RD SHARMA Solutions for Class 10 Maths Chapter 8 - Circles

Chapter 8 - Circles Exercise 8.48

Question 1

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) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

radius = 5 cm

So, OP = 5 cm

OQ = 12 cm

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

So, the correct option is (d).

Question 2

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

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

PQ is a tangent to the circle

So, OP+ PQ2 = OQ2

OP= OQ- PQ2

      = (25)- (24)2

      = 49

OP = 7

So, the correct option is (a).

Question 3

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) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

(b) 7 cm

(c) 5 cm

(d) 25 cm

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Given OP = 3 cm

         PA = 4 cm

Hence, OA= OP2 + PA2

OA2 = 3+ 42

      = 25

OA = 5 cm

So, the correct option is (c).

Question 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Chapter 8 - Circles Exercise 8.49

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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= AB+ BC2

             = 82 + 6

             = 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).

Question 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 6

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

Solution 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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).

Question 7

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

(a) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

(b) Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

(c) 10 cm

(d) 5 cm

Solution 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Given OQ = 8 cm

         OP = 6 cm

OP+ PQ2 = OQ2

6+ PQ2 = 82

       PQ= 64 - 36

             = 28

PQ = Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

So, the correct option is (b).

Question 8

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

Solution 8

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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).

Question 9

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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

Solution 9

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).

Question 10

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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

Solution 10

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Chapter 8 - Circles Exercise 8.50

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 3

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

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

AP = PQ    ....(1)

and OA= OP + PA2

      (15)= (9)+ AP2

       AP= 225 - 81

             = 144

       AP = 12

AP + AQ = 2AP

             = 24 cm

So, the correct option is (c).

Question 4

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

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Chapter 8 - Circles Exercise 8.51

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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

Solution 1

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).

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

In figure, if AP = PB, then

(a) AC = AB

(b) AC = BC

(c) AQ = QC

(d) AB = BC

Solution 2

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).

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

In figure, if AP = 10 cm, then BP = 

 Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 3

AP = 10 cm

AO = 6 cm

OB = 3 cm

AP2 + OA= OP2

OP= 102 + 62

OP= 136

Also OB+ BP2 = OP

        32 + BP= 136

        BP2 = 136 - 9

        Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

So, the correct option is (b).

Question 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Chapter 8 - Circles Exercise 8.52

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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

(a) PQ

(b) QR

(c) PR

(d) PS

Solution 1

PA = PD

AQ = QB

and PQ = PA + AQ

PQ = PD + QB

Hence PD + QB = PQ

So, the correct option is (a).

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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

Solution 2

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).

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Chapter 8 - Circles Exercise 8.53

Question 1

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

In figure, PR =

(a) 20 cm

(b) 26 cm

(c) 24 cm

(d) 28 cm

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

radius of circle 1 = 3 cm

radius of circle 2 = 5  cm

OP= OQ+ QP        and    O'S+ SR2 = O'R

OP= 4+ 32                         O'R= 5+ 122     

      = 16 + 9                          O'R2 = 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).

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - CirclesRd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 3

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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

Solution 3

OB = OC = OA = 5 cm

OQ = OP = 3 cm

OB= OQ+ BQ2

BQ2 = OB- OQ2

      = 5- 32

      = 16

BQ = 4 cm

also BQ = BP

BP = 4 cm

In ΔOPC,

OP2 + PC2 = OC2

PC = OC2 - OP2

       = 5- 32

       = 16

PC = 4 cm

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

So, the correct option is (c).

Question 4

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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

Solution 4

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

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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

(a) 11 cm

(b) 7 cm

(c) 5 cm

(d) 3 cm

Solution 1

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).

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - CirclesRd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Chapter 8 - Circles Exercise 8.55

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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

Solution 1

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).

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - CirclesRd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 2

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).

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - CirclesRd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Chapter 8 - Circles Exercise 8.56

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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).

Question 2

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

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

AB = 5 cm

BC = 12 cm

AB + BC2 = AC2

AC= 52 + 122

      = 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).

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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

Solution 5

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

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

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

Solution 1

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).

