RD Sharma Solutions for CBSE Class 10 Mathematics chapter 8 - Circles

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Chapter 8 - Circles Excercise Ex. 8.2

Question 1

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 1

 

 

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 2

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

Solution 2

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 3

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.

  

Solution 3

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 4

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.

  

Solution 4

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 5

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.

  

Solution 5

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)

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

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

Solution 9

Question 10

Solution 10

Question 11

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 11

Question 12

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 12

 

Question 13

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

Solution 13

 

Question 14

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

Solution 14



Question 15

Solution 15

Question 16

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


Solution 16



Question 17

Solution 17

Question 18

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 18

Question 19

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 19

Question 20

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.

Solution 20



Question 21

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 21

Question 22

Solution 22

Question 23

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

Solution 23

Question 24

Solution 24

Question 25

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 25

  

 

 

 

 

 

 

 

 

Question 26

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.

Solution 26


Question 27

Solution 27

Question 28


Solution 28



Question 29

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.

 

 

Solution 29

Question 30

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.

Solution 30

Question 31

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.

Solution 31

Question 32

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.

Solution 32

Question 33

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.

 

Solution 33

Question 34

Solution 34

 

 



Question 35

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.

  

Solution 35

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 36

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

Solution 36

 

Question 37

Solution 37

Question 38

Solution 38

Question 39

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 39

 

 

Question 40

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

Solution 40



Question 41

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 41

 

 

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 42

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

Solution 42

 

Question 43

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 43

 

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 cm2

 

Side PR = (12 + 16) = 28 cm

Side PQ = (12 + 14) = 26 cm

Question 44

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.

 

 

Solution 44

Question 45

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.

Solution 45

 

 

Question 46

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.

 

  

 

Solution 46

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

 

 

Question 47

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

  

Solution 47

AOQ=58° (given)

In right BAT,

ABT + BAT + ATB=180°

29° + 90° + ATB=180° 

ATB = 61° 

that is, ATQ = 61° 

Question 48

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

Solution 48

Question 49

Solution 49

Question 50

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.

 

Solution 50

Question 51

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

Solution 51

Question 52

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

Solution 52

 

 

Chapter 8 - Circles Excercise 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

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 8 - Circles Excercise 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) begin mathsize 12px style square root of 119 cm end style

Solution 1

radius = 5 cm

So, OP = 5 cm

OQ = 12 cm

begin mathsize 12px style so space in space triangle space OPQ
OP squared space plus space PQ squared space equals space OQ squared
PQ squared space equals space OQ squared space minus space OP squared
space space space space space space space space space space equals space 12 squared space minus space 5 squared
space space space space space space space space space space equals space 119
PQ space equals space square root of 119 end style

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

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) begin mathsize 12px style square root of 7 space cm end style

(b) 7 cm

(c) 5 cm

(d) 25 cm

Solution 3

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

begin mathsize 12px style If space tangents space PA space and space PB space from space straight a space point space straight P space to space straight a space circle space with space centre space straight O space are space inclined space to space each space other space at space an space angle space of space 80 degree space then space angle POA space is space equal space to
left parenthesis straight a right parenthesis space 50 degree
open parentheses straight b close parentheses space 60 degree
open parentheses straight c close parentheses space 70 degree
open parentheses straight d close parentheses space 80 degree end style

Solution 4

begin mathsize 12px style We space know space tangents space from space same space point space to space circle space are space inclined space at space same space angle.
Hnece space angle APO space equals space angle BPO
space space space space space space space space space space space space space angle APO space equals space 1 half angle straight P
space space space space space space space space space space space space space space angle APO space equals space 40 degree
In space triangle OAP
angle straight A space plus space angle POA space plus space angle APO space equals space 180 degree
space space space space space space space space space space space space space angle POA space equals space 180 degree space minus space 90 degree space minus space 40 degree
space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 50 degree
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Chapter 8 - Circles Excercise 8.49

