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This mock test of Test: Angle Subtended By The Chord for Class 9 helps you for every Class 9 entrance exam.
This contains 20 Multiple Choice Questions for Class 9 Test: Angle Subtended By The Chord (mcq) to study with solutions a complete question bank.
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QUESTION: 1

Any angle whose vertex is at the centre of the circle is called

Solution:

QUESTION: 2

In the given figure, if ∠OAB = 40^{o} then ∠ACB is equal to:

Solution:

QUESTION: 3

If the angles subtended by the chords of a circle at the centre are equal, then the chords are

Solution:

QUESTION: 4

Name the chord that subtends angle FOE at the centre.

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QUESTION: 5

In the given figure name the angle subtended by the chord PQ on the major arc

Solution:

QUESTION: 6

In figure, O and O” are centres of two circles intersecting at B and C. ACD is a straight line, then the value of x is:

Solution:

QUESTION: 7

The angles subtended by arcs XY and MN on the centre of the circle are 72° each. The length of chord XY = 15 cm, then the length of chord MN is:

Solution:

QUESTION: 8

In a circle with center O and a chord BC, points D and E lie on the same side of BC. Then, if∠BDC=80°, then ∠BEC =

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QUESTION: 9

In a circle with center O and a chord BC, the point D lies on the same side BC as O. If ∠ BOC = 50°,then ∠ BDC =

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QUESTION: 10

A regular octagon is inscribed in a circle. The angle that each side of the octagon subtends at the centre is

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QUESTION: 11

In the given figure, what is the measurement of YXZ?

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QUESTION: 12

In figure, O is the centre of the circle and PS bisects ∠QPR. Find ∠QRS.

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QUESTION: 13

An equilateral triangle ABC is inscribed in a circle with centre O. Angle BOC will be

Solution:

QUESTION: 14

In figure arc PQ = arc PR, ∠POQ = 30° and ∠QOS = 70°. Find ∠ROS.

Solution:
Sol^ :-
Join RP and PQ.
then,
RP=PQ [ Since *arc RP=arcPQ* ]
Now since segment RP=PQ
Then, âˆ POR =âˆ POQ = 30* [ Angles of same or equal segment ]
Now,
âˆ ROS = âˆ POQ+âˆ POR+âˆ QOS
âˆ ROS = 30*+30*+70*
âˆ ROS = 130*
ã€Š* represents DEGREEã€‹

QUESTION: 15

Chord AB subtends ∠AOB = 60° at centre. If OA = 5 cm then length of AB (in cm) is :

Solution:

QUESTION: 16

In the figure, ∠M = 82°, then ∠O = ?

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QUESTION: 17

If ∠OAB = 40° then ∠ACB is equal to:

Solution:

In ΔQAB, OA = OB [both are the radius of a circle]

∠OAB = ∠OBA ⇒ ∠OBA = 40°

[angles opposite to equal sides are equal] Also, ∠AOB + ∠OBA + ∠BAO = 180°

[by angle sum property of a triangle]

∠AOB + 40° + 40° = 180°

⇒ ∠AOB = 180° – 80° = 100°

We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.

∠AOB = 2 ∠ACB ⇒ 100° =2 ∠ACB

∠ACB = 100°/2 = 50°

QUESTION: 18

In the figure, if AB is the diameter of the circle, then the value of x is:

Solution:

QUESTION: 19

An equilateral ΔABC is inscribed in a circle with centre O. The measure of ∠BOC is:

Solution:

QUESTION: 20

If the regular polygon is inscribed in a circle the sum of all the angles are at the the centre is

Solution:

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