ML Aggarwal Solution Class 10 Chapter 15 Circles MCQs

 MCQs

Question 1

In the given figure, O is the centre of the circle. If ∠ABC = 20°, then ∠AOC is equal to

(a) 20°

(b) 40°

(c) 60°

(d) 10°









Sol :
In the given figure,
Arc AC subtends ∠AOC at the centre
and ∠ABC at the remaining part of the circle
∠AOC = 2∠ABC = 2 × 20° = 40° (b)

Question 2

In the given figure, AB is a diameter of the circle. If AC = BC, then ∠CAB is. equal to

(a) 30°

(b) 60°

(c) 90°

(d) 45°










Sol :
In the given figure,
AB is the diameter of the circle and AC = BC
∠ACB = 90° (angle in a semi-circle)
AC = BC
A=B

But A+B=90

A=B=902=45

CAB=45

Ans (d)


Question 3

In the given figure, if ∠DAB = 60° and ∠ABD = 50° then ∠ACB is equal to

(a) 60°

(b) 50°

(c) 70°

(d) 80°










Sol :
In the given figure,
∠DAB = 60°, ∠ABD = 50°
In ∆ADB, ∆ADB = 180° – (60° + 50°)
= 180° – 110° = 70°
∠ACB = ∠ADB
(angles in the same segment) = 70°
Ans (c)

Question 4

In the given figure, O is the centre of the circle. If ∠OAB = 40°, then ∠ACB is equal to

(a) 50°

(b) 40°

(c) 60°

(d) 70°









Sol :
In the given figure, O is the centre of the circle.
In ∆OAB,
∠OAB = 40°
But ∠OBA = ∠OAB = 40°

(OA=OB radii of the same circle )

AOB=180(40+40)

=18080=100

But arc AB subtends AOB at the centre and ACB at the remaining part of the circle

ACB=12AOB=12×100=50

Ans (a)


Question 5

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140°, then ∠BAC is equal to

(a) 80°

(b) 50°

(c) 40°

(d) 30°

Sol :

ABCD is a cyclic quadrilateral,

AB is the diameter of the circle circumscribing it

∠ADC = 140°, ∠BAC = Join AC











ADC+ABC=180
(opposite angles of the cyclic quadrilateral) 
140+ABC=180

ABC=180140=40
Now in ΔABC 
ACB=90
(angle in a semi-circle)

BAC=90ABC
=9040=50

Ans (b)

Question 6

In the given figure, O is the centre of the circle. If ∠BAO = 60°, then ∠ADC is equal to

(a) 30°

(b) 45°

(c) 60°

(d) 120°












Sol :
In the given figure, O is the centre of the circle ∠BAO = 60°











InΔABO,˙OA=OB
(radii of the same circle)

ABO=BAO=60

Ext. AOC=BAO+ABO

=60+60=120

Now arc AC subtends AOC at the centre and ADC at the remaining part of the circle

AOC=2ADC2ADC=120

ADC=1202=60

Ans (c)


Question 7

In the given figure, O is the centre of the circle. If ∠AOB = 90° and ∠ABC = 30°, then ∠CAO is equal to

(a) 30°

(b) 45°

(c) 90°

(d) 60°











Sol :
In the given figure, O is the centre of the circle










In ΔAOB

AOB=90,ABC=30
In AOB,AOB=90
OA=OB (radii of the same circle)

OAB=OBA=902=45

Arc AB subtends AOB at the centre and ACB at the remaining part of the circle

ACB=12AOB=12×90=45

Now in ΔACB,ABC=30,ACB=45

BAC=180(30+45)

=18075=105

But OAB=45

∠CAO = 105° – 45° = 60° 

Ans (d)


Question 8

In the given figure, O is the centre of a circle. If the length of chord PQ is equal to the radius of the circle, then ∠PRQ is

(a) 60°

(b) 45°

(c) 30°

(d) 15°












Sol :
In the given figure, O is the centre of the circle
Chord PQ = radius of the circle
∆OPQ is an equilateral triangle
∴∠POQ = 60°
Arc PQ subtends ∠POQ at the centre and
∴∠PRQ at the remaining part of the circle
PRO=12POQ=12×60=30
Ans (c)


Question 9

In the given figure, if O is the centre of the circle then the value of x is

(a) 18°

(b) 20°

(c) 24°

(d) 36°












Sol :
In the given figure, O is the centre of the circle.
Join OA.











