Class 9 Maths Chapter 10 Circles MCQs

Class 9 Maths Chapter 10 (Circles) MCQs are available online, here with answers. These are the objective questions prepared, as per the CBSE syllabus (2020-2021) and NCERT curriculum. The multiple choice questions are given here chapter-wise, with detailed explanations. Also check Important Questions for Class 9 Maths.

MCQs on Class 9 Circles

Solve the MCQs on circles given here with four multiple options and choose the right answer.

1) The center of the circle lies in______ of the circle.

a. Interior

b. Exterior

c. Circumference

d. None of the above

Answer: a

2) The longest chord of the circle is:

a. Radius

b. Arc

c. Diameter

d. Segment

Answer: c

3) Equal _____ of the congruent circles subtend equal angles at the centers.

a. Segments

b. Radii

c. Arcs

d. Chords

Answer: d

Explanation: See the figure below:

Class 9 Circles MCQs 1

Let ΔAOB and ΔCOD are two triangles inside the circle.

OA = OC and OB = OD (radii of the circle)

AB = CD (Given)

So, ΔAOB ≅ ΔCOD (SSS congruency)

∴ By CPCT rule, ∠AOB = ∠COD.

Hence, this prove the statement.

4) If chords AB and CD of congruent circles subtend equal angles at their centres, then:

a. AB = CD

b. AB > CD

c. AB < AD

d . None of the above

Answer: a

Explanation: Take the reference of the figure from above question.

In triangles AOB and COD,

∠AOB = ∠COD (given)

OA = OC and OB = OD (radii of the circle)

So, ΔAOB ≅ ΔCOD. (SAS congruency)

∴ AB = CD (By CPCT)

5) If there are two separate circles drawn apart from each other, then the maximum number of common points they have:

a. 0

b. 1

c. 2

d. 3

Answer: a

6) The angle subtended by the diameter of a semi-circle is:

a. 90

b. 45

c. 180

d. 60

Answer: c

Explanation: The semicircle is half of the circle, hence the diameter of the semicircle will be a straight line subtending 180 degrees.

7) If AB and CD are two chords of a circle intersecting at point E, as per the given figure. Then:
Class 9 Maths Circles MCQs 2

a.∠BEQ > ∠CEQ

b. ∠BEQ = ∠CEQ

c. ∠BEQ < ∠CEQ

d. None of the above

Answer: b

Explanation:

OM = ON (Equal chords are always equidistant from the centre)

OE = OE (Common)

∠OME = ∠ONE (perpendiculars)

So, ΔOEM ≅ ΔOEN (by RHS similarity criterion)

Hence, ∠MEO = ∠NEO (by CPCT rule)

∴ ∠BEQ = ∠CEQ

8) If a line intersects two concentric circles with centre O at A, B, C and D, then:

a. AB = CD

b. AB > CD

c. AB < CD

d. None of the above

Answer: a

Explanation: See the figure below:

Class 9 Maths Circles MCQs 3

From the above fig., OM ⊥ AD.

Therefore, AM = MD — 1

Also, since OM ⊥ BC, OM bisects BC.

Therefore, BM = MC — 2

From equation 1 and equation 2.

AM – BM = MD – MC

∴ AB = CD

9) In the below figure, the value of ∠ADC is:

Class 9 Maths Circles MCQs 4

a. 60°

b. 30°

c. 45°

d. 55°

Answer: c

Explanation: ∠AOC = ∠AOB + ∠BOC

So, ∠AOC = 60° + 30°

∴ ∠AOC = 90°

An angle subtended by an arc at the centre of the circle is twice the angle subtended by that arc at any point on the rest part of the circle.

So,

∠ADC = 1/2∠AOC

= 1/2 × 90° = 45°

10) In the given figure, find angle OPR.

Class 9 Maths Circles MCQs 5

a. 20°

b. 15°

c. 12°

d. 10°

Answer: d

Explanation: The angle subtended by major arc PR at the centre of the circle is twice the angle subtended by that arc at point, Q, on the circle.

So, ∠POR = 2 × ∠PQR, here ∠POR is the exterior angle

We know the values of angle PQR as 100°

So, ∠POR = 2 × 100° = 200°

∴ ∠ROP = 360° – 200° = 160°   [Full rotation: 360°]

Now, in ΔOPR,

OP and OR are the radii of the circle

So, OP = OR

Also, ∠OPR = ∠ORP

By angle sum property of triangle, we know:

∠ROP + ∠OPR + ∠ORP = 180°

∠OPR + ∠OPR = 180° – 160°

As, ∠OPR = ∠ORP

2∠OPR = 20°

Thus, ∠OPR = 10°

5 Comments

  1. Jaikrish Vetriselvi

    This is awesome where I come to know the correct answer and too with explanation but in the last sum after fine the value of ANGLE POR as 200. The why did you subtracted with 360 l don’t know can you kindly clarify my doubt please

    1. ∠POR is the angle formed by major arc PR with center o.
      Therefore, it is equal to twice of ∠PQR
      Now, ∠POR or ∠ROP with minor arc PR, will be:
      ∠ROP = 360° – 200° = 160° ; 360° is the full rotation at center o.

    2. we canceled 360 degree as it is forming a complete angle and we and we have to find the angle OPR. First we used complete angle theorem and then angle sum property

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  3. Great content!!!

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