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CBSE Class 7  >  Class 7 Test  >  Mathematics Olympiad   >   Olympiad Test: Triangle - 2 - Class 7 MCQ

Olympiad Test: Triangle - 2 - Free MCQ with solutions Class 7 Mathematics


MCQ Practice Test & Solutions: Olympiad Test: Triangle - 2 (10 Questions)

You can prepare effectively for Class 7 Mathematics Olympiad Class 7 with this dedicated MCQ Practice Test (available with solutions) on the important topic of " Olympiad Test: Triangle - 2". These 10 questions have been designed by the experts with the latest curriculum of Class 7 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 20 minutes
  • - Number of Questions: 10

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Olympiad Test: Triangle - 2 - Question 1
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In a ∆ABC, ∠A – ∠B = 33° and ∠B – ∠C = 18°, then ∠B = ?

Detailed Solution: Question 1

∠A = 33° + ∠B and ∠C = ∠B – 18°
∴ (33° + ∠B) + ∠B + (∠B – 18°) = 180° 
⇒ 3∠B = 165° 17. (d) ⇒ ∠B = 55°

Olympiad Test: Triangle - 2 - Question 2
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The sum of all angles of a triangle is

Detailed Solution: Question 2

The sum of all the angles of a triangle is 180°.

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Olympiad Test: Triangle - 2 - Question 3
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In the given figure, what value of x will make AOB a straight line?

Detailed Solution: Question 3

To make AOB a straight line, the total angle at point O must be 180°
From the figure, we can see three angles on the straight line:
∠AOC = 55º
∠BOD = 45º
∠COD = x
∠AOC + ∠COD + ∠BOD = 180º (Linear Pair)
55º + x + 45º = 180º
x + 100º = 180º
x = 180º - 100º
x = 80º

So, Correct Option:(B)

Olympiad Test: Triangle - 2 - Question 4
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In the given figure, AOB is a straight line,
∠AOC = (3x – 8)°, ∠COD = 50° and ∠BOD = (x + 10)°. The value of x is

Detailed Solution: Question 4

Here, (3x – 8)° + 50° + (x + 10)° = 180° 
⇒ 4x + 52 = 180° 
⇒ 4x = 180 – 52
⇒ 
⇒ x = 32

Olympiad Test: Triangle - 2 - Question 5
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In ∆ABC, ∠B = 90°, AB = 5cm and AC = 13cm then BC = ? 

Detailed Solution: Question 5

By Pythagoras theorem,
AB2 + BC2 = AC2 ⇒ 52 + BC2 = 132

= 12 cm 

Olympiad Test: Triangle - 2 - Question 6
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∆ABC is an isosceles triangle with ∠C = 90° and AC = 5 cm then AB = ?

Detailed Solution: Question 6

∆ABC is an isosceles right triangle with ∠C = 90° and AC = 5 cm.

In an isosceles right triangle, the two sides forming the right angle are equal.
So, AC = BC = 5 cm

By Pythagoras theorem,
AB2 = AC2 + BC2
AB2 = 52 + 52
AB2 = 25 + 25
AB2 = 50

AB = √50 = 5√2 cm

Answer: a) 5√2 cm

Olympiad Test: Triangle - 2 - Question 7
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The angles of a triangle are (3x)°, (2x – 7)° and (4x – 11)°. Then x = ?

Detailed Solution: Question 7

According to question
3x + (2x – 7) + (4x – 11) = 180°
⇒ 9x – 18 = 180
⇒ 9x = 198
⇒ x = 22

Olympiad Test: Triangle - 2 - Question 8
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The supplement of 45° is

Olympiad Test: Triangle - 2 - Question 9
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An angle is one – fifth of its supplement. The measure of the angle is

Detailed Solution: Question 9

Let the angle be x° and its supplement be 180° − x°.

Given x = one-fifth of its supplement
So, x = (1/5)(180 − x).

Multiply both sides by 5:
5x = 180 − x.

Bring like terms together:
5x + x = 180
6x = 180.

So x = 180 ÷ 6 = 30°.
This corresponds to option D.

Olympiad Test: Triangle - 2 - Question 10
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In the given figure, AOB is a straight line, ∠AOC = 68° and ∠BOC = x°, the value of x is.

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About this Class 7 Practice Test

This test contains 10 MCQs on Olympiad Test: Triangle - 2 with detailed solutions for each question. It is part of the Mathematics Olympiad Class 7 course on EduRev, designed to align with the current Class 7 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.

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