RD SHARMA SOLUTION CHAPTER- 10 Circles| CLASS 10TH MATHEMATICS-EDUGROWN

Chapter 10 – Circles Exercise Ex. 10.1

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 90^o .Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 10 – Circles Exercise Ex. 10.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

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

Question 5

Solution 5

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

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

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

Question 9

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

Solution 9

Question 10

Solution 10

Question 11

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

Solution 11

Question 12

Solution 12

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

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

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.

Solution 15

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

Question 17

Solution 17

Question 18

Two tangent segments PA and PB are drawn to a circle with centre O such that APB = 120^o. Prove that OP = 2 AP.Solution 18

Question 19

Solution 19

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 =BDSolution 20

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.

Solution 21

Question 22

Solution 22

Question 23

Solution 23

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

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 cmQuestion 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.

Solution 25

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.

Solution 26

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.

Solution 27

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.

Solution 28

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.

Solution 29

Question 30

Solution 30

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.

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°

⇒2∠PQR = 150°

⇒∠PQR = 75°

⇒∠SRQ =∠PQR = 75° (alternate angles)

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

In ∆RSQ,

∠RSQ + ∠SRQ + ∠RQS = 180°

⇒75° + 75° + ∠RQS = 180°

⇒ ∠RQS =30° 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

Question 33

Solution 33

Question 34

Solution 34

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

Question 36

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

Solution 36

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

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

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

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 cmQuestion 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.

Solution 40

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.

Solution 41

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.

Solution 42

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 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.

Solution 43

∠AOQ=58° (given)

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.

Solution 44

Question 45

Solution 45

Question 46

In Fig., BC is a tangent to the circle with centre O. OE bisects AP. Prove that ∆AEO ~ ∆ABCSolution 46

In ∆AOP,

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

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

In ∆AOE and ∆ABC,

∠ABC = ∠OEA = 90^o

∠A is common.

∆AEO ~ ∆ABC…(AA test)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.

Solution 47

Question 48

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

Solution 48

Question 49

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.Solution 49

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.

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 cmQuestion 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.

Solution 51

In DPQR,

∠POR is an external angle.

So,

∠POR = ∠PQO + ∠OPQ

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

∠PQO = 90^o

130˚ = 90˚ + ∠OPQ

∠OPQ = 40^o

∠1 = 40^o

Now,

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

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

∠2 = 65˚

∠1 + ∠2 =105^˚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.

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)

Chapter 10 – Circles Exercise 10.48

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) Solution 1

radius = 5 cm

So, OP = 5 cm

OQ = 12 cm

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 cmSolution 2

PQ is a tangent to the circle

So, OP² + PQ² = OQ²

OP² = OQ² – PQ²

      = (25)² – (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) 

(b) 7 cm

(c) 5 cm

(d) 25 cmSolution 3

Given OP = 3 cm

         PA = 4 cm

Hence, OA² = OP² + PA²

OA² = 3² + 4²

      = 25

OA = 5 cm

So, the correct option is (c).Question 4

Solution 4

Chapter 10 – Circles Exercise 10.49

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

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 AC² = AB² + BC²

             = 8² + 6² 

             = 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 9

Solution 9

Question 10

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 + DBSolution 10

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 11

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

(a) 

(b) 

(c) 10 cm

(d) 5 cmSolution 11

Given OQ = 8 cm

         OP = 6 cm

OP² + PQ² = OQ²

6² + PQ² = 8²

       PQ² = 64 – 36

             = 28

PQ = 

So, the correct option is (b).Question 12

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 cmSolution 12

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 13

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 + CASolution 13

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 14

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 cmSolution 14

Chapter 10 – Circles Exercise 10.50

Question 15

Solution 15

Question 16

Solution 16

Question 17

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 cmSolution 17

AP = PQ    ….(1)

and OA² = OP²  + PA²

      (15)² = (9)² + AP²

       AP² = 225 – 81

             = 144

       AP = 12

AP + AQ = 2AP

             = 24 cm

So, the correct option is (c).Question 18

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 cmSolution 18

Question 19

Solution 19

Chapter 10 – Circles Exercise 10.51

Question 20

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 cmSolution 20

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 21

In figure, if AP = PB, then

(a) AC = AB

(b) AC = BC

(c) AQ = QC

(d) AB = BCSolution 21

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 22

In figure, if AP = 10 cm, then BP = 

Solution 22

AP = 10 cm

AO = 6 cm

OB = 3 cm

AP² + OA² = OP²

OP² = 10² + 6²

OP² = 136

Also OB² + BP² = OP² 

        3² + BP² = 136

        BP² = 136 – 9

So, the correct option is (b).Question 23

Solution 23

Chapter 10 – Circles Exercise 10.52

Question 24

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

(a) PQ

(b) QR

(c) PR

(d) PSSolution 24

PA = PD

AQ = QB

and PQ = PA + AQ

PQ = PD + QB

Hence PD + QB = PQ

So, the correct option is (a).Question 25

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 cmSolution 25

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 26

Solution 26

Question 27

Solution 27

Chapter 10 – Circles Exercise 10.53

Question 28

In figure, PR =

(a) 20 cm

(b) 26 cm

(c) 24 cm

(d) 28 cmSolution 28

radius of circle 1 = 3 cm

radius of circle 2 = 5  cm

OP² = OQ² + QP²         and    O’S² + SR² = O’R² 

OP² = 4² + 3²                         O’R² = 5² + 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 29

Solution 29

Question 30

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 cmSolution 30

OB = OC = OA = 5 cm

OQ = OP = 3 cm

OB² = OQ² + BQ²

BQ² = OB² – OQ²

      = 5² – 3²

      = 16

BQ = 4 cm

also BQ = BP

BP = 4 cm

In ΔOPC,

OP² + PC² = OC²

PC²  = OC² – OP²

       = 5² – 3²

       = 16

PC = 4 cm

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

So, the correct option is (c).Question 31

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 cmSolution 31

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 10 – Circles Exercise 10.54

Question 32

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

(a) 11 cm

(b) 7 cm

(c) 5 cm

(d) 3 cmSolution 32

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 33

Solution 33

Question 34

Solution 34

Chapter 10 – Circles Exercise 10.55

Question 35

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 cmSolution 35

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 36

Solution 36

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 37

Solution 37

Chapter 10 – Circles Exercise 10.56

Question 38

Solution 38

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 39

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) 1Solution 39

AB = 5 cm

BC = 12 cm

AB²  + BC² = AC²

AC² = 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 40

Solution 40

Question 41

Solution 41

Question 42

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) 1.9Solution 42

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 10 – Circles Exercise 10.57

Question 43

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) 7Solution 43

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 44

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 44

Question 45

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 45

Question 46

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 cmSolution 46

Question 47

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 cmSolution 47

Question 48

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 cm²Solution 48

Question 49

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 49

Question 50

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

a. 10 cm

b. 7.5 cm

c. 5 cm

d. 2.5 cmSolution 50

Chapter 10 – Circles Exercise 10.58

Question 51

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

a. 45°

b. 35°

c. 55°

d. 62°

Solution 51

Question 52

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 52