Explore Exams
Login Signup
Sign in Open App
CBSE Class 9  >  Class 9 Test  >  Mathematics Olympiad  >  Math Olympiad Test: Circles- 1 - Class 9 MCQ

Math Olympiad Test: Circles- 1 - Free MCQ with solutions Class 9 Mathematics


MCQ Practice Test & Solutions: Math Olympiad Test: Circles- 1 (10 Questions)

You can prepare effectively for Class 9 Mathematics Olympiad for Class 9 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Math Olympiad Test: Circles- 1". These 10 questions have been designed by the experts with the latest curriculum of Class 9 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 10

Sign up on EduRev for free to attempt this test and track your preparation progress.

Math Olympiad Test: Circles- 1 - Question 1
Save

The largest chord of a circle is called its:

Detailed Solution: Question 1

The largest chord of a circle is called diameter of the circle.

Any chord which crosses the centre of the circle is the largest chord of the circle, which is the diameter of the circle.

Math Olympiad Test: Circles- 1 - Question 2
Save

AB and CD are two parallel chords of a circle such that AB = 8 cm, and CD = 6cm. If the chords are on the opposite sides of the center and the distance between them is 7cm, then the diameter of the circle is:

Detailed Solution: Question 2

Let the radius of the circle be r, and the length OP be x

∴ In ∆OPB, 
OP2 + PB2 = OB2 = r2
⇒ x2 + (4)2 = r2   [∵ PB = AB/2]
⇒ x2 = r2 – 16 ….(i) 
In ∆OCQ,
OQ2 + CQ2 = r2
⇒  (7 – x)2 + (3)2 = r2
[∵ OQ = PQ - OP = 7 - x]
⇒ (7 – x)2 = r2 – 9 ….(ii)
Subtracting (ii) from (i), we get
x2 – (7 – x)2 = – 7
⇒ (x – 7 + x) (x + 7 – x) = – 7 
⇒    (2x – 7) (7) = – 7  
⇒  (2x – 7) = – 1 
⇒ x = 3
∴ r2 = x2 + 16 [using (i)]
= (3)2 + 16 = 25
⇒ r = 5cm 
∴ d = 10 cm. 

Clear all your doubts with EduRev
Math Olympiad Test: Circles- 1 - Question 3
Save

Two circles of radii 13cm and 15cm intersect and the length of the common chord is 24 cm, then the distance between their centers is: 

Detailed Solution: Question 3

In ∆OPA, 


Similarly, 
In ∆O′PA,

= 9cm. 
∴ OO′ = OP + O′P = 5 + 9 = 14 cm.

Math Olympiad Test: Circles- 1 - Question 4
Save

Bisector AD of ∠BAC of ∆ABC passes through the center of the circumcircle of ∆ABC, then,

Detailed Solution: Question 4

∵ AD is the bisector of ∠BAC

∴ AD is the ⊥ bisector of BC. 
∵ O is the circumcentre of ∆ABC.
∴ AB = AC [∵ ∠ABD = ∠ACB]
(By using ∆ABD ≅ ∆ACD)

Math Olympiad Test: Circles- 1 - Question 5
Save

Find the measure of ∠ABC.

Detailed Solution: Question 5

Reflex ∠AOC = 360° – (110° + 120°) 
= 130° 

Math Olympiad Test: Circles- 1 - Question 6
Save

The measure of ∠BOC is

Detailed Solution: Question 6

∠BOC = 2 × ∠BAC [Angle subtended at center is double the angle subtended the circle]
= 2 × 30° = 60°

Math Olympiad Test: Circles- 1 - Question 7
Save

P is the center of the circle, and ∠XPZ = 120°, ∠XZY = 35°, then the measure of ∠YXZ is: 

Detailed Solution: Question 7


∠XPY = 2∠XOP
∵ ∠XOP = ∠XZY
(angles in the same segment)
∴ ∠XPY = 2∠XZY …(i)
Similarly, 
∠YPZ = 2∠YXZ …(ii)
Using (i) and (ii)
∠XPZ = 2 (∠XYZ + ∠YXZ)

Math Olympiad Test: Circles- 1 - Question 8
Save

PQRS is a cyclic quadrilateral such that PR is the diameter of the circle. If ∠QPR = 64° and ∠SPR = 31°, then, ∠R =?

Detailed Solution: Question 8


∠P = ∠QPR + ∠SPR
= 64 ° + 31° 
= 95°
∵ ∠Q and ∠S are angles of the semicircle
∴ ∠Q = ∠S = 90° 
∵ PQRS is a cyclic quadrilateral
∴ ∠P + ∠R = 180°
⇒ ∠R = 180° - 95° = 85°

Math Olympiad Test: Circles- 1 - Question 9
Save

If A, B, C are three points on a circle with center O such that ∠AOB = 90° and ∠BOC = 120°, then ∠ABC =?

Detailed Solution: Question 9


Reflex ∠AOC = 360° – (90° + 120°) = 150° 

Math Olympiad Test: Circles- 1 - Question 10
Save

O is the center of the circle, with AC = 30 cm and DA = 10√5 cm, then the measure of DC is 

Detailed Solution: Question 10

In ∆ACD, 
∠ADC = 90° [∵∠ACD is angle in semicircle]
∴  AC2 = DA2 + DC2
⇒ (30)2 = (10√5 )2 + DC2
⇒ DC2 = 900 – 500 
⇒ DC = √400 = 20 cm. 

23 videos|105 docs|45 tests
Join Course
Information about Math Olympiad Test: Circles- 1 Page
In this test you can find the Exam questions for Math Olympiad Test: Circles- 1 solved & explained in the simplest way possible. Besides giving Questions and answers for Math Olympiad Test: Circles- 1, EduRev gives you an ample number of Online tests for practice
Download as PDF

About this Class 9 Practice Test

This test contains 10 MCQs on Math Olympiad Test: Circles- 1 with detailed solutions for each question. It is part of the Mathematics Olympiad for Class 9 course on EduRev, designed to align with the current Class 9 syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

Explore more Class 9 Study Resources

If you found this Math Olympiad Test: Circles- 1 MCQ Test helpful, you can further enhance your Class 9 preparation with in-depth courses on EduRev, for all the subjects and topics available in Mathematics Olympiad for Class 9 course. The course offers:
  • Video Lectures: covering all the topics of Mathematics Olympiad for Class 9
  • Notes for Revision: offering revision notes, important questions, summaries, flashcards, cheatsheets & more for Class 9 preparation
  • MCQ Test: variety of questions as per the pattern of Class 9 exam
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup