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Test: Circles- 2 - Grade 9 MCQ

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25 Questions MCQ Test - Test: Circles- 2

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Test: Circles- 2 - Question 1

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AB is a chord of a circle with radius ‘r’. If P is any point on the circle such that ∠APB is a right angle , then AB is equal to

A.
3r
B.
r
C.
2r
D.
r²

Detailed Solution for Test: Circles- 2 - Question 1

- When ∠APB is 90°, point P lies on the circle's circumference forming a right angle with chord AB.
- According to the Thales' theorem, AB must be the diameter of the circle.
- The diameter of a circle is twice its radius.
- Therefore, AB equals 2r.

Answer: c

Test: Circles- 2 - Question 2

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In the given figure, AC is a diameter of the given circle and ∠BCD = 75^o. Then, ∠EAF−∠ABC is equal to

A.
15^o
B.
25^o
C.
20^o
D.
10^o

Detailed Solution for Test: Circles- 2 - Question 2

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Test: Circles- 2 - Question 3

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The region between chord and either of the arc is called

A.
a sector
B.
a semicircle
C.
a segment
D.
a quarter circle

Test: Circles- 2 - Question 4

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P is a point on the diameter AB of a circle and CD is a chord perpendicular to AB. If AP = 4 cm and PB = 16 cm, the length of chord CD is

A.
16 cm
B.
20 cm
C.
10 cm
D.
8 cm

Detailed Solution for Test: Circles- 2 - Question 4

AP+PB=AB [diameter of circle]
4+16=20[diameter]
so r=AO = CO=10
so clearly PO= OA-AP=10-4=6
PO=6 OC=10 So just apply pythagoreus theorem in triangle PCO
so CP will come 8 So chord CD = 2x8 = 16

Test: Circles- 2 - Question 5

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Arc ABC subtends an angle of 130^o at the centre O of the circle. AB is extended to P. Then ∠CBP equals :

A.
60^o
B.
70^o
C.
130^o
D.
65^o

Detailed Solution for Test: Circles- 2 - Question 5

Test: Circles- 2 - Question 6

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Chords AB and CD intersect at right angles. If ∠BAC = 40^o, then ∠ABD is equal to

A.
45^o
B.
50^o
C.
60^o
D.
none of these

Test: Circles- 2 - Question 7

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In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 40^o and 30^o respectively. Then, the measure of ∠AOC is

A.
170^o
B.
120^o
C.
110^o
D.
140^o

Detailed Solution for Test: Circles- 2 - Question 7

(Angle at the centre is double the angle at the circumference subtended by the same chord)

Test: Circles- 2 - Question 8

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In the given, AB is side of regular five sided polygon and AC is a side of a regular six sided polygon inscribed in the circle with centre O. AO and CB intersect at P, then ∠APB is equal to

A.
72^o
B.
98^o
C.
90^o
D.
96^o

Test: Circles- 2 - Question 9

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In the given figure, if ∠AOB = 80^o and ∠ABC = 30^o , then ∠CAO is equal to

A.
30^o
B.
60^o
C.
80^o
D.
40^o

Detailed Solution for Test: Circles- 2 - Question 9

2ACB=AOB
ACB=40
CAB+ACB+ABC=180
CAB=180-70
CAB=110
since,OA=OB(radius)
OAB=OBA

AOB+OAB+OBA=180
2OBA=100
OBA=50

CAO=CAB-OBA=110-5O
CAO=60^o

Test: Circles- 2 - Question 10

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In the figure, O is the centre of eh circle and ∠AOB = 80^o. The value of x is :

A.
40^o
B.
160^o
C.
60^o
D.
30^o

Detailed Solution for Test: Circles- 2 - Question 10

the angle subtend at center is double of the angle at the arc

so: 2x=80

so x=40

Test: Circles- 2 - Question 11

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In the given figure if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to

A.
4 cm
B.
3 cm
C.
5 cm
D.
2 cm

Test: Circles- 2 - Question 12

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If ∠OAB = 40^o, then the measure of ∠ACB is

A.
20^o
B.
50^o
C.
80^o
D.
40^o

Test: Circles- 2 - Question 13

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Chords AD and BC intersect each other at right angles at point P. ∠DAB = 35^o, then ∠ADC is equal to

A.
55^o
B.
45^o
C.
35^o
D.
65^o

Test: Circles- 2 - Question 14

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In the given figure, AB is a diameter of the circle APBR. APQ and RBQ are straight lines. If ∠A = 35^o and , then the measure of ∠PBR is

A.
165^o
B.
115^o
C.
155^o
D.
135^o

Test: Circles- 2 - Question 15

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In the figure, O is the center of the circle. If ∠OAB = 40^o, then ∠ACB is equal to :

A.
60^o
B.
70^o
C.
50^o
D.
40^o

Test: Circles- 2 - Question 16

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X is a point on a circle with centre O. If X is equidistant from the two radii OP, OQ, then arc PX : arc PQ is equal to

A.
it is 1 : 2
B.
it is 1 : 1
C.
it is 2 : 1
D.
it is 2 : 3

Detailed Solution for Test: Circles- 2 - Question 16

Test: Circles- 2 - Question 17

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BC is a diameter of the circle and ∠BAO = 60^o . Then ∠ADC is equal to

A.
45^o
B.
30^o
C.
60^o
D.
90^o

Test: Circles- 2 - Question 18

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What fraction of the whole circle is minor arc RP in the given figure ?

A.
1/5 of the circle
B.
1/2 of the circle
C.
1/4 of the circle
D.
1/3 of the circle

Test: Circles- 2 - Question 19

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In the given figure, AD is the diameter of the circle and AE = DE. If ∠ABC = 115^o, then the measure of ∠CAE is

A.
70^o
B.
90^o
C.
80^o
D.
60^o

Test: Circles- 2 - Question 20

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In the figure, if ∠DAB = 60^o, ∠ABD = 50^o, then ∠ACB is equal to :

A.
60^o
B.
50^o
C.
70^o
D.
80^o

Test: Circles- 2 - Question 21

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If ABCD is a cyclic trapezium in which AD ║ BC and ∠B = 60^o, then ∠BCD is equal to

A.
60^o
B.
80^o
C.
100^o
D.
None of the above

Test: Circles- 2 - Question 22

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In the given figure, O is the centre of the circle and ∠AOC = 130^o. Then ∠ABC is equal to

A.
115^o
B.
130^o
C.
65^o
D.
165^o

Detailed Solution for Test: Circles- 2 - Question 22

360^o - 130^o =230^o

Therefore, angle ABC =1/2reflex angle AOC

Angle ABC =1/2 230^o

So, angle ABC =115^o

Test: Circles- 2 - Question 23

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In the given circle, O is the centre and ∠BDC = 42^o. Then, ∠ACB is equal to

A.
58^o
B.
42^o
C.
52^o
D.
48^o

Detailed Solution for Test: Circles- 2 - Question 23

In ∆ BDC and ∆ BAC
Angle BAC = BDC
(angle made on same segment BC)
Since ABC is making right angle (90)
So,
In ∆ABC
ABC +BAC+ACB=180
(angle sum property of triangle)
90+42+ACB=180
ACB=180-132
ACB=48^o

Test: Circles- 2 - Question 24

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AOB is the diameter of the circle. If ∠AOE = 150^o, then the measure of ∠CBE is

A.
120^o
B.
125^o
C.
105^o
D.
115^o

Test: Circles- 2 - Question 25

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In the given figure, a circle is centred at O. The value of x is :

A.
55^o
B.
70^o
C.
110^o
D.
125^o

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