CBSE Class 9 Maths MCQ Questions of Circles are furnished here with answers. These questions 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. Along these lines, if you need to get ready for your Maths test in a simple and successful way at that point attempt to solve every one of the questions given here. You can likewise check here some other significant assets to help you in preparations for your CBSE Class 9 Maths Exam.
Free Online MCQ Questions for Class 9 Maths with Answers was Prepared Based on the Latest Exam Pattern for the Academic Session. students can solve NCERT Class 9 Maths: Circles Multiple Choice Questions with Answers to realize their arrangement level. CBSE Class 9 Maths Circles MCQ Questions with answers accessible. The MCQ Questions for Class 9 Circle with answers have been set up according to the most recent syllabus, NCERT books, and proposed in class 9 by CBSE, NCERT. Multiple Choice Questions for a significant part of tests for class 9 Circle and whenever practiced appropriately can assist you with getting better marks.
Tackle the given underneath MCQ Questions for Class 9 score more in Class Exam.
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:
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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