RBSE Solutions for Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Additional Questions is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 5 Plane Geometry and Line and Angle Additional Questions.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Plane Geometry and Line and Angle |

Exercise |
Additional Questions |

Number of Questions Solved |
24 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 5 Plane Geometry and Line and Angle Additional Questions

**Multiple Choice Questions**

Question 1.

In the figure, PQ || RS, ∠QPR = 70°, ∠ROT = 20° then the (RBSESolutions.com) value of x is:

(A) 20°

(B) 70°

(C) 110°

(D) 50°

Solution:

D

Question 2.

Two angles measures (30 – a)° and (125 + 2a)°. If each one is the (RBSESolutions.com) supplement of the other, then ‘a’ is:

(A) 35°

(B) 25°

(C) 65°

(D) 45°

Solution:

B

Question 3.

In the given figure, if ∠COA = 62°, then x is:

(A) 108°

(B) 128°

(C) 118°

(D) 32°

Solution:

C

Question 4.

The value of the variable (in the given figure) is (RBSESolutions.com) equal to:

(A) 6

(B) 7

(C) 8

(D) 9

Solution:

C

Question 5.

In the given figure, what (RBSESolutions.com) value of x would make AOB a straight line?

(A) 12°

(B) 14°

(C) 16°

(D) 18°

Solution:

C

Question 6.

In the given figure, the value of y is:

(A) 9°

(B) 18°

(C) 36°

(D) 54°

Solution:

B

Question 7.

In the given figure, if l || m then the (RBSESolutions.com) value of x is:

(A) 30°

(B) 40°

(C) 35°

(D) 45°

Solution:

A

Question 8.

In the given figure, if AB || CD, then the (RBSESolutions.com) value of x is:

(A) 30°

(B) 20°

(C) 60°

(D) 45°

Solution:

A

Question 9.

In the given figure, the (RBSESolutions.com) value of x is equal to:

(A) 25°

(B) 30°

(C) 20°

(D) 40°

Solution:

B

**Very Short Answer Type Questions**

Question 1.

If the angles (3x – 20)° and (2x – 40)° are complementary angles. Find x.

Solution.

As we know that sum of the measure of (RBSESolutions.com) an angle and its complement is equal to 90°

(3x – 20)° + (2x – 40)° = 90°

⇒ 5x – 60° = 90°

⇒ 5x = 150°

⇒ x = 30°

Question 2.

An angle is 25° less than its complement. What is its measure?

Solution.

Let an angle be x

Its complement = 90° – x

According to question

x = 90° – x – 25°

⇒ 2x = 65°

⇒ x = 32[latex]\frac { { 1 }^{ 0 } }{ 2 }[/latex]

Hence, measurement of the required angle be 32[latex]\frac { { 1 }^{ 0 } }{ 2 }[/latex].

Question 3.

In the given figure, POS is a line, find (RBSESolutions.com) the measure of ∠QOR.

Solution.

Here we are given that POS is a straight line

∠POQ + ∠QOR + ∠ROS = 180° (straight angle)

⇒ 60° + 5x – 20° + 40° = 180°

⇒ 5x + 80° = 180°

⇒ 5x = 100°

⇒ x = 20°

∠QOR = 5x – 20° = 5 x 20° – 20° = 80°

Question 4.

In the given figure, if ∠1 + ∠2 = 100°, then (RBSESolutions.com) find the measure of ∠4.

Solution.

∠1 + ∠2 = 100° (given)

But ∠1 = ∠2 (vertically opposite angles)

⇒ 2∠2 = 100°

⇒ ∠2 = 50°

Also ∠2 + ∠4 = 180° (linear pair)

⇒ ∠4 = 180° – 50° = 130°

Question 5.

In the given figure, (RBSESolutions.com) find the value of x.

Solution.

Here ∠ACB + ∠ACD = 180° (Linear pair of angles)

⇒ 5x + 7y = 180°

⇒ 12y = 180°

⇒ y = 15°

Also x + 3y = 7y (Exterior angle property)

⇒ x = 4y

x = 4 x 15° ⇒ x = 60°

Question 6.

In the given (RBSESolutions.com) figure, find x and also find ∠POS.

Solution.

∠POS + ∠SOR + ∠ROQ = 180° (Straight angle)

(x + 45°) + x + (x + 30°) = 180°

⇒ 3x + 75° = 180°

⇒ 3x = 105°

⇒ x = 35°

∠POS = (x + 45°) = (35° + 45°) = 80°

Question 7.

In the given figure, if lines EF || GH, then find (RBSESolutions.com) the value of x.

Solution.

EF || GH

115° + (x + 32°)= 180° (The sum of the interior angles on the same side of a transversal is supplementary)

⇒ x + 147° = 180°

⇒ x = 180° – 147° = 33°

Question 8.

