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Class 9 Exam  >  Class 9 Tests  >  Math Olympiad Test: Lines and Angles- 2 - Class 9 MCQ

Math Olympiad Test: Lines and Angles- 2 - Class 9 MCQ


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15 Questions MCQ Test - Math Olympiad Test: Lines and Angles- 2

Math Olympiad Test: Lines and Angles- 2 for Class 9 2025 is part of Class 9 preparation. The Math Olympiad Test: Lines and Angles- 2 questions and answers have been prepared according to the Class 9 exam syllabus.The Math Olympiad Test: Lines and Angles- 2 MCQs are made for Class 9 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Math Olympiad Test: Lines and Angles- 2 below.
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Math Olympiad Test: Lines and Angles- 2 - Question 1
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Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the larger angle is:

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

Let the measure of angle be x.
∴ Its complement = (90° - x)
∴  2x = 3(90° - x)
⇒  5x = 3 × 90°
⇒ x = 54°
and  (90° - x ) = 36°
∴ Measure of larger angle = 54°

Math Olympiad Test: Lines and Angles- 2 - Question 2
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In the figure, OA and OB are opposite rays ∠AOC + ∠BOD = 63°. The measure of angle ∠COD is;

Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 2

∵ OA and OB are opposite rays.
∴ ∠AOB is a straight angle.
⇒ ∠AOB = 180°
⇒ (∠AOC + ∠BOD) + ∠COD = 180°
⇒ 63° + ∠COD = 180°
⇒ ∠COD = 117°

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Math Olympiad Test: Lines and Angles- 2 - Question 3
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Determine the value of x from the given figure;

Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 3

∠AOB = 180°
⇒ ∠AOP + ∠POB = 180°
⇒ ∠AOP + ∠POQ + ∠BOQ = 180°
⇒ 90° + 3x + x = 180°
⇒  4x = 90° 
⇒ x = 90°/4 = 22.5°

Math Olympiad Test: Lines and Angles- 2 - Question 4
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b = a + 20°, then a =
Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 4

∵ Sum of angles around a point = 360°
∠AOB + ∠BOC = 360°
⇒ 90° + a + b = 360°
⇒ a + b = 270° ...(i)  and 
b = a + 20° ...(ii)
Using (ii) in (i)
a + (a + 20°) = 270°
⇒ 2a = 250°
⇒ a = 125°

Math Olympiad Test: Lines and Angles- 2 - Question 5
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The value of m is
Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 5

∠m = ∠x  [Vertically opposite ∠s]
∵ ∠AOB = 180°
⇒ ∠BOF + ∠COF + ∠AOC = 180°
⇒ ∠BOF + ∠DOE +∠AOC = 180°
⇒ x° + 2x° + 3x° = 180°
⇒  6x° = 180°
⇒ x = 30° 
∴ m = 30°

Math Olympiad Test: Lines and Angles- 2 - Question 6
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Find x from the figure.
Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 6

Here ∠COQ = ∠POD   [vertically opposite ∠s]
∵ ∠AOB = 180°   (AOB is a straight line)
⇒ ∠POA + ∠POD + ∠BOD = 180°
⇒ 2x° + 3x° + 20° + 3x° = 180°
⇒ 8x = 160°
⇒ x = 20°

Math Olympiad Test: Lines and Angles- 2 - Question 7
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x = 3y = 6/7z, then, find the value of y.
Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 7

x = 3y, z = 21/6 y = 7/2y.
∴ x + y + z = 180°
⇒ 3y + y + 7/2 y = 180° 
⇒ 4y + 7/2 y = 180°
⇒ 15y = 180° × 2
⇒ y = 24°

Math Olympiad Test: Lines and Angles- 2 - Question 8
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Find x from the adjoining figure.
Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 8

∠AOD = ∠BOC [vertically opposite ∠s]
⇒ y = 60° + x   ...(i) and,
∵ ∠DOC = 180°
⇒ 60° + x + 100° = 180°
⇒ x = 20°

Math Olympiad Test: Lines and Angles- 2 - Question 9
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In the figure, l ∥ m, Find k.
Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 9

It is clear from the figure that,
105° + 5x = 180°
⇒   5x = 75°
⇒ x = 15°

Math Olympiad Test: Lines and Angles- 2 - Question 10
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AB || CD, and ∠RQB = 115°, and ∠PRQ = 30°. The measure of ∠APC is:
Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 10

Here ∠RQB + ∠RQP =180°   (∵ AB is a straight line)
⇒ ∠RQP = 180° - 115° = 65°
Now ∠PRQ = 30°
∵ ∠PRQ, ∠RQP and ∠APQ are the ∠S of D
∴ ∠PRQ + ∠RQP + ∠APQ = 180°
⇒ ∠APQ = 180°- 65°- 30° = 85°
∠APC = ∠APQ [Vertically opposite ∠S]
∴ ∠APC = 85°

Math Olympiad Test: Lines and Angles- 2 - Question 11
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If x : y = 2 : 3, then the value of y is equal to:
Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 11

∵  X : Y = 2 : 3 

∴ Let the angles x and y be 2k and 3k respectively
∴ 2k + 3k = 180° 
[Sum of ∠S in the interior of transversal]
⇒ 5k = 180° ⇒ k = 36°
∴ y = 3k = 3 × 36° = 108°

Math Olympiad Test: Lines and Angles- 2 - Question 12
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AB || CD || EF and GH || KL. The measure of ∠HKL is
Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 12

The figure below is an extended version of the question figure, where we have added extra points and lines to make the angle calculation easier.

Extending GH to M, we have,

∠CHG = ∠APH = 60° [Corresponding ∠S]
∠APH = ∠MPD = 60°  [Vertically opposite ∠S]
∠APH = ∠MPD, then,
∠MPD + ∠LKP = 180° [Sum of interior ∠S]
⇒ ∠LKP = 180° - 60° = 120°, also
∠KHD = ∠PKH = 25° (Alternate ∠S)
∴ ∠HKL = ∠LKP + ∠PKH
= 120° + 25° = 145° 

Math Olympiad Test: Lines and Angles- 2 - Question 13
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Find x from the given figure (CP || DQ):
Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 13

Construct a line l || AB,

∠a = 25° [ Alternate ∠S]
∠c = 105°
∴ ∠b = 105° [Vertically opposite ∠S]
∴ x = a + b = 25° + 105° = 130°

Math Olympiad Test: Lines and Angles- 2 - Question 14
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AB || CD and PQ || RS, then x - y =
Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 14

∠DD1S = ∠B1D1C1 = 2x   [Vertically opposite ∠S]
∠B1D1C1 + ∠A1B1D1 = 180° [Interior ∠S]
⇒  2x + 2y = 180°
⇒ x + y = 90° ...(i)
Also,
∠AA1C1 + ∠A1B1D1  [Corresponding ∠S]
⇒ 30° = 2y
⇒ y = 15° ...(ii)
Using (ii) in (i), we get
x = 90° - y = 90° - 15° = 75° 

Math Olympiad Test: Lines and Angles- 2 - Question 15
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The value of y, if AB || PQ is
Detailed Solution for Math Olympiad Test: Lines and Angles- 2 - Question 15

 From the figure,

(2y + y + y) + 35° = 180° (Interior ∠S)
⇒ 4y = 180° - 35°
⇒ 4y = 145°
⇒ y = 145° ÷ 4
⇒ y = 36.25°

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