Question 1

begin mathsize 12px style If space TP space and space TQ space are space two space tangents space to space straight a space circle space with space centre space straight O space so space that space angle POQ space equals space 110 degree comma space then comma space angle PTQ space is space equal space to
open parentheses straight a close parentheses space 60 degree
open parentheses straight b close parentheses space 70 degree
open parentheses straight c close parentheses space 80 degree
open parentheses straight d close parentheses space 90 degree end style

Solution 1

begin mathsize 12px style PTQO space is space straight a space quadrilateral space sum space of space all space angles space equals space 360 degree
angle straight P space plus space angle straight Q space plus space angle POQ space plus space angle PTQ space equals space 360 degree
90 degree space plus space 90 degree space plus space 110 degree space plus space angle PTQ space equals space 360 degree
angle PTQ space plus space 290 degree space equals space 360 degree
angle PTQ space equals space 70 degree
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 2

begin mathsize 12px style PQ space is space straight a space tangent space to space straight a space circle space with space centre space straight O space at space the space point space straight P. space If space triangle OPQ space is space an space isosceles space triangle comma space then space angle OQP space is space equal space to
open parentheses straight a close parentheses space 30 degree
open parentheses straight b close parentheses space 45 degree
open parentheses straight c close parentheses space 60 degree
open parentheses straight d close parentheses space 90 degree end style

Solution 2

begin mathsize 12px style triangle OPQ space is space an space isoceles space triangle
so comma space OP space equals space PQ
and space angle straight P space equals space 90 degree
and space angle straight O space equals space angle straight Q
also space angle straight P space plus space angle straight O space plus space angle straight Q space equals space 180 degree
2 angle straight Q space equals space 180 degree space minus space 90 degree
angle straight Q space equals space 45 degree
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 3

begin mathsize 12px style Two space equal space circles space touch space each space other space externally space at space straight C space and space AB space is space straight a space common space tangent space to space the space circles. space Then comma space angle ACB space equals
left parenthesis straight a right parenthesis space 60 degree
left parenthesis straight b right parenthesis space 45 degree
left parenthesis straight c right parenthesis space 30 degree
left parenthesis straight d right parenthesis space 90 degree end style

Solution 3

begin mathsize 12px style AO subscript 1 space equals space straight O subscript 1 straight C space and space straight O subscript 2 straight B space equals space CO subscript 2
so space angle CBO subscript 2 space equals space angle BCO subscript 2 space equals space 45 degree space space space space.... open parentheses 1 close parentheses
and space angle CAO subscript 1 space equals space angle straight O subscript 1 CA space equals space 45 degree space space space.... open parentheses 2 close parentheses
also space angle straight O subscript 1 CO subscript 2 space equals space 180 degree space space space space space space space space space space space..... open parentheses 3 close parentheses
so space from space open parentheses 1 close parentheses comma space open parentheses 2 close parentheses comma space open parentheses 3 close parentheses
angle straight O subscript 1 CA space plus space angle straight O subscript 2 CB space plus space angle ACB space equals space 180 degree
45 degree space plus space 45 degree space plus space angle ACB space equals space 180 degree
angle ACB space equals space 90 degree
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 4

begin mathsize 12px style ABC space is space straight a space right space angled space triangled comma space right space angled space at space straight B space such space that space BC space equals space 6 space cm space and space AB space equals space 8 space cm. space straight A space circle space with space centre space straight O space is space inscribed space in space triangle ABC. space The space radius space of space the space circle space is
open parentheses straight a close parentheses space 1 space cm
open parentheses straight b close parentheses space 2 space cm
open parentheses straight c close parentheses space 3 space cm
open parentheses straight d close parentheses space 4 space cm end style

Solution 4

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

begin mathsize 12px style PQ space is space straight a space tangent space drawn space from space point space straight P space to space straight a space circle space with space centre space straight O space and space QOR space is space straight a space diameter space of space the space circle space such space that space angle POR space equals space 120 degree comma space then space angle OPQ space is
open parentheses straight a close parentheses space 60 degree
open parentheses straight b close parentheses space 45 degree
left parenthesis straight c right parenthesis space 30 degree
open parentheses straight d close parentheses space 90 degree space end style

