Table of Contents
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 90o .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 = 120o. 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 = 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.
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 = 90o.
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 = 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.
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
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