ML Aggarwal Solution Class 10 Chapter 15 Circles MCQs

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29

 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)