In the figure, ∠POM and ∠QOM form (RBSESolutions.com) a linear pair. If x – 2y = 30°, find x and y.

Solution.

∠POM and ∠QOM form a linear pair.

x + y = 180° …(i)

Also x – 2y = 30° (given) …(ii)

Solving (i) and (ii), we get x = 130° and y = 50°

Question 9.

In the figure, a : b : c = 4 : 3 : 5. If AOB is a straight line, find the (RBSESolutions.com) value of a, b and c.

Solution.

We have, a : b : c = 4 : 3 : 5

⇒ ∠a = 4x, ∠b = 3x and ∠c = 5x

But ∠a + ∠b + ∠c = 180° (Straight angle)

⇒ 4x + 3x + 5x = 180°

⇒ 12x = 180°

⇒ x = 15°

Angles are 4 x 15° = 60°, 3 x 15° = 45° and 5 x 15° = 75°

Question 10.

If ray OC stands on line AB such that ∠AOC = ∠COB, then (RBSESolutions.com) show that ∠AOC = 90°.

Solution.

Since ray OC stands on line AB. Therefore

∠AOC + ∠COB = 180° (Linear pair of angles)

But ∠AOC = ∠COB (given)

⇒ 2∠AOC = 180°

⇒ ∠AOC = 90°

**Short Answer Type Questions**

Question 1.

In the given figure, AC ⊥ AB. If ∠BAP = x, ∠PAQ = (x + 5°) and ∠CAQ = (2x + 5°), find the (RBSESolutions.com) value of x and reflex ∠PAQ.

Solution.

We are given that AC ⊥ AB

⇒ ∠CAB = 90° (given)

⇒ ∠BAP + ∠PAQ + ∠CAQ = 90°

⇒ x + (x + 5°) + (2x + 5°) = 90°

⇒ 4x + 10° = 90°

⇒ 4x = 80°

⇒ x = 20°

⇒ ∠PAQ = (x + 5°) = 20° + 5° = 25°

Reflex ∠PAQ = 360° – 25° = 335°

Question 2.

In the given figure, AB || CD, if ∠OAB = 58° and ∠OCD = 45° then (RBSESolutions.com) find the value of x.

Solution.

We are given that AB || CD and ∠OAB = 58° and ∠OCD = 42°.

Construction: Draw a line PQ || AB || CD

To find: x

PQ || AB (by construction)

⇒ ∠AOP = ∠OAB = 58° (alternate angle)

∴ ∠COP = ∠OCD = 42° (alternate angle)

∴ ∠AOC = ∠AOP + ∠COP = 58° + 42°

⇒ ∠AOC = 100°

Thus, reflex ∠AOC

x = 360° – ∠AOC = 360° – 100° = 260°

Hence, x = 260°

Question 3.

In the figure, if EF || CD. Prove that AB || CD.

Solution.

We are given that EF || CD

EF || CD

∠FEC + ∠ECD = 180° (Sum of the interior angles (RBSESolutions.com) on the same side of a transversal is 180°)

⇒ 125° = ∠ECD = 180°

⇒ ∠ECD = 180° – 125° = 55°

⇒ ∠BCD = ∠BCE + ∠ECD = 25° + 25° = 80°

∠ABC = ∠BCD = 80°

These are alternate interior angles and they are equal.

Hence, AB || CD.

Question 4.

“If a transversal intersects two parallel lines, then each (RBSESolutions.com) pair of interior angles are supplementary”. Prove it.

Solution.

Given: AB and CD are two parallel lines and EF is a transversal which intersect them at M and N respectively forming two pairs of interior angles ∠1, ∠3 and ∠2, ∠4.

To Prove: (i) ∠1 + ∠3 = 180° or ∠2 + ∠4 = 180°

Proof: Since ray ND stands on line EF,

∠3 + ∠5 = 180° …(i) (linear pair of angles)

But ∠1 = ∠5 …(ii) (Corresponding angles as AB || CD)

From (i) and (ii), we get

∠1 + ∠3 = 180° …(iii)

Again ray CN stands on EF,

∠2 + ∠6 = 180° (Linear pair of angles)

But ∠4 = ∠6 (Corresponding angles as AB || CD)

⇒ ∠2 + ∠4 = 180° …(iv)

Hence, we can say if a transversal intersects two parallel lines, then (RBSESolutions.com) each pair of interior angles are supplementary

i.e. ∠1 + ∠3 = 180° or ∠2 + ∠4 = 180°

We hope the given RBSE Solutions for Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Additional Questions will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Additional Questions, drop a comment below and we will get back to you at the earliest.

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