RBSE Solutions for Class 7 Maths Chapter 7 Lines and Angles Additional Questions is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 7 Lines and Angles Additional Questions.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 7 |

Chapter Name |
Lines and Angles |

Exercise |
Ex 7.2 |

Number of Questions |
19 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 7 Lines and Angles Additional Questions

**Multiple Choice Questions**

Question 1

Supplementary (RBSESolutions.com) angle of 87° is :

(A) 103°

(B) 93°

(C) 180°

(D) 90°

Question 2

Sum of two supplementary angles will be :

(A) 90°

(B) 180°

(C) 270°

(D)none

Question 3

Sum of two (RBSESolutions.com) complementary angles will be :

(A) 90°

(B) 180°

(C) 270°

(D) 360°

Question 4

An angle is 60° then its complementary angle will be :

(A) 60°

(B) 120°

(C) 30°

(D) 90°

Question 5

Sum of internal angles made by (RBSESolutions.com) a transversal on its one side will be :

(A) 180°

(B) 90°

(C) 270°

(D) 360°

Question 6

The value of x° in the (RBSESolutions.com) following figure :

(A) 150°

(B) 130°

(C) 150°

(D) 90°

Answers:

1. (B), 2. (B), 3. (A), 4. (C), 5. (A), 6. (A)

**Fill in the blanks**

(i) Two straight lines (RBSESolutions.com) intersect each other at ……… point.

(ii) A traversal line intersects two or more straight lines at ……….. points.

(iii) All straight lines intersecting at one point are called …………. lines.

(iv) If ∠2 and ∠4 are corresponding angles then both angles are ………….

Answers:

(i) one

(ii) different

(iii) concurrent

(iv) equal

True/False:

(i) Sum of complementary (RBSESolutions.com) angles are 180°.

(ii) Sum of supplementary angles are 180°

(iii) Two adjesent angles are supplementary.

(iv) An acute angle is not an adjecent angle of obtuse angle.

Answers:

(i) F

(ii) T

(iii) T

(iv) F

**Very Short Answers Type Questions**

Question 1

What is the measure of complement (RBSESolutions.com) angle of each of the following angles :

(i) 45°

(ii) 65°

(iii) 41°

(iv) 54°

Solution:

We know that sum of two complementary angles are 90°. Therefore

(i) Measure of complementary angle of 45°

= (90° – 45°) = 45°

(ii) Measure of complementary angle of 65°

= (90° – 65°) = 25°

(iii) Measure of complementary angle of 41°

= (90° – 41°) = 49°

(iv) Measure of complementary angle of 54°

= (90° – 54°) = 36°

Question 2

What will be the (RBSESolutions.com) measure of supplement angle of the following angles :

(i) 100°

(ii) 90°

(iii) 55°

(iv) 125°

Solution:

We know that sum of two supplementary angles are 180°. Therefore

(i) Supplementary angle of 100°

= (180° – 100°) = 80°

(ii) Supplementary angle of 90°

= (180° – 90°) = 90°

(iii) Supplementary angle of 55°

= (180° – 55°) = 125°

(iv) Supplementary angle of 125°

= (180° – 125°) = 55°

Question 3

What can you say about the internal (RBSESolutions.com) opposite angles in each case if external angle is a :

(i) a right angle

(ii) an obtuse angle

(iii) an acute angle

Solution:

(i) If external angle is a right angle then sum of internal opposite angle will also be 90°. it means each angle will be ap acute angle.

(ii) If external angle is an obtuse angle then out of two internal opposite angles either both are acute angle or one is obtuse angle and other is acute angle.

(iii) If external angle is an acute angle than each internal opposite angle will be acute angle.

**Short Answer Type Questions**

Question 1

Find the example from (RBSESolutions.com) your surroundings where lines interseced at right angles.

Solution:

There are following examples from our surroundings where lines intersecting at right angles are:

(i) Edge of the black board.

(ii) Angle between top of the table with its legs.

(iii) Edges of the carrum board.

(iv) Edge of the papersheet.

Question 2

Find the measure of angles made by intersecting lines on (RBSESolutions.com) the vertices of an equilateral triangle.

Solution:

Let ABC is an equilateral triangle.

∵ All angles of equilateral triangle are equal.

∴ ∠A =∠B =∠C = x (Let)

⇒ x + x + x = 180°

⇒ 3x = 180°

⇒ x = 60°

∴ ∠A =∠B =∠C = 60°

Question 3

Find the complement angle of (RBSESolutions.com) the following angles :

Solution:

Sum of complements of each other is 90°

(i) Complement angle of 20° = 90°- 20° = 70°

(ii) Complement angle of 63° = 90° – 63° = 27°

(iii) Complement angle of 57° = 90° – 57° = 33°

**Long Answer Type Questions**

Question 1

Find the value of x, y and z in each (RBSESolutions.com) of the following :

Solution:

(i) Two lines intersects (RBSESolutions.com) each other

∴ ∠x = ∠55°, (Vertically opposite angles)

∴ x + z = 180° (Linear pair angle)

⇒ 55° + ∠z = 180°

⇒ ∠z = 180°- 55° = 125°

Clearly ∠z = ∠y, (Vertically opposite angles)

⇒ ∠y = 125°

∠x = 55°, ∠y = 125°

and ∠z = 125°.

(ii) 40° + ∠x + 25° = 180° (Straight angle)

⇒ 65 + ∠x = 180°

⇒ ∠x = 180° – 65° = 115°

and ∠y + 40° = 180°, (Linear pair angle)

⇒ ∠y= 180° – 40° = 140°

and ∠y + ∠z = = 180°, (Linear pair angle)

∠z = 180°- ∠y

= 180°- 140° = 40°

∴ ∠x = 115°, ∠y = 140° and ∠z = 40°

Question 2

Find the unknown angle (RBSESolutions.com) from the following figure if p ॥ q.

Solution:

∠e + 125° = 180°, (Linear pair angle)

⇒ ∠e = 180° – 125° = 55°,

∠e = ∠f = 55°, (Vertically opposite angles)

In figure, p ॥ q and t is a transversal

∠a = ∠f, (Alternate angle)

∠a = 55°, (∵ ∠f = 55°)

∠d = 125°, (Corresponding angels)

∠c = ∠a = 55°, (Vertically opposite angles)

and ∠b = ∠d = 125°, (Vertically opposite angles)

Thus, ∠a = 55°, ∠b = 125°, ∠c = 55°, ∠d = 125°, ∠e = 55° and ∠f= 55°

Question 3

If l ॥ m then find the value (RBSESolutions.com) of in each of ∠x the following figure:

Solution:

(i) In figure l ॥ m and a is transversal line ∠x = 110°

(Corresponding angle)

(ii) In figure l ॥ m and a is (RBSESolutions.com) transversal line

∴ ∠x = 100°

(Corresponding angle)

Question 4

(i) Is l ॥ m? why?

(ii) If l ॥ m and t is transveral (RBSESolutions.com) than find the value of x.

(iii) Is l ॥ m and why?

Solution:

(i) In figure, alternate (RBSESolutions.com) angles are equal

∴ l ॥ m

(ii) In figure, l ॥ m and t is a transversal

∴ x + 70° = 180°,

Sum of angles made by a transversal on its one side is 180°,

∴ x = 180°- 70° = 110°

(iii) In figure, x = 180° – 130° = 50°

∴ Corresponding (RBSESolutions.com) angles are equal

∴ l ॥ m

We hope the RBSE Solutions for Class 7 Maths Chapter 7 Lines and Angles Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 7 Lines and Angles Additional Questions, drop a comment below and we will get back to you at the earliest.

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