Olympiad Test: Lines And Angles - 1 - Class 7 MCQ

# Olympiad Test: Lines And Angles - 1 - Class 7 MCQ

Test Description

## 10 Questions MCQ Test Mathematics Olympiad Class 7 - Olympiad Test: Lines And Angles - 1

Olympiad Test: Lines And Angles - 1 for Class 7 2024 is part of Mathematics Olympiad Class 7 preparation. The Olympiad Test: Lines And Angles - 1 questions and answers have been prepared according to the Class 7 exam syllabus.The Olympiad Test: Lines And Angles - 1 MCQs are made for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Olympiad Test: Lines And Angles - 1 below.
Olympiad Test: Lines And Angles - 1 - Question 1

### What is the supplement of 64°?

Detailed Solution for Olympiad Test: Lines And Angles - 1 - Question 1

180° – 64° = 116°

Olympiad Test: Lines And Angles - 1 - Question 2

### What is the value of x in the given figure?

Detailed Solution for Olympiad Test: Lines And Angles - 1 - Question 2

64 + x + 93 = 180
⇒ 157 + x = 180°
⇒ x = 180° – 157° = 23°

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Olympiad Test: Lines And Angles - 1 - Question 3

### In the given figure two straight lines AB and PQ intersect at a point O. If ∠AOC = 42° what is the measure of ∠BOD?

Detailed Solution for Olympiad Test: Lines And Angles - 1 - Question 3

∠AOC = 42°
∠BOD = ∠AOC [ Vertically opposite ]
∠BOD = 42°

Olympiad Test: Lines And Angles - 1 - Question 4

What is the complement of 67°?

Olympiad Test: Lines And Angles - 1 - Question 5

​In the following figure, if AB ∥ CD, ∠ BAP = 108° and ∠ PCD = 120°, what is the measure of ∠ APC?

Detailed Solution for Olympiad Test: Lines And Angles - 1 - Question 5

AB ∥ PQ
∴ ∠APQ = 180° – 108° = 72°
PQ ∥CD
∠ CPQ = 180° –120° = 60°
∠ APC= ∠APQ+ ∠CPQ = 72° + 60°
= 132°

Olympiad Test: Lines And Angles - 1 - Question 6

In the given figure MN ∥ PQ ∠MNE = 120°, ∠EPQ = 100°.What is the value of x?

Detailed Solution for Olympiad Test: Lines And Angles - 1 - Question 6

Given, MN ∥PQ
∠PON = 100°
∴ ∠NOE = 180° – 100 = 80°
∠ENO = 180° – 120° = 60°
x = 180° – (80° + 60°)
= 180° –140° = 40°

Olympiad Test: Lines And Angles - 1 - Question 7

In the given figure AB ∥ CD and EF is transversal. If ∠1 and ∠2 are in the ratio 5:7 what is the measure of ∠8?

Detailed Solution for Olympiad Test: Lines And Angles - 1 - Question 7

5x + 7x = 180°
⇒ 12x = 180° ⇒
∠1 = 5x = 5×15 = 75°
∠2 = 7x = 7×15 = 105°
∠1 = ∠5 = 75°
∠8 = 180° – 75° = 105°

Olympiad Test: Lines And Angles - 1 - Question 8

What is the value of x in the following figure?

Detailed Solution for Olympiad Test: Lines And Angles - 1 - Question 8

Here, 2x –10° + 3x + 20° = 180°
⇒ 5x + 10° = 180°
⇒  5x = 170°

Olympiad Test: Lines And Angles - 1 - Question 9

Find the angle which is equal to its complement

Detailed Solution for Olympiad Test: Lines And Angles - 1 - Question 9

Olympiad Test: Lines And Angles - 1 - Question 10

In the given figure, ABC is a straight line. What is the value of y?

Detailed Solution for Olympiad Test: Lines And Angles - 1 - Question 10

In ∆ABP
∠ ABP = 180° –(26°+38°)
= 180° – 64 = 116°
∠PBC = 180° – 116° = 64°
PB = PC
∠PBC = 64°
y = 180° – (64° + 64°)
= 180° –128°
= 52°

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