RBSE Solutions for Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Miscellaneous Exercise 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 Miscellaneous Exercise.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Plane Geometry and Line and Angle |

Exercise |
Miscellaneous Exercise |

Number of Questions Solved |
16 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 5 Plane Geometry and Line and Angle Miscellaneous Exercise

**Multiple Choice Questions (Q1 to Q7)**

Question 1.

In the adjoining figure, if AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then (RBSESolutions.com) the value of ∠QRS.

(A) 85°

(B) 135°

(C) 145°

(D) 110°

Question 2.

In figure, for what value of x, POQ is a line?

(A) 20°

(B) 25°

(C) 30°

(D) 35°

Question 3.

In figure, if OP || RS, ∠OPQ =110° and ∠QRS = 130° then ∠PQR is equal to

(A) 40°

(B) 50°

(C) 60°

(D) 70°

Question 4.

In figure, the (RBSESolutions.com) reflex ∠AOB is equal to:

(A) 60°

(B) 120°

(C) 300°

(D) 360°

Question 5.

In figure, two lines AB and CD intersect at O. The angles at the (RBSESolutions.com) point O are marked in the figure. Then ∠x – ∠y is equal to:

(A) 56°

(B) 118°

(C) 62°

(D) 180°

Question 6.

In figure, which of the following pairs of (RBSESolutions.com) angles are not corresponding angles:

(A) ∠1, ∠5

(B) ∠2, ∠6

(C) ∠3, ∠7

(D) ∠3, ∠5

Question 7.

A transversal n intersects two parallel lines l and m at (RBSESolutions.com) points G and H respectively. The angles so formed have been marked in the figure, if ∠1 is an acute angle, which of the following statement is false:

(A) ∠1 + ∠2 = 180°

(B) ∠2 + ∠5 = 180°

(C) ∠3 + ∠8 = 180°

(D) ∠2 + ∠6 = 180°

Question 8.

In figure, find the value of x.

Solution.

∠x + 140° = 180° (linear pair)

⇒ x = 40°

Question 9.

In figure, AB || CD, find ∠x and ∠y from the (RBSESolutions.com) angles given in the figure.

Solution.

Here it is given that AB || CD

∠x + 80° = ∠BCD (alt. angles)

⇒ ∠x + 80° = 116°

⇒ ∠x = 116° – 80°

⇒ ∠x = 36°

Now at point B

∠y + ∠x + 80° = 180° (straight angle)

⇒ ∠y = 180° – 116°

⇒ ∠y = 64°

Hence, ∠x = 36° and ∠y = 64°

Question 10.

In figure, lines l and m are parallel. Find ∠x and give (RBSESolutions.com) reasons for your answer.

Solution.

From figure, we (RBSESolutions.com) can write

x + x + ∠BPQ = 180° (straight angle)

⇒ 2x + ∠BPQ = 180°

∠BPQ = 180° – 2x …(i)

l || m

∠BPQ = 68° (alt. angle) …(ii)

From equations (i) and (ii), we have

68° = 180° – 2x

⇒ 2x = 180° – 68°

⇒ 2x = 112°

⇒ x = 56°

Question 11.

In figure, which of the lines are parallel to (RBSESolutions.com) each other and why? Give reasons?

Solution.

As we know that if a number of lines meet at a point (RBSESolutions.com) then the sum of all the angles so formed = 360°

Here

∠PAQ + ∠QAB + reflex ∠PAB = 360°

⇒ 55° + ∠QAB + 255° = 360°

⇒ ∠QAB = 360° – 310°

⇒ ∠QAB = 50°

⇒ ∠QAB = ∠ABR = 50° (alt. angle)

QA || BR

Also ∠PAB = ∠ABS = 105° (alt. angles)

PA || BS

Hence QA || BR and PA || BS.

Question 12.

In the given figure, AC || PQ and AB || RS, then find ∠y. Also give (RBSESolutions.com) reasons in support of your answer.

Solution.

PQ || AC ( given)

∠y = ∠BAC (alt. angles) …(i)

Again AB || RS (given)

∠BAC = ∠LMC …(ii) (corresponding angles)

From (i) and (ii), we get

∠LMC = ∠y

In ΔLMC, exterior angle is equal to sum (RBSESolutions.com) of opposite interior angles

⇒ 80° = ∠y + 30°

⇒ ∠y = 80° – 30°

⇒ ∠y = 50°

Question 13.

In figure, AB || CD, PQ || EF, then find the value of x.

Solution.

According to question

∠FEC + 52° + 60° = 180° (straight angle)

∠FEC = 180° – 112°

⇒ ∠FEC = 68°

Also AB || CD (given)

∠x = ∠CST (corresponding angles)

Also QP || EF (given)

∠QSE + ∠FES = 180° (The sum of the (RBSESolutions.com) interior angles on the same side of a transversal is 180°)

⇒ ∠QSE + 68° = 180°

⇒ ∠QSE = 180° – 68°

⇒ ∠QSE = 112° = ∠CST (vertically opposite angles)

x = ∠CST = 112° (corresponding angle).

Question 14.

In figure, find which of the lines l, m, n, p, q and r are parallel and why?

Solution.

∠PAC + ∠QCA = 75° + 105° = 180° (Sum of the interior angles)

n || p

Also ∠ACD = ∠CDS = 114° (alternate angles)

m || r

Hence n || p and m || r

Question 15.

In figure, the two lines intersect (RBSESolutions.com) each other. If ∠1 + ∠2 + ∠3 = 230°, then find the value of ∠1 and ∠4.

Solution.

According to question

∠1 + ∠2 + ∠3 + ∠4 = 360° (Angles formed at a point is 360°)

But ∠1 + ∠2 + ∠3 = 230° (given)

(∠1 + ∠2 + ∠3) + ∠4 = 360°

230° + ∠4 = 360°

∠4 = 360° – 230°

∠4 = 130°

But ∠1 + ∠4 = 180° (linear pair axiom)

∠1 + 130°= 180°

∠1 = 180°- 130° = 50°

Question 16.

Two plane mirrors PQ and QR are joint (RBSESolutions.com) together at Q with an angle of 30°. The incident ray AB parallel to the mirror RQ, then find the value of ∠BCQ, ∠CBQ and ∠BDC.

Solution.

BL || CD

∠1 = ∠2 i.e. ∠LBC = ∠BCD

Also CD ⊥ PQ

⇒ ∠CDB = ∠CDQ = 90°

⇒ ∠DCQ = 180° – 120° = 60° = ∠DCB

as CD is bisector ∠BCQ

⇒ ∠BCQ = ∠BCD + ∠DCQ = 60° + 60° = 120°

In ∆CBD,

∠CBD + 90° + 60° = 180°

⇒ ∠CBD or ∠CBQ = 180° – 150° = 30°

We hope the given RBSE Solutions for Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Miscellaneous Exercise 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 Miscellaneous Exercise, drop a comment below and we will get back to you at the earliest.