# Class 9 Maths Chapter 10 Circles MCQs

## Trigonometry # 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 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 Maths Chapter 10 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

2) The longest chord of the circle is:

b. Arc

c. Diameter

d. Segment

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

a. Segments

c. Arcs

d. Chords

Explanation: See the figure below: 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

d . None of the above

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

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

a. 90

b. 45

c. 180

d. 60

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: a.∠BEQ > ∠CEQ

b. ∠BEQ = ∠CEQ

c. ∠BEQ < ∠CEQ

d. None of the above

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

Explanation: See the figure below: 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: a. 60°

b. 30°

c. 45°

d. 55°

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,

= 1/2 × 90° = 45°

10) In the given figure, find angle OPR. a. 20°

b. 15°

c. 12°

d. 10°

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°

11) In the given figure, ∠AOB = 90º and ∠ABC = 30º, then ∠CAO is equal to: (a) 30º

(b) 45º

(c) 60º

(d) 90º

Explanation:

Given that ∠AOB = 90º and ∠ABC = 30º

OA = OB (Radii of the circle)

Let x= ∠OAB = ∠OBA = x

In the triangle OAB,

∠OAB + ∠OBA + ∠AOB = 180° (By using the angle sum property of triangle)

⇒ x + x + 90° = 180°

⇒ 2x = 180° – 90°

⇒ x =  90°/ 2 = 45°

Therefore, ∠OAB = 45° and ∠OBA = 45°

By using the theorem, “ the angles subtended by arcs at the centre of the circle double the angle subtended at the remaining part of the circle”, we can write

∠AOB = 2∠ACB

This can also be written as,

∠ACB =  ½  ∠AOB =  (½) × 90° = 45°

Now, apply the angle sum property of triangle on the triangle ABC,

∠ACB + ∠BAC + ∠CBA = 180°

∠ACB + [∠BAO + ∠CAO] + ∠CBA = 180° (As,∠BAC = ∠BAO + ∠CAO)

Now, substitute the known values, we get

45° + (45° + ∠CAO) + 30° = 180°

∠CAO = 180°- (30° + 45° + 45°)

∠CAO = 180°-120°

∠CAO = 60°

Hence, ∠CAO is equal to 60°.

12) ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to:

(a) 30º

(b) 40º

(c) 50º

(d) 80º

Explanation: We know that the sum of opposite angles of a cyclic quadrilateral is 180°.

Hence, ∠ADC + ∠ABC = 180°

Now, substitute ∠ADC = 140º in the above equation, we get

140° + ∠ABC = 180°

∠ABC = 180° – 140° = 40°

since the angle subtended by a diameter at the circumference of the circle, is 90°

Hence,  ∠ACB = 90°

By using the angle property of triangle in the triangle, ABC,

∠CAB + ∠ABC + ∠ACB = 180°

∠CAB + 40° + 90° = 180°

∠CAB = 180° – 90° – 40°

∠CAB = 50°

Therefore, ∠CAB or ∠BAC =50°.

13) In the given figure, if ∠OAB = 40º, then ∠ACB is equal to (a) 40º

(b) 50º

(c) 60º

(d) 70º

Explanation:

Given that ∠OAB = 40º,

In  the triangle OAB,

Since, the angles opposite to equal sides are equal,

∠OAB = ∠OBA (i.e.) ∠OBA = 40°

Now, by using the angle sum property of triangle, we can write

∠AOB + ∠OBA + ∠BAO = 180°

Now, substitute the known values,

∠AOB + 40° + 40° = 180°

∠AOB = 180 – 80° = 100°

∠AOB = 2 ∠ACB  (Since, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle)

∠ACB = ½ ∠AOB

Hence, ∠ACB = 100°/2 = 50°.

14) In the given figure, if ∠ABC = 20º, then ∠AOC is equal to: (a) 10º

(b) 20º

(c) 40º

(d) 60º

Explanation:

Given that, ∠ABC = 20º.

∠AOC = 2∠ABC (since the angle subtended by an arc at the centre of the circle is double the angle subtended at the remaining part.)

Now, substitute the values, we get

∠AOC = 2 × 20°

Therefore, ∠AOC = 40°.

15) In the given figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to: (a) 2 cm

(b) 3 cm

(c) 4 cm

(d) 5 cm

Explanation:

From the given diagram, we can observe that OC is perpendicular to chord AB. Therefore, OC bisects the chord AB Hence. AC=CB

Also,

AC+CB=AB

AC+CB=8

2AC = 8 (Since, AC = CB)

AC = 8/2 = 4 cm

As, the triangle OCA is a right-angled triangle, by using Pythagoras theorem, we can write

AO2=AC2+OC2

52=42+OC2

52−42=OC2

OC2=9

OC=3 cm

As, OD is the radius of the circle, OA=OD=5cm

CD=OD-OC

CD = 5-3 = 2 cm

Hence, the value of CD is equal to 2cm.

16) In the given figure, BC is the diameter of the circle and ∠BAO = 60º. Then ∠ADC is equal to (a) 30º

(b) 45º

(c) 60º

(d) 120º

Explanation:

Given that ∠BAO = 60°

Since OA = OB,∠OBA = 60°

Then ∠ADC = 60° (As, the angles in the same segment are equal).

17) In the given figure, if ∠DAB = 60º, ∠ABD = 50º, then ∠ACB is equal to: (a) 50º

(b) 60º

(c) 70º

(d) 80º

Explanation:

Given that, ∠DAB = 60º, ∠ABD = 50º

By using the angle sum property in the triangle ABD

∠ADB=∠ACB (Since the angles subtended at the circumference by the same arc are equal)

We know that the angles subtended at the circumference by the same arc are equal.

∠ACB =70º

18) In the given figure, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to: (a) 30º

(b) 45º

(c) 60º

(d) 90º

Explanation:

We know that the angle at circumference subtended by the diameter of the circle is the right angle.

Hence, ∠ACB = 90°

Also, given that AC = BC

Therefore, ∠CAB = ∠CBA (As, the angles opposite to equal sides are also equal)

Now, by using the angle sum property of triangle in ∆ACB, we can write

∠CAB + ∠ABC + ∠BCA = 180°

∠CAB + ∠CAB + 90° = 180°

2∠CAB = 180° – 90°

∠CAB = 45°

Therefore, ∠CAB is equal to 45°.

19) If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is:

(a) 6 cm

(b) 8 cm

(c) 10 cm

(d) 12 cm

Explanation:

Given that AB = 12 cm, BC = 16 cm and AB is perpendicular to BC. Hence, AC is the diameter of the circle passing through points A, B and C.

Hence, ABC is a right-angled triangle.

Thus by using the Pythagoras theorem:

AC2 = (CB)2 + (AB)2

⇒ AC2 = (16)2 + (12)2

⇒ AC2 = 256 + 144

⇒ AC2 = 400

Hence, the diameter of the circle, AC = 20 cm.

Thus, the radius of the circle is 10 cm.

20) AD is the diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is

(a) 4 cm

(b) 8 cm

(c) 15 cm

(d) 17 cm

Explanation:

Given that, Diameter, AD = 34 cm.

Chord, AB = 30 cm. Hence, the radius of the circle, OA = 17 cm

Now, consider the figure.

From the centre “O”. OM is perpendicular to the chord AB.

(i.e) OM ⊥ AM

AM =  ½  AB

AM =  ½ (30) = 15 cm

Now by using the Pythagoras theorem in the right triangle AOM,

AO2 = OM2 + AM2

OM2 = AO2– AM2

OM2= 172 – 152

OM2 = 64

OM = √64

OM = 8 cm

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