Question

Class 9 Maths MCQ Questions of Circles with Answers?

Answer 1

Class 9 Maths MCQ Questions of Circles are provided here with answers. These MCQ Questions of Circles are exceptionally valuable to amend terrifically significant ideas given in the section. Every one of the questions is exceptionally simple and basic which are intended to test your comprehension of the ideas and properties identified with circles. 

Free Online MCQ Questions for Class 9 Maths with Answers have been set up according to the most Latest syllabus and Exam Pattern for the Academic Session.  Multiple Choice Questions for a significant part of tests for class 9 Circle and whenever practiced appropriately can assist you with getting better marks.

Practice MCQ Questions for Class 9 Maths

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

(a) 45°
(b) 75°
(c) 90°
(d) 60°

2. Greatest chord of a circle is called its

(a) chord
(b) secant
(c) radius
(d) diameter

3. 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

4. A line that intersects a circle in two distinct points is a

(a) Secant
(b) Chord
(c) Radius
(d) Diameter

5. The center of the circle lies in______ of the circle.

(a) Interior
(b) Exterior
(c) Circumference
(d) None of the above

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

(a) Segments
(b) Radii
(c) Arcs
(d) Chords

7. 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

8. 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

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

(a) 90
(b) 45
(c) 180
(d) 60

10. Segment of a circle is the region between an arc and ………..of the circle.

(a) perpendicular
(b) radius
(c) chord
(d) secant

11. The region between chord and either of the arc is called

(a) a sector
(b) a semicircle
(c) a segment
(d) a quarter circle

12. The region between an arc and the two radii joining the centre of the end points of the arc is called a:

(a) Segment
(b) Semi circle
(c) Minor arc
(d) Sector

13.  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

14. 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

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

(a) 60°
(b) 30°
(c) 45°
(d) 55°

16. In the given figure, find angle OPR.

(a) 20°
(b) 15°
(c) 12°
(d) 10°

17. The degree measure of a semicircle is

(a) 0°
(b) 90°
(c) 360°
(d) 180°

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

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

(a) 50º
(b) 60º
(c) 70º
(d) 80º

20. A chord of a circle which is twice as long as its radius is a ____ of the circle

(a) Diameter
(b) perpendicular
(c) arc
(d) secant

Answer 2

1. Answer: (a) 45°

2. Answer: (d) diameter

3. Answer: (b) 8 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 =  1/2  AB 

AM =  1/2 (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

4. Answer: (a) Secant

5. Answer: (a) Interior

6. Answer: (d) Chords

Explanation: See in the given figure:

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.

7. Answer: (a) AB = CD

Explanation: 

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)

8. Answer: (a) 0

9. Answer: (c) 180

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

10. Answer: (c) chord

11. Answer: (c) a segment

12. Answer: (d) Sector

13. Answer: (b) ∠BEQ = ∠CEQ

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

14. Answer: (a) AB = CD

Explanation: In the given figure: 

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

15. Answer: (c) 45°

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°

16. Answer: (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°

17. Answer: (d) 180°

18. Answer: (a) AB = CD

19. Answer: (c) 70º

Explanation: Given that, ∠DAB = 60º, ∠ABD = 50º

By using the angle sum property in the triangle ABD

∠DAB+∠ABD+∠ADB=180º

60º+50º+∠ADB=180º

∠ADB=180º-110º

∠ADB=70º

∠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

20. Answer: (a) Diameter

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