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

$\therefore \angle \mathrm{A}=\angle \mathrm{B}$

But $\angle \mathrm{A}+\angle \mathrm{B}=90^{\circ}$

$\therefore \angle A=\angle B=\frac{90^{\circ}}{2}=45^{\circ}$

$\therefore \angle C A B=45^{\circ}$

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°

$(\because \mathrm{OA}=\mathrm{OB}$ radii of the same circle $)$

$\therefore \angle A O B=180^{\circ}-\left(40^{\circ}+40^{\circ}\right)$

$=180^{\circ}-80^{\circ}=100^{\circ}$

But arc AB subtends $\angle \mathrm{AOB}$ at the centre and $\angle \mathrm{ACB}$ at the remaining part of the circle

$\therefore \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOB}=\frac{1}{2} \times 100^{\circ}=50^{\circ}$

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

$\therefore \angle \mathrm{ADC}+\angle \mathrm{ABC}=180^{\circ}$

(opposite angles of the cyclic quadrilateral) 

$140^{\circ}+\angle \mathrm{ABC}=180^{\circ}$

$\therefore \angle \mathrm{ABC}=180^{\circ}-140^{\circ}=40^{\circ}$

Now in $\Delta \mathrm{ABC}$ 

$\angle \mathrm{ACB}=90^{\circ}$

(angle in a semi-circle)

$\therefore \angle \mathrm{BAC}=90^{\circ}-\angle \mathrm{ABC}$

$=90^{\circ}-40^{\circ}=50^{\circ}$

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°

$\operatorname{In} \Delta \mathrm{ABO}, \dot{\mathrm{OA}}=\mathrm{OB}$

(radii of the same circle)

$\therefore \angle \mathrm{ABO}=\angle \mathrm{BAO}=60^{\circ}$

Ext. $\angle \mathrm{AOC}=\angle \mathrm{BAO}+\angle \mathrm{ABO}$

$=60^{\circ}+60^{\circ}=120^{\circ}$

Now arc AC subtends $\angle \mathrm{AOC}$ at the centre and $\angle \mathrm{ADC}$ at the remaining part of the circle

$\therefore \angle A O C=2 \angle A D C \Rightarrow 2 \angle A D C=120^{\circ}$

$\Rightarrow \angle \mathrm{ADC}=\frac{120^{\circ}}{2}=60^{\circ}$

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 $\Delta \mathrm{AOB}$

$\angle \mathrm{AOB}=90^{\circ}, \angle \mathrm{ABC}=30^{\circ}$

In $\triangle \mathrm{AOB}, \angle \mathrm{AOB}=90^{\circ}$

$\mathrm{OA}=\mathrm{OB} $ (radii of the same circle)

$\therefore \angle \mathrm{OAB}=\angle \mathrm{OBA}=\frac{90^{\circ}}{2}=45^{\circ}$

Arc AB subtends $\angle A O B$ at the centre and $\angle \mathrm{ACB}$ at the remaining part of the circle

$\therefore \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOB}=\frac{1}{2} \times 90^{\circ}=45^{\circ}$

Now in $\Delta \mathrm{ACB}, \angle \mathrm{ABC}=30^{\circ}, \angle \mathrm{ACB}=45^{\circ}$

$\therefore \angle B A C=180^{\circ}-\left(30^{\circ}+45^{\circ}\right)$

$=180^{\circ}-75^{\circ}=105^{\circ}$

But $\angle \mathrm{OAB}=45^{\circ}$

∠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

$\therefore \angle \mathrm{PRO}=\frac{1}{2} \angle \mathrm{POQ}=\frac{1}{2} \times 60^{\circ}=30^{\circ}$

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.