Solution 5

begin mathsize 12px style angle POR space equals space 120 degree
so space angle POQ space equals space 180 degree space minus space angle POR
space space space space space space space space space space space space space space space space space space space space equals space 60 degree
angle OPQ space plus space angle straight Q space plus space angle POQ space equals space 180 degree
angle OPQ space plus space 90 degree space plus space 60 degree space equals space 180 degree
angle OPQ space equals space 30 degree
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

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

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) begin mathsize 12px style square root of 7 cm end style

(b) begin mathsize 12px style 2 square root of 7 cm end style

(c) 10 cm

(d) 5 cm

Solution 7

Given OQ = 8 cm

         OP = 6 cm

OP+ PQ2 = OQ2

6+ PQ2 = 82

       PQ= 64 - 36

             = 28

PQ = begin mathsize 12px style 2 square root of 7 end style

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

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

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

 

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

begin mathsize 12px style SQ space is space diameter
and space SQ space equals space 6 space cm
so space OQ space equals space SQ over 2 space equals space 3 space cm
and space QR equals space 4 space cm
OR squared space equals space OQ squared space plus space QR squared
space space space space space space space space space space equals space 4 squared space plus space 3 squared
space space space space space space space space space space equals space 25
OR space equals space 5 space cm
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Chapter 8 - Circles Excercise 8.50

Question 1

begin mathsize 12px style In space figure comma space the space perimeter space of space triangle ABC space is
open parentheses straight a close parentheses space 30 space cm
left parenthesis straight b right parenthesis space 60 space cm
left parenthesis straight c right parenthesis space 45 space cm
left parenthesis straight d right parenthesis space 15 space cm end style

Solution 1

begin mathsize 12px style AQ space equals space 4 space cm space space
PC space equals space 5 space cm space space
BR space equals space 6 space cm space space
AQ space and space AR space are space tangent space to space circle space from space same space point. space space
Hence space AQ space equals space AR space equals space 4 space cm space space space..... left parenthesis 1 right parenthesis space space
similarly space space BR space equals space BP space equals space 6 space cm space space space space space..... left parenthesis 2 right parenthesis space space
and space PC space equals space CQ space equals space 5 space cm space space space space...... left parenthesis 3 right parenthesis space space
perimeter space of space triangle ABC space equals space AB space plus space BC space plus space CA
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space left parenthesis AR space plus space RB right parenthesis space plus space left parenthesis BP space plus space PC right parenthesis space plus space open parentheses CQ space plus space QA close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses AQ space plus space AR close parentheses space plus space open parentheses RB space plus space BP close parentheses space plus space open parentheses CP space plus space CQ close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses 4 space plus space 4 close parentheses space plus space open parentheses 6 space plus space 6 close parentheses space plus space open parentheses 5 space plus space 5 close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 8 space plus space 12 space plus space 10
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 30 space cm
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 2

begin mathsize 12px style In space figure comma space AP space is space straight a space tangent space to space the space circle space with space centre space straight O space such space that space OP space equals space 4 space cm space and space angle OPA space equals space 30 degree. space Then comma space AP space equals
open parentheses straight a close parentheses space 2 square root of 2 space cm
open parentheses straight b close parentheses space 2 space cm
open parentheses straight c close parentheses space 2 square root of 3 space cm
open parentheses straight d close parentheses space 3 square root of 2 space cm end style

Solution 2

begin mathsize 12px style OP space equals space 4 space cm
AP space is space tangent space to space circle
So space OA space perpendicular space AP
In space triangle OAB
cos space 30 degree space equals space AP over OP
AP space equals space OP space cos space 30 degree
space space space space space space space equals space 4 space cross times space fraction numerator square root of 3 over denominator 2 end fraction
space space space space space space space equals space 2 square root of 3 space cm
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

