO = 6 cm

OB = 3 cm

AP^{2} + OA^{2 }= OP^{2}

OP^{2 }= 10^{2} + 6^{2}

OP^{2 }= 136

Also OB^{2 }+ BP^{2} = OP^{2 }

3^{2} + BP^{2 }= 136

BP^{2} = 136 - 9

So, the correct option is (b).

## Chapter 8 - Circles Exercise 8.52

In Figure, if quadrilateral PQRS circumscribes a circle, then PD + QB =

(a) PQ

(b) QR

(c) PR

(d) PS

PA = PD

AQ = QB

and PQ = PA + AQ

PQ = PD + QB

Hence PD + QB = PQ

So, the correct option is (a).

In figure, two equal circles touch each other at T, if QP = 4.5 cm, then QR =

(a) 9 cm

(b) 18 cm

(c) 15 cm

(d) 13.5 cm

PQ = PT .....(1)

and PT = PR .....(2)

so from (1) & (2)

PQ = PR

PQ = PR = 4.5 cm

QR = PQ + PR

= 4.5 + 4.5 = 9 cm

So, the correct option is (a).

## Chapter 8 - Circles Exercise 8.53

In figure, PR =

(a) 20 cm

(b) 26 cm

(c) 24 cm

(d) 28 cm

radius of circle 1 = 3 cm

radius of circle 2 = 5 cm

OP^{2 }= OQ^{2 }+ QP^{2 } and O'S^{2 }+ SR^{2} = O'R^{2 }

OP^{2 }= 4^{2 }+ 3^{2} O'R^{2 }= 5^{2 }+ 12^{2}

= 16 + 9 O'R^{2} = 169

= 25 O'R' = 13 cm

OP = 5 cm

OO' = OK + KO'

= 3 + 5

= 8 cm

PR = PO + OK + KO' + O'R

= 5 + 3 + 5 + 13

= 26 cm

So, the correct option is (b).

Two concentric circles of radii 3 cm and 5 cm are given. Then length of chord BC which touches the inner circle at P is equal to

(a) 4 cm

(b) 6 cm

(c) 8 cm

(d) 10 cm

OB = OC = OA = 5 cm

OQ = OP = 3 cm

OB^{2 }= OQ^{2 }+ BQ^{2}

BQ^{2} = OB^{2 }- OQ^{2}

= 5^{2 }- 3^{2}

= 16

BQ = 4 cm

also BQ = BP

BP = 4 cm

In ΔOPC,

OP^{2} + PC^{2} = OC^{2}

PC^{2 } = OC^{2} - OP^{2}

= 5^{2 }- 3^{2}

= 16

PC = 4 cm

BC = BP + PC = 4 + 4 = 8 cm

So, the correct option is (c).

In figure, there are two concentric circles with centre O. PR and PQS are tangents to the inner circle from point plying on the outer circle. If PR = 7.5 cm, then PS is equal to

(a) 10 cm

(b) 12 cm

(c) 15 cm

(d) 18 cm

Given PR = 7.5 cm

so PR = PQ

PQ = 7.5 cm

PS is the chord to the larger circle. We know that, perpendicular drawn from centre bisect the chords.

Hence PQ = QS

PS = PQ + QS

= 2PQ

= 2 × 7.5

= 15 cm

So, the correct option is (c).

## Chapter 8 - Circles Exercise 8.54

In figure, if AB = 8 cm and PE = 3 cm, then AE =

(a) 11 cm

(b) 7 cm

(c) 5 cm

(d) 3 cm

AC = AB .....(1)

BD = DP ......(2)

PE = EC ......(3)

AB = 8 so AC = 8 cm

PE = 3 so EC = 3 cm

AE = AC - EC = 8 - 3 = 5 cm

So, the correct option is (c).

## Chapter 8 - Circles Exercise 8.55

In figure, the sides AB, BC and CA of a triangle ABC, touch a circle at P,Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then length of BC is

(a) 11 cm

(b) 10 cm

(c) 14 cm

(d) 15 cm

PA = AR

AR = 4 cm

BP = BQ and QC = RC

BQ = 3 cm

Given AC = 11

AR + RC = 11

4 + RC = 11

RC = 7

so QC = 7 cm

BC = BQ + QC

= 3 + 7

= 10 cm

So, the correct option is (b).