$\angle \mathrm{ADB}=\angle \mathrm{ACB}=2 x$

(Angles in the same segment)

arc $\mathrm{AB}$ subtends $\angle \mathrm{AOB}$ at the centre and $\angle \mathrm{ADB}$ at the remaining part of the circle

$\therefore \angle \mathrm{AOB}=2 \angle \mathrm{ADB}=2 \times 2 x=4 x$

$\operatorname{In} \Delta \mathrm{OAB}$

$\angle \mathrm{OAB}=\angle \mathrm{OBA}=3 x(\mathrm{OA}=\mathrm{OB})$

Sum of angles of $\Delta \mathrm{OAB}=180^{\circ}$ $\Rightarrow 3 x+3 x+4 x=180^{\circ}$

$\Rightarrow 10 x=180^{\circ}$

$\Rightarrow x=\frac{180^{\circ}}{10}=18^{\circ}$

$\therefore x=18^{\circ}$

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

$\because \mathrm{PQ}$ is tangent and OP is radius 

$\therefore \mathrm{OP} \perp \mathrm{PQ}$

Now in right ΔOPQ

$\mathrm{OQ}^{2}=\mathrm{OP}^{2}+\mathrm{PQ}^{2}$

$(25)^{2}=\mathrm{OP}^{2}+(24)^{2}$

$\Rightarrow \mathrm{OP}^{2}=25^{2}-24^{2}=625-576$

$\Rightarrow \mathrm{OP}^{2}=49=(7)^{2} \Rightarrow \mathrm{OP}=7 \mathrm{~cm}$

∴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

$\because P Q$ is tangent and $O Q$ is the radius $\therefore P Q^{2}=O P^{2}-O Q^{2}=13^{2}-5^{2}=169-25$

$\quad=144=(12)^{2}$

$\therefore P Q=12 \mathrm{~cm}$

Now area of $\Delta \mathrm{OPQ}=\frac{1}{2} \mathrm{PQ} \times \mathrm{OQ}$

$\left(\frac{1}{2}\right.$ base $\times$ height $)$

$=\frac{1}{2} \times 12 \times 5=30 \mathrm{~cm}^{2}$

$\therefore$ area of quad. $\mathrm{PQOR}=2 \times 30=60 \mathrm{~cm}^{2}$

(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

$\because$ OA radius and $A P$ is tangent to the circle

$\therefore \angle \mathrm{OAP}=90^{\circ}$

Similarly $\angle \mathrm{OBP}=90^{\circ}$

$\therefore \angle A O B+\angle A P B=180^{\circ}$

$\Rightarrow 130^{\circ}+\angle A P B=180^{\circ}$

$\Rightarrow \angle A P B=180^{\circ}-130^{\circ}=50^{\circ}$

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

$\angle \mathrm{POR}=55^{\circ}$

$\because$ OR is radius and $P R$ is tangent

$\therefore \mathrm{OR} \perp \mathrm{PR}$

$\therefore$ In $\Delta \mathrm{OPR}$

$\angle \mathrm{OPR}=90^{\circ}-55^{\circ}=35^{\circ}$

$\therefore \mathrm{QPR}=2 \times \angle \mathrm{OPR}=2 \times 35^{\circ}=70^{\circ}$

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°

$\because$ OA is radius and AP is tangent

$\therefore \angle \mathrm{OAP}=90^{\circ}$ and $\angle \mathrm{OPA}=\frac{1}{2} \angle \mathrm{APB}$

$=\frac{1}{2} \times 80^{\circ}=40^{\circ}$

$=\frac{1}{2} \times 80^{\circ}=40^{\circ}$

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

$\therefore \mathrm{OA} \perp \mathrm{PA}$

$\because \angle \mathrm{APB}=90^{\circ}$

$(\because \mathrm{PA} \perp \mathrm{PB})$

$\therefore \angle \mathrm{APO}=90^{\circ} \times \frac{1}{2}=45^{\circ}$

$\therefore \angle \mathrm{AOP}=90^{\circ}-45^{\circ}=45^{\circ}$

i.e. OA=AP=5 cm

$\therefore \mathrm{OP}=\sqrt{\mathrm{OA}^{2}+\mathrm{PA}^{2}}=\sqrt{5^{2}+5^{2}}$

$=\sqrt{25+25}=\sqrt{50}=\sqrt{2 \times 25}$

$=\sqrt{2 \times 5} \mathrm{~cm}=5 \sqrt{2} \mathrm{~cm}$

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 $\Delta \mathrm{OEC}$ $\quad \mathrm{OE}=8-5=3 \mathrm{~cm}$

$\mathrm{OC}=5 \mathrm{~cm}$

$\therefore \mathrm{CE}=\sqrt{\mathrm{OC}^{2}-\mathrm{OE}^{2}}=\sqrt{5^{2}-3^{2}}$

$=\sqrt{25-9}=\sqrt{16}=4 \mathrm{~cm}$

$\therefore \mathrm{CD}=2 \times \mathrm{CE}=2 \times 4=8 \mathrm{~cm}$

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

$\mathrm{OC}=4 \mathrm{~cm}, \mathrm{OA}=5 \mathrm{~cm}$

and $\mathrm{OC} \perp \mathrm{ACB}$

$\therefore$ In right $\Delta \mathrm{OAC}$

$\mathrm{OA}^{2}=\mathrm{OC}^{2}+\mathrm{AC}^{2} \Rightarrow 5^{2}=4^{2}+\mathrm{AC}^{2}$

$\Rightarrow 25=16+\mathrm{AC}^{2} \Rightarrow \mathrm{AC}^{2}=25-16=9=(3)^{2}$

$\therefore \mathrm{AC}=3 \mathrm{~cm}$

$\therefore$ Length of chord $\mathrm{AB}=2 \times \mathrm{AC}$

$=2 \times 3=6 \mathrm{~cm}$

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

$\angle \mathrm{RPQ}=50^{\circ}$

$\because$ OP is radius and $\mathrm{PR}$ is tangent to the circle

$\therefore$ OP $\perp P R$

But $\angle \mathrm{OPQ}+\angle \mathrm{RPQ}=90^{\circ}$

$\Rightarrow \angle \mathrm{OPQ}+50^{\circ}=90^{\circ}$

$\Rightarrow \angle \mathrm{OPQ}=90^{\circ}-50^{\circ}=40^{\circ}$

$\because \mathrm{OP}=\mathrm{OQ} \quad$ (radii of the same circle)

$\therefore \angle \mathrm{OQP}=\angle \mathrm{OPQ}=40^{\circ}$

and $\angle \mathrm{POQ}=180^{\circ}-(\angle \mathrm{OPQ}+\angle \mathrm{OQP})$

$=180^{\circ}-\left(40^{\circ}+40^{\circ}\right)$

$=180^{\circ}-80^{\circ}=100^{\circ}$

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°

$\angle \mathrm{OAB}=\frac{50^{\circ}}{2}=25^{\circ}$

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

$\because \mathrm{BD}$ and $\mathrm{BF}$ are tangents to the circle

$\therefore \mathrm{BF}=\mathrm{BD}=4 \mathrm{~cm}$

Similarly, $\mathrm{CD}=\mathrm{CE}=3 \mathrm{~cm}$

$\therefore \mathrm{AE}=\mathrm{AC}-\mathrm{CE}=8-3=5 \mathrm{~cm}$

and $\mathrm{AF}=\mathrm{AE}=5 \mathrm{~cm}$

Now $\mathrm{AB}=\mathrm{AF}+\mathrm{BF}=5+4=9 \mathrm{~cm}$

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

$\therefore$ AR and AQ are tangents to the circles 

$\therefore AQ=AR=4 \mathrm{~cm}$

Similarly CQ=CP=5 cm

and BP=BR=6 cm

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

$\mathrm{BC}=\mathrm{BP}+\mathrm{CP}=6+5=11 \mathrm{~cm}$

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

$\therefore$ Perimeter of the $\Delta \mathrm{ABC}=\mathrm{AB}+\mathrm{BC}+\mathrm{CA}$

=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.