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

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

begin mathsize 12px style radius space equals space 5 space cm
OP space perpendicular space XY space and space XY space parallel to space AB
so space OQ space perpendicular space AB
and space PE space equals space 8 space cm
also space AB space equals space AE space plus space EB
space space space space space space space space space space space space space space space space equals space 2 AE space space space space space space space space space space space space space space space space space space open curly brackets AE space equals space EB close curly brackets
PE space equals space 8 space cm
OP space equals space 5 space cm
OE space equals space 3 space cm space and space OA space equals space 5 space cm
OA squared space equals space OE squared space plus space AE squared
AE squared space equals space 5 squared space minus space 3 squared space space space space rightwards double arrow space AE space equals space 4 space cm
Hence space AB space equals space 8 space cm
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 5

begin mathsize 12px style If space PT space is space tangent space drawn space from space straight a space point space straight P space to space straight a space circle space touching space it space at space straight T space and space straight O space is space the space centre space of space the space circle comma space then space angle OPT space plus space angle POT space equals
open parentheses straight a close parentheses space 30 degree
open parentheses straight b close parentheses space 60 degree
open parentheses straight c close parentheses space 90 degree
open parentheses straight d close parentheses space 180 degree end style

Solution 5

begin mathsize 12px style In space triangle OPT
angle POT space plus space angle OPT space plus space angle straight T space equals space 180 degree
angle POT space plus space angle OPT space equals space 180 degree space minus space angle straight T
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 180 degree space minus space 90 degree
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 90 degree
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Chapter 8 - Circles Excercise 8.51

Question 1

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

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

In figure, if AP = 10 cm, then BP = 

 begin mathsize 12px style open parentheses straight a close parentheses space square root of 91 space cm
open parentheses straight b close parentheses space square root of 127 space cm
open parentheses straight c close parentheses space square root of 119 space cm
open parentheses straight d close parentheses space square root of 109 space cm end style

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

        begin mathsize 12px style BP space equals space square root of 127 end style

So, the correct option is (b).

Question 4

begin mathsize 12px style In space Figure comma space if space PR space is space tangent space to space the space circle space at space straight P space and space straight Q space is space the space centre space of space the space circle comma space then space angle POQ space equals
open parentheses straight a close parentheses space 110 degree
open parentheses straight b close parentheses space 100 degree
open parentheses straight c close parentheses space 120 degree
open parentheses straight d close parentheses space 90 degree end style

Solution 4

begin mathsize 12px style angle RPO space equals space 90 degree
given space angle RPQ space equals space 60 degree
so space angle QPO space equals space angle RPO space minus space angle RPQ
space space space space space space space space space space space space space space space space space space space space space equals space 90 degree space minus space 60 degree
space space space space space space space space space space space space space space space space space space space space space equals space 30 degree
In space triangle OPQ comma space OP comma space is space equal space to space OQ
Hence space angle OPQ space equals space angle OQP... left parenthesis since space OP space equals space OQ space which space are space radii space of space the space same space circle right parenthesis
space space space space space space space space space space space space space angle OPQ space equals space angle OQP space equals space 30 degree
In space triangle OPQ
angle straight O space plus space angle OPQ space plus space angle OQP space equals space 180 degree
angle POQ space equals space 180 degree space minus space 30 degree space minus space 30 degree
space space space space space space space space space space space space space space space equals space 120 degree
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Chapter 8 - Circles Excercise 8.52

Question 1

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

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

begin mathsize 12px style In space Figure comma space APB space is space straight a space tangent space to space straight a space circle space with space centre space straight O space at space point space straight P. space If space angle QPB space equals space 50 degree comma space then space the space measure space of space angle POQ space is
left parenthesis straight a right parenthesis space 100 degree
left parenthesis straight b right parenthesis space 120 degree
left parenthesis straight c right parenthesis space 140 degree
open parentheses straight d close parentheses space 150 degree end style