Question 2

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°

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 3

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

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles 

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 4

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

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 5

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

Solution 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 6

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 cm2

b. 65 cm2

c. 30 cm2

d. 32.5 cm2

Solution 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 7

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°

Solution 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 8

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

Solution 8

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Chapter 8 - Circles Exercise 8.58

Question 1

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

a. 45°

b. 35°

c. 55°

d. 62°

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 2

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°, thenRd-sharma Solutions Cbse Class 10 Mathematics Chapter - CirclesAQB is equal to

a. 20°

b. 40°

c. 35°

d. 45°

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Chapter 8 - Circles Exercise Ex. 8.1

Question 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 ....... .

Solution 1

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 90o .

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Chapter 8 - Circles Exercise Ex. 8.2

Question 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 1

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 3

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 4

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

Solution 4

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 5

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 6

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.

Solution 6

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 7

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.

Solution 7

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 8

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

Solution 8

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 9

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

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 9

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 10

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 10

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 11

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


Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 11

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 12

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 12

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 13

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. 

Solution 13

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 14

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.

Solution 14

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 15

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.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 15

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 16

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.

Solution 16

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 17

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 17

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 18

Two tangent segments PA and PB are drawn to a circle with centre O such that Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - CirclesAPB = 120o. Prove that OP = 2 AP.

Solution 18

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 19

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 19

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 20

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

Solution 20

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles 

 

 

 

 

 

 

 

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 21

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.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 21

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 22

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 22

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 23

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 23

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 24

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.

Solution 24

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

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. AO2 = AP2 + PO2

132 = AP2 + 52

AP2 = 144

AP = 12

Perimeter of ∆ABC = 24 cm

Question 25

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.

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Solution 25

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 26

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.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 26

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 27

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.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 27

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 28

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.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 28

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 29

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.

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 29

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 30

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 30

 

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles



Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 31

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.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles 

Solution 31

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°

2PQR = 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° 

Question 32

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

Solution 32

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 33

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 33

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 34

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 34

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 35

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.

Solution 35

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 36

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

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 36

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 37

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.

Solution 37

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

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

Hence, by applying the Pythagoras Theorem, we have,

OA2 = OB2 + AB2

152 = 92 + AB2

AB2 = 152 - 92

AB2 = 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.

Question 38

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

Solution 38

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 39

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 cm2, find the sides PQ and PR.

Solution 39

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

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

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

= x + 30

 

Area of PQR

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Area of PQR = 336 cm2

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Side PR = (12 + 16) = 28 cm

Side PQ = (12 + 14) = 26 cm

Question 40

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.

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Solution 40

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 41

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.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 41

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 42

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 cm2.

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles 

 

Solution 42

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

 

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles 

 

 

Tangents drawn from an external point to a circle are equal

AM=AN

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

 

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 43

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

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles 

Solution 43

AOQ=58° (given)

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

In right BAT,

ABT + BAT + ATB=180°

29° + 90° + ATB=180° 

ATB = 61° 

that is, ATQ = 61° 

Question 44

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

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Solution 44

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 45

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 45

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 46

In Fig., BC is a tangent to the circle with centre O. OE bisects AP. Prove that ∆AEO ~ ∆ABCRd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 46

In ∆AOP,

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

In an isosceles triangle,the median drawnOEA = 90o

In ∆AOE and ∆ABC,

ABC = OEA = 90o

A is common.

∆AEO ~ ∆ABC…(AA test)

Question 47

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.

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 47

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles
Question 48

In fig., O is the centre of the circle and BCD is tangent to it at C. Prove that Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - CirclesBAC + Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - CirclesACD = 90o.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Solution 48

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 49

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

Solution 49

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

 

 

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles

Question 50

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.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles 

Solution 50

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

Question 51

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.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles 

Solution 51

In DPQR,

POR is an external angle.

So,

POR = PQO + OPQ

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

PQO = 90o

130˚ = 90˚ + OPQ

OPQ = 40o

1 = 40o

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˚

Question 52

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.

Rd-sharma Solutions Cbse Class 10 Mathematics Chapter - Circles 

Solution 52

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)