ADB=ACB=2x

(Angles in the same segment)

arc AB subtends AOB at the centre and ADB at the remaining part of the circle

AOB=2ADB=2×2x=4x

InΔOAB

OAB=OBA=3x(OA=OB)

Sum of angles of ΔOAB=180 3x+3x+4x=180

10x=180

x=18010=18

x=18

Ans (a)


Question 10

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

Sol :

From Q, length of tangent PQ to the circle = 24 cm

and QO = 25 cm










PQ is tangent and OP is radius 

OPPQ

Now in right ΔOPQ
OQ2=OP2+PQ2
(25)2=OP2+(24)2
OP2=252242=625576
OP2=49=(7)2OP=7 cm

∴Radius of the circle=7 cm

Ans (a)


Question 11

From a point which is at a distance of 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

Sol :

Let point P is 13 cm from O, the centre of the circle

Radius of the circle (OQ) = 5 cm

PQ and PR are tangents from P to the circle

Join OQ and OR











PQ is tangent and OQ is the radius PQ2=OP2OQ2=13252=16925
=144=(12)2
PQ=12 cm

Now area of ΔOPQ=12PQ×OQ

(12 base × height )

=12×12×5=30 cm2

area of quad. PQOR=2×30=60 cm2

(a)


Question 12

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

(a) 90°

(b) 50°

(c) 70°

(d) 40°

Sol :

Angles between two radii OA and OB = 130°

From A and B, tangents are drawn which meet at P









OA radius and AP is tangent to the circle
OAP=90
Similarly OBP=90

AOB+APB=180

130+APB=180

APB=180130=50

Ans (b)


Question 13

In the given figure, PQ and PR are tangents from P to a circle with centre O. If ∠POR = 55°, then ∠QPR is

(a) 35°

(b) 55°

(c) 70°

(d) 80°











Sol :

In the given figure,
PQ and PR are the tangents to the circle from a point P outside it
POR=55
OR is radius and PR is tangent
ORPR
In ΔOPR
OPR=9055=35
QPR=2×OPR=2×35=70
Ans (c)


Question 14

If tangents PA and PB from an exterior point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to

(a) 50°

(b) 60°

(c) 70°

(d) 100°

Sol :

Length of tangents PA and PB to the circle from a point P

outside the circle with centre O, and inclined an angle of 80°











OA is radius and AP is tangent

OAP=90 and OPA=12APB

=12×80=40

=12×80=40

Ans (a)


Question 15

In the given figure, PA and PB are tangents from point P to a circle with centre O. If the radius of the circle is 5 cm and PA ⊥ PB, then the length OP is equal to

(a) 5 cm

(b) 10 cm

(c) 7.5 cm

(d) 5√2 cm










Sol :
In the given figure,
PA and PB are tangents to the circle with centre O.
Radius of the circle is 5 cm, PA ⊥ PB.










OA is radius and PA is tangent to the circle

OAPA
APB=90

(PAPB)

APO=90×12=45

AOP=9045=45

i.e. OA=AP=5 cm

OP=OA2+PA2=52+52

=25+25=50=2×25

=2×5 cm=52 cm

Ans (d)


Question 16

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

Sol :

AB is the diameter of a circle with radius 5 cm

At A, XAY is a tangent to the circle

CD || XAY at a distance of 8 cm from A

Join OC












In right ΔOEC OE=85=3 cm
OC=5 cm
CE=OC2OE2=5232
=259=16=4 cm
CD=2×CE=2×4=8 cm

Ans (d)


Question 17

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 is

(a) 3 cm

(b) 6 cm

(c) 9 cm

(d) 1 cm

Sol :

Radii of two concentric circles are 4 cm and 5 cm

AB is a chord of the bigger circle

which is tangent to the smaller circle at C.