$\Delta \mathrm{OPQ}$ is an isosceles triangle 

OP=PQ

$\because \mathrm{OP}$ is radius and $\mathrm{PQ}$ is tangent to the circle 

$\therefore \mathrm{OP} \perp \mathrm{PQ}$ i.e., $\angle \mathrm{OPQ}=90^{\circ}$

$\because \mathrm{OP}=\mathrm{PQ}(\because \Delta \mathrm{OPQ}$ is an isosceles triangle $)$

$\therefore \angle \mathrm{QOP}=\angle \mathrm{PQO}=\frac{90^{\circ}}{2}=45^{\circ}$

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

$\therefore \mathrm{OT} \perp \mathrm{TP}$

$\therefore$ In $\Delta \mathrm{OPT}$

$\quad \angle \mathrm{TOP}+\angle \mathrm{OPT}=90^{\circ}$

$\Rightarrow \angle \mathrm{TOP}+25^{\circ}=90^{\circ}$

$\angle \mathrm{TOP}=90^{\circ}-25^{\circ}=65^{\circ}$

But $\angle \mathrm{TOP}+x=180^{\circ} \quad$ (Linear pair)

$65^{\circ}+x=180^{\circ} \Rightarrow x=180^{\circ}-65^{\circ}=115^{\circ}$

$\therefore x=115^{\circ}$

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°

$\Rightarrow 40^{\circ}+\angle \mathrm{AOB}=180^{\circ}$

$\Rightarrow \angle \mathrm{AOB}=180^{\circ}-40^{\circ}=140^{\circ}$

Now arc AB subtends $\angle A O B$ at the centre and $\angle \mathrm{ACB}$ is on the remaining part of the circle

$\therefore \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOB}=\frac{1}{2} \times 140^{\circ}=70^{\circ}$

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

$\because P B$ and $P A$ are the tangents to the first circle

$\therefore P B=P A=3.8 \mathrm{~cm}$

Similarly PC and PA are tangents to the second circle

$\therefore \mathrm{PA}=\mathrm{PC}=3.8 \mathrm{~cm}$

$\mathrm{BC}=\mathrm{PB}+\mathrm{PC}=3.8+3.8=7.6 \mathrm{~cm}$

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)

$\because \mathrm{AB} \| \mathrm{PQR}$

$\therefore \angle \mathrm{BQR}=\angle \mathrm{B}$ (alternate angles)

$\therefore \angle A=\angle B=70^{\circ}$

$\therefore \angle \mathrm{AQB}+\angle \mathrm{A}+\angle \mathrm{B}=180^{\circ}$

(Angles of a triangle)

$\Rightarrow \angle A Q B+70^{\circ}+70^{\circ}=180^{\circ}$

$\Rightarrow \angle A Q B+140^{\circ}=180^{\circ}$

$\therefore \angle A O B=180^{\circ}-140^{\circ}=40^{\circ}$

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 $\Delta \mathrm{APC}$ and $\Delta \mathrm{BPD}$

$\angle \mathrm{P}=\angle \mathrm{P}$ (common)

$\angle \mathrm{A}=\angle \mathrm{BDP}$

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

$\therefore \Delta \mathrm{APC} \sim \Delta \mathrm{BPD}$  (AA axiom)

$\frac{\mathrm{PA}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{PB}}$

$\mathrm{PA} \cdot \mathrm{PB}=\mathrm{PC} \cdot \mathrm{PD}$

$12 \cdot \mathrm{PB}=15 \times 8$ (PD=15-7=8)

$\mathrm{PB}=\frac{15 \times 8}{12}=10$

$\therefore A B=A P-P B=12-10=2 \mathrm{~cm}$

Ans (a)