Lines and Angles MCQs

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Class 9 Maths Chapter 6 Lines and Angles MCQs are provided below for the students as per the latest exam pattern of CBSE (2021-22) syllabus and NCERT Curriculum. Students can improve their problem-solving ability by practising the given objective-type questions. Students will also be able to score good marks by solving chapter-wise questions.

Some objective-type questions (MCQs) have been provided below from the chapter –Lines and Angles. Every question has four different answers. Students will have to find the correct answer out of the four. Students can cross-check their answer from their answer from the explanation and solved answers given below.

Also Read: NCERT Solutions for Class 7 Mathematics Chapter 5: Lines and Angles

Question:.When two lines intersect with each other, the vertically opposite angles formed are:

  1. Unequal
  2. Equal
  3. Cannot be determined
  4. None of the above

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Answer: b. Equal

Explanation: While two different lines are intersecting each other, then the angles formed on the opposite sides of the point of intersection are always equal.

Question:.When two lines are parallel, they:

  1. Intersect at one point
  2. Intersect at two points
  3. Intersect at three points
  4. Does not intersect at any point

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Answer: d. does not intersect at any point

Explanation: if two lines are parallel to each other, they will never intersect at any point. These lines are therefore, known as non-intersecting lines.

Question:.In the given figure, if ∠GED = 135° and AB || CD, EF CD, AGE will be:

  1. 120°
  2. 90°
  3. 135°
  4. 140°

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Answer: c. 135°

Explanation: Given, GE is transversal and AB is parallel to CD

∠GED = 135°

 ∠GED and ∠AGE are alternate interior angles; they will have same value.

So, ∠GED = ∠AGE = 135°

Question:.In a triangle, if the measure of an exterior angle is 105° and its opposite interior angles are equal. Find the value of these equal angles

  1. 72 ½ °
  2. 52 ½ °
  3. 75°
  4. 37°

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Answer: b. 52 ½

Explanation: Given, exterior angle = 105°

Let us consider the interior angles as x

Using the exterior angle theorem,

[Sum of the interior opposite angles = exterior angle]

2x = 105°

x = 52 ½ °. Hence, each of the interior opposite angles measure 52 ½ °.

Question:.Find the value of x, if AB is parallel to DE in the given figure:

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  1. 45°
  2. 25°
  3. 55°
  4. 35°

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Answer: d. 35°

Explanation: If BCF is considered a triangle and ∠B = 180°- 120° = 60°

∠F = 180°- 95° = 85°

∠C = x

Since sum of the three angles of a triangle is 180°,

Therefore, 60° + x + 85° = 180°

x = 180° - 60°

x = 35°

So, the value of x is 35°

Question:.If, ∠ MXQ = 135 ° and ∠ MYR = 40 °. PQ is parallel to RS. Find ∠ XMY

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  1. 90°
  2. 60°
  3. 85°
  4. 55°

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Answer: c. 85°

Explanation: Sum of interior angles on same side is 180°.

Hence, ∠MXQ + ∠ XMB = 180 °

∠ XMB = 180 ° - 135°

∠ XMB = 45°

∠BMY = ∠ MYR = 40°  [alternate interior angles are equal]

So, ∠ XMB + ∠BMY = ∠XMY

= 45° + 40°

= 85°

Hence, the value of ∠ XMY = 85°

Question: If the ratio of the angles 5:3:7. The triangle will be:

  1. An isosceles triangle
  2. A right triangle
  3. An obtuse-angled triangle
  4. An acute- angled triangle

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Answer: c. an acute-angled triangle.

Explanation: It is known that, the sum of the interior angles of a triangle is 180°

Let is consider the angles as 5x, 3x and 7x

Hence, 5x + 3x +7x = 180°

15x = 180°

x = 180°/5 = 12°

So, 5x = 5 x 12° = 60°

3x = 3 x 12° =36°

7x = 7 x 12° = 84 °

Here, all angles are less than 90°. Hence, it is clear that the triangle is an acute angled triangle.

Question: If the ratio of the angles 2:4:3. The value of the smallest angle will be:

  1. 40°
  2. 80°
  3. 60°
  4. 20 °

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Answer: a. 40°

Explanation: let us consider 2:4:3 as 2x, 4x and 3x

So, 2x + 4x +3x = 180° [the sum of the interior angles of a triangle is 180°]

9x = 180°

x = 20°

Hence, the value of:

2x = 2(20°) = 40°

4x = 4(20°) = 80°

3x = 3 (20°) = 60°

So, the smallest angle is 40°.

Question: Find the value of x from the given figure, where POQ is a line.

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  1. 20°
  2. 30°
  3. 25°
  4. 35° 

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Answer: a. 20°

Explanation:  Given POQ is a line, which means POQ = 180°.

40° + 4x + 3x = 180°

40° + 7x = 180°

7x = 180° - 40°

7x = 140°

 x = 140°/7

x = 20°

So, x = 20°

Question:  If AOB is a line then the measure of BOC, COD and DOA respectively in the given figure, are:

MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with Answers 12

  1. 36°, 54°, 90°
  2. 90°, 54°, 36°
  3. 90°, 36°, 54°
  4. 36°, 90°, 54° 

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Answer: a. 36°, 54°, 90°

Explanation:  ∠AOD + ∠ DOC + ∠ COR = 180°   [sum of the interior angles of a triangle is 180°]

5y + 3y + 2y = 180°   

10y =180°   

y = 180° / 10

y = 18°   

So, the values of 5y, 3yand 2y are:

5y = 5 (18°) = 90°

3y = 3 (18°) = 54°

2y = 2 (18°) = 36°

Question:  Find the value of x, if B lies on AC. Given AB = x + 3, AC = 4x – 5 and BC = 2x.

  1. 5
  2. 8
  3. 3
  4. 2

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Answer: b. 8

Explanation: Given that B lies on AC

Hence, AB + BC = AC

 (x + 3) + (2x) = 4x – 5

3x + 3 = 4x – 5

3x – 4x = -5 +3

-x = - 8

x = 8

Hence, the value of x is 8. 

Question: The interior opposite angles of a triangle are in the ratio 1: 3. Given that, an exterior angle of the triangle is 80°. Find the value of interior opposite angles.

  1. 40°, 120°
  2. 30°, 60°
  3. 20°, 60°
  4. 90°, 60°

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Answer: c. 20°, 60°

Explanation: Let the third angle of the triangle be y

Hence, y + 80° = 180°

          y = 100°

Let the other two angles (1:3) be 1x and 3x

1x + 3x + 100° =180°    …..[sum of all interior angles is 180°]

4 x = 180° - 100°

4 x = 80°

x = 80°/4

x = 20°

Hence, the value of 3x = 3(20°) = 60°

So, the angles are 20° and 60°

Also Check:

CBSE X Related Questions

1.

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

Length (in mm)

Number of leaves

118 - 126

3

127 - 135 

5

136 - 144

9

145 - 153

12

154 - 162

5

163 - 171

4

172 - 180

2

Find the median length of the leaves. 
(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

      2.
      If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos2 A – sinA or not

          3.
          A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

              4.

              A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

                  5.
                  An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
                  An umbrella has 8 ribs which are equally spaced

                      6.
                      Find the sums given below :
                      1. \(7 + 10\frac 12+ 14 + ....... + 84\)
                      2. \(34 + 32 + 30 + ....... + 10\)
                      3. \(–5 + (–8) + (–11) + ....... + (–230)\)

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