Join OA, OC











OC=4 cm,OA=5 cm

and OCACB

In right ΔOAC

OA2=OC2+AC252=42+AC2

25=16+AC2AC2=2516=9=(3)2

AC=3 cm

Length of chord AB=2×AC

=2×3=6 cm

Ans (b)


Question 18

In the given figure, AB is a chord of the circle such that ∠ACB = 50°. If AT is tangent to the circle at the point A, then ∠BAT is equal to

(a) 65°

(b) 60°

(c) 50°

(d) 40°












Sol :
In the given figure, AB is a chord of the circle
such that ∠ACB = 50°
AT is tangent to the circle at A
AT is tangent and AB is a chord
∠ACB = ∠BAT = 50°
(Angles in the alternate segments)
Ans (c)

Question 19

In the given figure, O is the centre of a circle and PQ is a chord. If the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is
(a) 100°
(b) 80°
(c) 90°
(d) 75°











Sol :
In the given figure, O is the centre of the circle.
PR is tangent and PQ is chord ∠RPQ = 50°
OP is radius and PR is tangent to the circle

RPQ=50

OP is radius and PR is tangent to the circle

OP PR

But OPQ+RPQ=90

OPQ+50=90

OPQ=9050=40

OP=OQ (radii of the same circle)

OQP=OPQ=40

and POQ=180(OPQ+OQP)

=180(40+40)

=18080=100

Ans (a)


Question 20

In the given figure, PA and PB are tangents to a circle with centre O. If ∠APB = 50°, then ∠OAB is equal to

(a) 25°

(b) 30°

(c) 40°

(d) 50°












Sol :
In the given figure,
PA and PB are tangents to the circle with centre O.
∠APB = 50°
But ∠AOB + ∠APB = 180°
∠AOB + 50° = 180°
⇒ ∠AOB = 180° – 50° = 130°
In ∆OAB,
OA = OB (radii of the same circle)
∠OAB = ∠OBA
But ∠OAB + ∠OBA = 180° – ∠AOB
= 180° – 130° = 50°
OAB=502=25

Ans (a)


Question 21

In the given figure, sides BC, CA and AB of ∆ABC touch a circle at point D, E and F respectively. If BD = 4 cm, DC = 3 cm and CA = 8 cm, then the length of side AB is

(a) 12 cm

(b) 11 cm

(c) 10 cm

(d) 9 cm












Sol :
In the given figure,
sides BC, CA and AB of ∆ABC touch a circle at D, E and F respectively.
BD = 4 cm, DC = 3 cm and CA = 8 cm













BD and BF are tangents to the circle

BF=BD=4 cm

Similarly, CD=CE=3 cm

AE=ACCE=83=5 cm

and AF=AE=5 cm

Now AB=AF+BF=5+4=9 cm

Ans (d)


Question 22

In the given figure, sides BC, CA and AB of ∆ABC touch a circle at the points P, Q and R respectively. If PC = 5 cm, AR = 4 cm and RB = 6 cm, then the perimeter of ∆ABC is

(a) 60 cm

(b) 45 cm

(c) 30 cm

(d) 15 cm











Sol :
In the given figure, sides BC, CA and AB of ∆ABC
touch a circle at P, Q and R respectively
PC = 5 cm, AR = 4 cm, RB = 6 cm

AR and AQ are tangents to the circles 
AQ=AR=4 cm
Similarly CQ=CP=5 cm

and BP=BR=6 cm

Now AB=AR+BR=4+6=10 cm

BC=BP+CP=6+5=11 cm

AC=AQ+CQ=4+5=9 cm

Perimeter of the ΔABC=AB+BC+CA

=10+11+9=30 cm

Ans (c)


Question 23

PQ is a tangent to a circle at point P. Centre of circle is O. If ∆OPQ is an isosceles triangle, then ∠QOP is equal to

(a) 30°

(b) 60°

(c) 45°

(d) 90°

Sol :

PQ is tangent to the circle at point P centre of the circle is O.