EK = 9 cm

and EK = EM

Hence EM = 9 cm .....(1)

Also EK = ED + DK

and DK = DH

EK = ED + HD .......(2)

EM = EF + FM

and FM = FH

EM = EF + FH ......(3)

(2) + (3)

EK + EM = ED + EF + DH + HF

18 = ED + DF + EF

perimeter = 18 cm

So, the correct option is (a).

## Chapter 8 - Circles Exercise 8.56

AB = 29 cm

AD = 23

DS = 5 cm

DS = DR

so DR = 5 cm

AR = AD - DR

= 23 - 5

= 18 cm

AR = AQ

AQ = 18 cm

BQ = AB - AQ

= 29 - 18

BQ = 11 cm

As OP || BQ and OQ || PB

Hence, OP = BQ

OP = 11 cm

So, the correct option is (a).

In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is

(a) 4

(b) 3

(c) 2

(d) 1

AB = 5 cm

BC = 12 cm

AB^{2 } + BC^{2} = AC^{2}

AC^{2 }= 5^{2} + 12^{2}

= 169

AC = 13 cm

Let BQ = x

AQ = AR = 5 - x

CR = AC - AR

= 13 - (5 - x)

= x + 8

And CP = CR = x + 8

so BP = BC - PC

= 12 - (x + 8)

= 4 - x

But BP = BQ = x

4 - x = x

x = 2

and BQ || OP and OQ || PB

so BQ = PO

PO = 2 cm

So, the correct option is (c).

In figure, QR is a common tangent to the given circles touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm, then the length of QR (in cm) is

(a) 3.8

(b) 7.6

(c) 5.7

(d) 1.9

PT = 3.8 cm

We know

PQ = PT and PT = PR

Hence PQ = 3.8 cm and PR = 3.8 cm

Now, QR = QP + PR

= 3.8 + 3.8

QR = 7.6 cm

So, the correct option is (b).

## Chapter 8 - Circles Exercise 8.57

In figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, then x =

(a) 10

(b) 9

(c) 8

(d) 7

AB = x cm

BC = 7 cm

CR = 3 cm

AS = 5 cm

CR = CQ

CQ = 3 cm

given BC = 7 cm

BQ = BC - QC

= 7 - 3

= 4 cm

And BQ = BP

so BP = 4 cm

Also AS = AP

Hence AP = 5 cm

AB = AP + BP

= 5 + 4

= 9 cm

x = 9 cm

So, the correct option is (b).

If angle between two radii of a circle is 130°, the angle between the tangents at the ends of radii is

a. 90°

b. 50°

c. 70°

d. 40°

If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to

If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is

a. 3 cm

b. 6 cm

c. 9 cm

d. 1 cm

At one end A of a diameter AB of a circle of radius 5 cm, Tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is

a. 4 cm

b. 5 cm

c. 6 cm

d. 8 cm

From a point P which is at a distance 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is

a. 60 cm^{2}

b. 65 cm^{2}

c. 30 cm^{2}

d. 32.5 cm^{2}

If PA, PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to

a. 25°

b. 30°

c. 40°

d. 50°

The pair of tangents AP and AQ drawn from an external point to a circle with centre O are perpendicular to each other and length of each tangent is 5 cm. The radius of the circle is

a. 10 cm

b. 7.5 cm

c. 5 cm

d. 2.5 cm

## Chapter 8 - Circles Exercise 8.58

In figure, if ∠AOB = 125° , then ∠COD is equal to

a. 45°

b. 35°

c. 55°

d. 62°

In figure, if PQR is tangent to a circle at Q whole centre is O , AB is a chord parallel to PR and ∠BQR = 70°, then∠AQB is equal to

a. 20°

b. 40°

c. 35°

d. 45°

### Other Chapters for CBSE Class 10 Mathematics

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Pairs of Linear Equations in Two Variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- Trigonometric Identities Chapter 12- Heights and Distances Chapter 13- Areas Related to Circles Chapter 14- Surface Areas and Volumes Chapter 15- Statistics Chapter 16- Probability### RD SHARMA Solutions for CBSE Class 10 Subjects

### Kindly Sign up for a personalised experience

- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions

#### Sign Up

#### Verify mobile number

Enter the OTP sent to your number

Change