Solution 3

begin mathsize 12px style APB space is space tangent
so space OP space perpendicular space APB
Hence space angle OPB space equals space 90 degree
angle OPQ space equals space 90 degree space minus space 50 degree
space space space space space space space space space space space space space space space equals space 40 degree
In space triangle OPQ
OP space equals space OQ
Hence space angle OPQ space equals space angle OQP space equals space 40 degree
so space angle POQ space equals space 180 degree space minus space angle OPQ space minus space angle OQP
space space space space space space space space space space space space space space space space space space space space space equals space 180 degree space minus space 40 degree space minus space 40 degree
space space space space space space space space space space space space space space space space space space space space space equals space 100 degree
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 4

begin mathsize 12px style In space figure comma space if space tangents space PA space and space PB space are space drawn space to space straight a space circle space such space that space angle APB space equals space 30 degree space and space chord space AC space is space drawn space parallel space to space the space tangent space PB comma space then space angle ABC space equals
open parentheses straight a close parentheses space 60 degree
open parentheses straight b close parentheses space 90 degree
open parentheses straight c close parentheses space 30 degree
open parentheses straight d close parentheses space None space of space these end style

Solution 4

begin mathsize 12px style AC space parallel to space PB
and space PA space equals space PB
In space triangle PAB comma
angle PAB space equals space angle PBA
angle PAB space plus space angle PBA space plus space angle APB space equals space 180 degree
angle PAB space equals space 75 degree
AC space parallel to space BP
By space alternative space interior space angle space property
angle CAB space equals space angle ABP space equals space 75 degree
OB space perpendicular space BP comma
Hence space angle OBA space equals space angle OBP space minus space angle ABP
space space space space space space space space space space space space space space space angle OBA space equals space angle OAB space equals space 15 degree
so space angle AOB space equals space 180 degree space minus space angle OAB space minus space angle OBA space equals space 150 degree
We space also space know space that space the space angle space made space by space any space chord space at space the space centre space is space twice space the space angle space made space by space same space chord space on space the space space circle.
Hence space angle AOB space equals space 2 angle ACB
space space space space space space space space space space space space space space space angle ACB space equals space 1 half cross times 150 degree
space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 75 degree
In space triangle ABC comma space angle ACB space equals space 75 degree space and space angle CAB space equals space 75 degree
angle ABC space equals space 180 degree space minus space angle ACB space minus space angle CAB
space space space space space space space space space space space space space equals space 30 degree
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Chapter 8 - Circles Excercise 8.53

Question 1

 

In figure, PR =

(a) 20 cm

(b) 26 cm

(c) 24 cm

(d) 28 cm

Solution 1

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

begin mathsize 12px style Two space circles space of space same space radii space straight r space and space centres space straight O space and space straight O apostrophe space touch space each space other space at space straight P space as space shown space in space figure. space If space OO apostrophe space is space produced space to space meet space the space circle space straight C open parentheses straight O apostrophe comma space straight r close parentheses space at space straight A space and space AT space is space straight a space tangent space to space the space circle space straight C left parenthesis straight O comma space straight r right parenthesis space such space that space straight O apostrophe straight Q space perpendicular space AT. space Then space AO space colon space AO apostrophe space equals
open parentheses straight a close parentheses space 3 divided by 2
open parentheses straight b close parentheses space 2
open parentheses straight c close parentheses space 3
open parentheses straight d close parentheses space 1 divided by 4 end style

Solution 2

begin mathsize 12px style AO space equals space AO apostrophe space plus space straight O apostrophe straight P space plus space PO
space space space space space space space space equals space straight r space plus space straight r space plus space straight r
space space space space space space space space equals space 3 straight r
AO apostrophe space equals space straight r
AO space colon space AO apostrophe space equals space fraction numerator 3 straight r over denominator straight r end fraction space equals space 3 space
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 3

 

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

 

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 Excercise 8.54

Question 1

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

begin mathsize 12px style In space figure comma space PQ space and space PR space are space tangents space drawn space from space straight P space to space straight a space circle space with space centre space straight O. space If space angle OPQ space equals space 35 degree comma space then
open parentheses straight a close parentheses space straight a space equals space 30 degree comma space straight b space equals space 60 degree
left parenthesis straight b right parenthesis space straight a space equals space 35 degree comma space straight b space equals space 55 degree
open parentheses straight c close parentheses space straight a space equals space 40 degree comma space straight b space equals space 50 degree
open parentheses straight d close parentheses space straight a space equals space 45 degree comma space straight b space equals space 45 degree end style