ΔOPQ is an isosceles triangle 
OP=PQ

OP is radius and PQ is tangent to the circle 

OPPQ i.e., OPQ=90

OP=PQ(ΔOPQ is an isosceles triangle )

QOP=PQO=902=45

Ans (c)


Question 24

In the given figure, PT is a tangent at T to the circle with centre O. If ∠TPO = 25°, then the value of x is
(a) 25°
(b) 65°
(c) 115°
(d) 90°









Sol :
In the given figure, PT is the tangent at T to the circle with centre O.
∠TPO = 25°
OT is the radius and TP is the tangent

OTTP
In ΔOPT

TOP+OPT=90
TOP+25=90
TOP=9025=65
But TOP+x=180 (Linear pair)
65+x=180x=18065=115
x=115

Ans (c)


Question 25

In the given figure, PA and PB are tangents at ponits A and B respectively to a circle with centre O. If C is a point on the circle and ∠APB = 40°, then ∠ACB is equal to
(a) 80°
(b) 70°
(c) 90°
(d) 140°










Sol :
In the given figure,
PA and PB are tangents to the circle at A and B respectively
C is a point on the circle and ∠APB = 40°
But ∠APB + ∠AOB = 180°

40+AOB=180
AOB=18040=140

Now arc AB subtends AOB at the centre and ACB is on the remaining part of the circle

ACB=12AOB=12×140=70

Ans (b)

Question 26

In the given figure, two circles touch each other at A. BC and AP are common tangents to these circles. If BP = 3.8 cm, then the length of BC is equal to
(a) 7.6 cm
(b) 1.9 cm
(c) 11.4 cm
(d) 5.7 cm








Sol :
In the given figure, two circles touch each other at A.
BC and AP are common tangents to these circles
BP = 3.8 cm
PB and PA are the tangents to the first circle

PB=PA=3.8 cm

Similarly PC and PA are tangents to the second circle

PA=PC=3.8 cm

BC=PB+PC=3.8+3.8=7.6 cm
Ans (a)

Question 27

In the given figure, if sides PQ, QR, RS and SP of a quadrilateral PQRS touch a circle at points A, B, C and D respectively, then PD + BQ is equal to
(a) PQ
(b) QR
(c) PS
(d) SR









Sol :
In the given figure,
sides PQ, QR, RS and SP of a quadrilateral PQRS
touch a circle at the points A, B, C and D respectively
PD and PA are the tangents to the circle
∴ PA = PD …(i)
Similarly, QA and QB are the tangents
∴ QA = QB …(ii)
Now PD + BQ = PA + QA = PQ (a)
[From (i) and (ii)]

Question 28

In the given figure, PQR is a tangent at Q to a circle. If AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to
(a) 20°
(b) 40°
(b) 35°
(d) 45°









Sol :
In the given figure, PQR is a tangent at Q to a circle.
Chord AB || PR and ∠BQR = 70°
BQ is chord and PQR is a tangent
∠BQR = ∠A

(Angles in the alternate segments)

ABPQR
BQR=B (alternate angles)
A=B=70
AQB+A+B=180
(Angles of a triangle)

AQB+70+70=180
AQB+140=180
AOB=180140=40

Ans (b)

Question 29

Two chords AB and CD of a circle intersect externally at a point P. If PC = 15 cm, CD = 7 cm and AP= 12 cm, then AB is
(a) 2 cm
(b) 4 cm
(c) 6 cm
(d) none of these








Sol :
In the given figure,
two chords AB and CD of a circle intersect externally at P.
PC = 15 cm, CD = 7 cm, AP = 12 cm
Join AC and BD










In ΔAPC and ΔBPD
P=P (common)
A=BDP

{ Ext. of a cyclic quad. is equal to its interior opposite angles\}

ΔAPCΔBPD  (AA axiom)
PAPD=PCPB
PAPB=PCPD
12PB=15×8 (PD=15-7=8)

PB=15×812=10

AB=APPB=1210=2 cm

Ans (a)

Comments

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2