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Class 7 Exam  >  Class 7 Tests  >  Mathematics Olympiad Class 7  >   Olympiad Test: Triangle - 2 - Class 7 MCQ

Olympiad Test: Triangle - 2 - Class 7 MCQ


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10 Questions MCQ Test Mathematics Olympiad Class 7 - Olympiad Test: Triangle - 2

Olympiad Test: Triangle - 2 for Class 7 2024 is part of Mathematics Olympiad Class 7 preparation. The Olympiad Test: Triangle - 2 questions and answers have been prepared according to the Class 7 exam syllabus.The Olympiad Test: Triangle - 2 MCQs are made for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test: Triangle - 2 below.
Solutions of Olympiad Test: Triangle - 2 questions in English are available as part of our Mathematics Olympiad Class 7 for Class 7 & Olympiad Test: Triangle - 2 solutions in Hindi for Mathematics Olympiad Class 7 course. Download more important topics, notes, lectures and mock test series for Class 7 Exam by signing up for free. Attempt Olympiad Test: Triangle - 2 | 10 questions in 20 minutes | Mock test for Class 7 preparation | Free important questions MCQ to study Mathematics Olympiad Class 7 for Class 7 Exam | Download free PDF with solutions
Olympiad Test: Triangle - 2 - Question 1
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In a ∆ABC, ∠A – ∠B = 33° and ∠B – ∠C = 18°, then ∠B = ?

Detailed Solution for Olympiad Test: Triangle - 2 - 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 for Olympiad Test: Triangle - 2 - 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?

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 for Olympiad Test: Triangle - 2 - 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 for Olympiad Test: Triangle - 2 - 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 = ?
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 for Olympiad Test: Triangle - 2 - 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
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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