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### Maths MCQ Class IX Ch-10 | Circle

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**Mathematics MCQ | Class 09 | Chapter 10**

# CIRCLE

**MCQ Based on the definition of circle.****MCQ Based on the properties of circle.****MCQ Based on the cyclic quadrilateral.**

**MCQ Based on the definition of circle.****MCQ Based on the properties of circle.****MCQ Based on the cyclic quadrilateral.**__Features__

__Features__

**In the MCQ given below you find the important MCQ which are strictly according to the CBSE syllabus and are very useful for the CBSE Examinations.****Solution Hints are also given to some difficult problems.****Each MCQ contains four options from which one option is correct.**

__Action Plan__

__Action Plan__

**First of all students should Learn and write all basic points and Formulas related to the Chapter 10 Circle.****Start solving the NCERT Problems with examples.****Solve the important assignments on the Chapter 10 Class IX.****Then start solving the following MCQ.**

**MCQ |Chapter 10 | Circle | Class IX**

Question 1

The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is calleda) square

b) triangle

c) rectangle

d) circle

Answer d

Question 2

The centre of the circle lies in______ of the circle.

a) Interior

b) Exterior

c) Circumference

d) None of the above

Answer: a

Question 3

A circle divides the plane into __________ parts.

a) two

b) three

c) four

d) five

Answer b

Question 4

The longest chord of the circle is:

a) Radius

b) Arc

c) Diameter

d) Segment

Answer: c

Question 5

From the diagram given below, the shaded area is __________

a) major segment

b) minor segment

c) major sector

d) minor segment

Answer b

Question 6

How many circles can pass through three points which are not collinear?

a) One

b) Two

c) Three

d) Four

Answer a

Question 7

Equal _____ of the congruent circles subtend equal angles at the centres.

a) Segments

b) Radii

c) Arcs

d) Chords

Answer: d

Explanation: See the figure below:

Let Î”AOB and Î”COD are two triangles inside the circle.

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

AB = CD (Given)

So, Î”AOB ≅ Î”COD (SSS congruency)

∴ ∠AOB = ∠COD ……….. By CPCT

Hence, this prove the statement.

Question 8

The region between an arc and the two radii, joining the centre to the end points of the arc is called __________

a) arc

b) chord

c) sector

d) area of circle

Answer c

Question 9

When does both segments and both sectors become equal?

a) When two arcs are equal

b) When radius = 2 x chord

c) When two arcs are unequal

d) When chord = radius

Answer a

Question 10

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:

In triangles AOB and COD,

∠AOB = ∠COD (given)

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

So, Î”AOB ≅ Î”COD. (SAS congruency)

∴ AB = CD (By CPCT)

Question 11

From the diagram given below, if O is the centre of the circle and OL is perpendicular to chord AB, then which of the following is true?

a) AL > LB

b) AL < LB

c) AL ≠ LB

d) AL = LB

Answer d

Question 1**2**

a) 0

b) 1

c) 2

d) 3

Answer: a

Question 13

What is the value of AC if radius of circle is 5cm and OB = 3cm?

a) 5cm

b) 8cm

c) 4cm

d) 2cm

Question 1**4**

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

a) 90^{o}

b) 45^{o}

c) 180^{o}

d) 60^{o}

Answer: c

Explanation: Angle in a semi circle is always right

Question 1**5**

Find the value of AB if radius of circle is 10cm and PQ = 12cm and RS = 16cm?

a) 14cmb) 16cm

c) 4cm

d) 2cm

Answer d

Question 1**6**

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

Answer: b

Explanation:

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

OE = OE …………….. (Common)

∠OME = ∠ONE …… (Each = 90^{o})

So, Î”OEM ≅ Î”OEN ……… (by RHS congruence rule)

Hence, ∠MEO = ∠NEO …….. (by CPCT)

∴ ∠BEQ = ∠CEQ

Question 17

Find the radius of circle if AB = 7cm, RS = 6cm and PQ = 8cm.

a) 5cm

b) 8cm

c) 4cm

d) 20cm

Answer a

Explanation:

Let r be the radius of circle.

Since perpendicular from centre to a chord bisects the chord,

AR = RS/2 = 3cm and BP = PQ/2 = 4cm

In Î”BOP, ∠OBP = 90°

By Pythagoras theorem, OP^{2} = OB^{2} + BP^{2}

⇒ r^{2} = x^{2} + (4)^{2}

⇒ r^{2} = x^{2} + 16 ——————- (i)

Similarly, In Î”AOR, ∠OAR = 90°

By Pythagoras theorem, OR^{2} = OA^{2} + AR^{2}

⇒ r^{2} = (7 – x)^{2} + (3)^{2}

⇒ r^{2} = (7 – x)^{2} + 9 ———————— (ii)

From equation i and ii, x^{2} + 16 = (7 – x)^{2} + 9

⇒ x^{2} + 16 = 7^{2} – 2 x 7 x x + x^{2} + 9

⇒ 2 x 7 x x = 49 – 16 + 9

⇒ x = 3

Substituting value of x in equation i, r^{2} = 3^{2} + 16

⇒ r = 5cm.