Solution 2

begin mathsize 12px style In space triangle space OPQ
angle OQP space equals space 90 degree
angle OPQ space equals space 35 degree
also comma space angle OQP space plus space angle OPQ space plus space straight b space equals space 180 degree
straight b space equals space 180 degree space minus space 90 degree space minus space 35 degree
space space space space equals space 55 degree
We space also space know comma space that space if space two space tangent space are space drawn space from space same space point space to space circle comma space touches space at space straight Q comma space straight R space then space angle QPR space is space bisected space by space OP.
Hence space angle OPQ space equals space angle OPR
space space space space space space space space space space space space space space space space space space space space 35 degree space equals space straight a
straight a space equals space 35 degree space and space straight b space equals space 55 degree
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 3

begin mathsize 12px style In space figure comma space if space TP space and space TQ space are space tangents space drawn space from space an space external space point space straight T space to space straight a space circle space with space centre space straight O space such space that space angle TQ straight P space equals space 60 degree comma space then space angle OPQ space equals
open parentheses straight a close parentheses space 25 degree
open parentheses straight b close parentheses space 30 degree
open parentheses straight c close parentheses space 40 degree
open parentheses straight d close parentheses space 60 degree end style

Solution 3

begin mathsize 12px style angle OQT space equals space 90 degree space as space OQ space perpendicular space to space TQ
given space angle PQT space equals space 60 degree
Hence space angle OQP space equals space angle OQT space minus space angle PQT
space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 30 degree
increment OQP space is space isoceles space with space OP space equals space OQ
Hence comma space angle OQP space equals space angle OPQ space equals space 30 degree
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Chapter 8 - Circles Excercise 8.55

Question 1

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

begin mathsize 12px style In space figure comma space straight a space circle space touches space the space side space DF space of space triangle DEF space at space straight H space and space touches space ED space and space EF space produced space at space straight K space and space straight M space respectively. space If space EK space equals space 9 space cm comma space then space the space perimeter space of space triangle EDF space is
open parentheses straight a close parentheses space 18 space cm
open parentheses straight b close parentheses space 13.5 space cm
open parentheses straight c close parentheses space 12 space cm
open parentheses straight d close parentheses space 9 space cm end style

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

begin mathsize 12px style In space figure comma space DE space and space DF space are space tangents space from space an space external space point space straight D space to space straight a space circle space with space centre space straight A. space If space DE space equals space 5 space cm space and space DE space perpendicular space DF comma space then space the space radius space of space the space circle space is
open parentheses straight a close parentheses space 3 space cm
open parentheses straight b close parentheses space 5 space cm
open parentheses straight c close parentheses space 4 space cm
open parentheses straight d close parentheses space 6 space cm end style

Solution 3

begin mathsize 12px style DE space equals space DF space equals space 5 space cm
We space know space AD space bisects space the space angle space EDF space
since space straight D space is space the space intersection space of space the space two space tangents space and space AD space is space the space line space segment space joining space the
centre space to space the space point space of space intersection space of space the space two space tangents.
rightwards double arrow angle EDA space equals space 45 degree
Hence space in space triangle AED comma
AE space is space radius space of space circle
Also space tan 45 degree space equals space AE over ED
AE space equals space ED space tan 45 degree
space space space space space space equals space ED
space space space space space space equals space 5 space cm space
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Chapter 8 - Circles Excercise 8.56