Question 18

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:

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

Question 19

Two circles of radius 13cm and 17cm intersect at A and B. What is the distance between the centres of the circles if AB = 24cm?

b) 25cm

c) 22cm

d) 14cm

Answer d

Explanation:

Since line joining the centres of the circles bisects the common chord, AC = BC = 12cm.

Also, In Î”AOC, ∠OCA = 90°

By Pythagoras theorem, OA^{2} = OC^{2} + AC^{2}

⇒ 15^{2} = OC^{2} + 12^{2}

⇒ OC^{2} = 225 – 144

⇒ OC = 9cm

Similarly, In Î”AO’C, ∠O’CA = 90°

By Pythagoras theorem, O’A^{2} = O’C^{2} + AC^{2}

⇒ 13^{2} = O’C^{2} + 12^{2}

⇒ OC^{2} = 169 – 144

⇒ OC = 5cm

Now, OO’ = OC + O’C

⇒ OO’ = 9 + 5 = 14cm.

Question 20

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

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°

Question 21

In the given figure, find angle OPR.

a) 20°

b) 15°

c) 12°

d) 10°

Answer: d

Explanation: Major arc PR subtend Reflex ∠POR at the centre of the circle and ∠PQR in the remaining part of the circle

So, Reflex ∠POR = 2 × ∠PQR,

We know the values of angle PQR as 100°

So, Reflex ∠POR = 2 × 100° = 200°

∴ ∠POR = 360° – 200° = 160°

Now, in Î”OPR,

OP and OR are the radii of the circle

So, OP = OR

Also, ∠OPR = ∠ORP = x

By angle sum property of triangle, we know:

∠POR + ∠OPR + ∠ORP = 180°

160° + x + x = 180°

2x = 180° – 160°

2x = 20° ⇒ x = 10°

Question 22

In the given figure, ∠AOB = 90Âº and ∠ABC = 30Âº, then ∠CAO is equal to:

a) 30Âº

b) 45Âº

c) 60Âº

d) 90Âº

Answer: c

Explanation:

Given that ∠AOB = 90Âº and ∠ABC = 30Âº

OA = OB (Radii of the circle)

Let ∠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°

45° + 30° + ∠BAC = 180°

75° + ∠BAC = 180° ⇒ ∠BAC = 180° - 75°

⇒ ∠BAC = 105°

∠CAO + ∠OAB = 105°

∠CAO + 45° = 105°

⇒ ∠CAO = 105° - 45° = 60°

Question 23

a) 30Âº

b) 40Âº

c) 50Âº

d) 80Âº

Answer: c

Explanation:

Given that ABCD is a cyclic quadrilateral and ∠ADC = 140Âº.

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

Hence, ∠ADC + ∠ABC = 180°

140° + ∠ABC = 180°

∠ABC = 180° – 140° = 40°

Now ∠ACB = 90° . . . . . . . Angles in the semi circle is right

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

∠CAB + ∠ABC + ∠ACB = 180°

∠CAB + 40° + 90° = 180°

∠CAB + 130° = 180°

∠CAB = 180° – 130°

∠CAB = 50°

Therefore, ∠CAB or ∠BAC =50°.

Question 24

** **In the given figure, if ∠OAB = 40Âº, then ∠ACB is equal to

a) 40Âº

b) 50Âº

c) 60Âº

d) 70Âº

Answer: b

Explanation:

Given that ∠OAB = 40Âº,

In the triangle OAB,

OA = OB (radii)

∠OAB = ∠OBA = 40° …… (∴ angles opposite to equal sides are equal)

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

∠AOB + ∠OBA + ∠BAO = 180°

∠AOB + 40° + 40° = 180°

∠AOB = 180 – 80° = 100°

Since, 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

∠ACB = ½ ∠AOB

Hence, ∠ACB = 100°/2 = 50°.

Question 25

a) 10Âº

b) 20Âº

c) 40Âº

d) 60Âº

Answer: c

Explanation:

Given that, ∠ABC = 20Âº.

since the angle subtended by an arc at the centre of the circle is double the angle subtended at the remaining part.

∠AOC = 2∠ABC

∠AOC = 2 × 20° = 40°.

Question 26

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

Answer: a

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

AO^{2} = AC^{2} + OC^{2}

5^{2} = 4^{2} + OC^{2}

5^{2} − 4^{2} = OC^{2}

OC^{2} = 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.

Question 27

b) 45°

c) 60°

d) 20°

Answer: d

Explanation: Since ABCD is a cyclic quadrilateral, Sum of either pair of opposite angles of cyclic quadrilateral is 180°

⇒ ∠1 + ∠2 = 180°

⇒ 3k + 6k = 180° ⇒ k = 20°

Now, x = ∠1 [Exterior angle formed when one side of cyclic quadrilateral is produced is equal to the interior opposite angle]

⇒ x = 3k ⇒ x = 60°.

Question 28

If PQRS is a cyclic quadrilateral and PQ is diameter, find the value of ∠PQS.

a) 45°

b) 110°

c) 20°

d) 80°

Answer: c

Explanation: Since PQRS is a cyclic quadrilateral, Sum of either pair of opposite angles of cyclic quadrilateral is 180°. ⇒ ∠QRS + ∠QPS = 180°

⇒ ∠QPS = 70° ————–(i)

Also, ∠PSQ = 90° [Angle in a semicircle] ———-(ii)

In Î”PSQ, ∠PSQ + ∠SPQ + ∠SQP = 180° [Angle sum property of triangle]

⇒ 90° + 70° + ∠PQS = 180° [from equation i and ii]

⇒ ∠PQS = 20°.

Question 29

In the given figure, BC is the diameter of the circle and ∠BAO = 60Âº. Then ∠ADC is equal to

a) 30Âº

b) 45Âº

c) 60Âº

d) 120Âº

Answer: c

Explanation:

Given that ∠BAO = 60°

Since OA = OB, ∠OBA = 60°

Then ∠ADC = 60° ………. (Angles in the same segment of a circle are equal).

Question 30

If RS is equal to the radius of circle, find the value of ∠PTQ.

a) 90°

b) 60°

c) 45°

d) 30°

Answer: b

Explanation: Join OS, OR and RQ

From figure, ∠PRQ = 90° as angle in a semicircle is right angle.

Now, ∠QRT = 180° – ∠PRQ [Linear Pair]

⇒ ∠QRT = 180° – 90° = 90° ————–(i)

In Î”ROS, OS = OR = RS ⇒ Î”ROS is an equilateral triangle.

⇒ ∠ROS = 60°

Now, ∠SQR = ½ ∠ROS [angle subtended by an arc of circle at the centre is twice the angle subtended by the arc on circumference]

⇒ ∠SQR = 30° —————-(ii)

In Î”TQR, ∠TRQ + ∠QTR + ∠TQR = 180° [Angle sum property of triangle]

⇒ 90° + ∠PQR + 30° = 180° [from equation i and ii]

⇒ ∠PQR = 60°.

Question 31

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

a) 50Âº

b) 60Âº

c) 70Âº

d) 80Âº

Answer: c

Explanation:

∠DAB = 60Âº, ∠ABD = 50Âº

By using the angle sum property in the triangle ABD

∠DAB + ∠ABD + ∠ADB = 180Âº

60Âº + 50Âº + ∠ADB = 180Âº

∠ADB = 180Âº - 110Âº

∠ADB = 70Âº

∠ADB = ∠ACB ……… (Angles in the same segment of a circle are equal).

∠ACB =70Âº

Question 32

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

a) 30Âº

b) 45Âº

c) 60Âº

d) 90Âº

Answer: b

Explanation:

We know that the angle in a semi circle is = 90°

Hence, ∠ACB = 90°

Also, given that AC = BC

∠CAB = ∠CBA = x ….. (let) (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°

x + x + 90° = 180°

2x = 180° – 90° ⇒ 2x = 90° ⇒ x = 45°

∠CAB = x = 45°

Question 33

Find the value of ∠ABC if O is the centre of circle.

a) 60°

b) 120°

c) 240°

d) 180°

Answer: a

Explanation: Reflex ∠AOC = 240°

⇒ ∠AOC = 360° – Reflex ∠AOC = 360° – 240° = 120°

Since angle subtended by an arc of circle at the centre is twice the angle subtended by the arc on circumference, ∠AOC = 2∠ABC

⇒ ∠ABC = ½ ∠AOC = 60°.

Question 34

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

Answer: c

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:

AC^{2} = (CB)^{2} + (AB)^{2}

⇒ AC^{2} = (16)^{2} + (12)^{2}

⇒ AC^{2} = 256 + 144

⇒ AC^{2} = 400

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

Thus, the radius of the circle is 10 cm.

Question 35

Find the value of ∠QOR if O is the centre of circle.

a) 50°

b) 65°

c) 130°

d) 30°

Answer c

Explanation: Join OP

In Î”POQ, OP = OQ [Radii of same circle]

⇒ ∠OPQ = ∠OQP = 20° [Angles opposite to equal sides are equal]

In Î”POR, OP = OR [Radii of same circle]

⇒ ∠OPR = ∠ORP = 45° [Angles opposite to equal sides are equal]

Now, ∠QPR = ∠OPR + ∠OPQ = 45° + 20° = 65°

Angle subtended by an arc of circle at the centre is twice the angle subtended by the arc on circumference ⇒ ∠ROQ = 2∠QPR = 2 x 65° = 130°.

Question 36

a) 4 cm

b) 8 cm

c) 15 cm

d) 17 cm

Answer: b

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,

AO^{2} = OM^{2} + AM^{2}

OM^{2} = AO^{2}– AM^{2}

OM^{2}= 17^{2} – 15^{2}

OM^{2} = 64

OM = √64

OM = 8 cm

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