Question 1

begin mathsize 12px style In space figure comma space straight a space circle space with space centre space straight O space is space inscribed space in space straight a space quadrilateral space ABCD space such space that comma space it space touches space sides space BC comma space AB comma space AD space and space CD space at space points space straight P comma space straight Q comma space straight R space and space straight S space respectively. space If space AB space equals space 29 space cm comma space AD space equals space 23 space cm comma space angle straight B space equals space 90 degree space and space DS space equals space 5 space cm comma space then space the space radius space of space the space circle space left parenthesis in space cm right parenthesis space is
open parentheses straight a close parentheses space 11
open parentheses straight b close parentheses space 18
open parentheses straight c close parentheses space 6
open parentheses straight d close parentheses space 15 end style

Solution 1

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

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

begin mathsize 12px style Two space circles space touch space each space other space externally space at space straight P. space AB space is space straight a space common space tangent space to space the space circle space touching space them space at space straight A space and space straight B. space The space value space of space angle APB space is
left parenthesis straight a right parenthesis space 30 degree
left parenthesis straight b right parenthesis space 45 degree
left parenthesis straight c right parenthesis space 60 degree
open parentheses straight d close parentheses space 90 degree end style

Solution 3

begin mathsize 12px style Let space angle OPA space equals space straight theta
We space know space angle OPA space equals space angle OAP
So comma
angle AOP space equals space 180 degree space minus space open parentheses angle OPA space plus space angle OAP close parentheses
angle AOP space equals space 180 degree space minus space 2 straight theta space space space space space.... open parentheses 1 close parentheses
Let space angle straight O apostrophe PB space equals space straight alpha
Hence space angle straight O apostrophe BP space equals space straight alpha
So comma space angle BO apostrophe straight P space equals space 180 degree space minus space open parentheses angle straight O apostrophe PB space plus space angle straight O apostrophe BP close parentheses
angle BO apostrophe straight P space equals space 180 degree space minus space 2 straight alpha space space space space space space space space.... open parentheses 2 close parentheses
In space quadrilateral space ABO apostrophe straight O comma
sum space of space all space angles space equals space 360 degree
angle BAO space plus space angle ABO apostrophe space plus space angle AOO apostrophe space plus space angle BO apostrophe straight O space equals space 360 degree
90 degree space plus space 90 degree space plus space 180 space minus space 2 straight theta space plus space 180 space minus space 2 straight alpha space equals space 360 degree
540 degree space minus space 2 open parentheses straight alpha space plus space straight theta close parentheses
straight alpha space plus space straight theta space equals space 90 degree space space space space... open parentheses 3 close parentheses
Also space angle OPO apostrophe space equals space 180 degree
angle APO space plus space angle APB space plus space angle BPO apostrophe space equals space 180 degree
space straight theta plus space angle APB space plus straight alpha space space equals space 180 degree
from space left parenthesis 3 right parenthesis space
angle APB space equals space 180 degree space minus space 90 degree
space space space space space space space space space space space space space space equals space 90 degree
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 4

begin mathsize 12px style In space figure comma space PQ space and space PR space are space two space tangents space to space straight a space circle space with space centre space straight O. space If space angle QPR space equals space 46 degree comma space then space angle QOR space equals
open parentheses straight a close parentheses space 67 degree
open parentheses straight b close parentheses space 134 degree
open parentheses straight c close parentheses space 44 degree
open parentheses straight d close parentheses space 46 space degree end style

Solution 4

begin mathsize 12px style angle ORP space equals space angle OQP space equals space 90 degree
Also space sum space of space all space angles space of space quadrilateral space PQOR space is space 360 degree
angle ORP space plus space angle OQP space plus space angle QOR space plus space angle QPR space equals space 360 degree
90 degree space plus space 90 degree space plus space angle QOR space plus space 46 degree space equals space 360 degree
angle QOR space equals space 360 degree space minus space open parentheses 180 degree space plus space 46 degree close parentheses
angle QOR space equals space 134 degree
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 5

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 Excercise 8.57

Question 1

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

 

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

Solution 3

  

 

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

 

 

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

 

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

 

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

 

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

 

Chapter 8 - Circles Excercise 8.58

Question 1

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

a. 45°

b. 35°

c. 55°

d. 62°

Solution 1

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°, then AQB is equal to

a. 20°

b. 40°

c. 35°

d. 45°

 

Solution 2