RBSE Solutions for Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Ex 5.1 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 Exercise 5.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Plane Geometry and Line and Angle |

Exercise |
Ex 5.1 |

Number of Questions Solved |
5 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 5 Plane Geometry and Line and Angle Ex 5.1

Question 1.

If angles of magnitude (2x + 4) and (x – 1) form (RBSESolutions.com) a linear pair. Find these angles.

Solution.

According to question

(2x + 4)° + (x – 1)° = 180°

(Linear pair axiom)

⇒ 3x + 3 = 180

⇒ 3x = 177

∴Angles are (2 x 59 + 4)° = 122° and (59 – 1)° = 58

Question 2.

From the given figure

(i) Find the magnitude of ∠BOD

(ii) Find the magnitude of ∠AOD

(iii) Write the pair (RBSESolutions.com) of vertically opposite angles.

(iv) Name the adjacent supplementary angles of ∠AOC

Solution.

(i) ∠AOC = 52° (given)

⇒ ∠BOD = 52°

(Vertically opposite angles)

(ii) ∠AOC + ∠AOD = 180°

(Linear pair axiom)

⇒ 52° + ∠AOD = 180°

⇒ ∠AOD = 180° – 52°

⇒ ∠AOD = 128°

(iii) (∠AOC, ∠BOD) and (∠AOD, ∠BOC) are vertically opposite pairs of angles.

(iv) The adjacent supplementary angles of ∠AOC are ∠AOD and ∠BOC.

Question 3.

In the given figure, ∠PQR = ∠PRQ then (RBSESolutions.com) prove that ∠PQS = ∠PRT.

Solution.

We are given that ∠PQR = ∠PRQ ,..(i)

From figure,

∠PQS + ∠PQR = 180°

(linear pair axiom) …(ii)

and ∠PRT + ∠PRQ = 180°

(linear pair axiom) …(iii)

Using (ii) and (iii)

∠PQS + ∠PQR = ∠PRT + ∠PRQ

Again using (i) i.e. ∠PQR = ∠PRQ, we get

⇒ ∠PQS = ∠PRT

Hence proved

Question 4.

In figure OP, OQ, OR and OS are (RBSESolutions.com) four rays. Prove that

∠PPQ + ∠QOR + ∠SOR + ∠POS = 360°.

Solution.

In figure, we need to produce (RBSESolutions.com) any of the rays OP, OQ, OR or OS backwards to a point.

Let us produce ray OQ backwards to a point T so that TOQ is a line see figure.

Now, ray OP stands on line TOQ

Therefore, ∠TOP + ∠POQ = 180° …(i)

(Linear pair axiom)

Similarly, ray OS stands (RBSESolutions.com) on line TOQ.

Therefore, ∠TOS + ∠SOQ = 180° …(ii)

But ∠SOQ = ∠SOR + ∠QOR

So, (ii) becomes

∠TOS + ∠SOR + ∠QOR = 180° …(iii)

Now adding (i) and (iii), we get

∠TOP + ∠POQ + ∠SOR + ∠QOR = 360° …(iv)

But ∠TOP + ∠TOS = ∠POS

Therefore (iv) becomes

∠POQ + ∠QOR + ∠SOR + ∠POS = 360°

Question 5.

In the given figure, if

∠x + ∠y = ∠p + ∠q, then (RBSESolutions.com) prove that AOB is a line.

Solution.

Method I: In the given figure, let us (RBSESolutions.com) assume that AOB is a straight line

∴ ∠x + ∠y = 180°

(linear pair of angles) …(i)

and ∠p + ∠q = 180°

(linear pair of angles) …(ii)

From (i) and (ii), we get

∠x + ∠y = ∠p + ∠q

Hence proved

Method II: As we know that sum of the (RBSESolutions.com) angles around a point is 360°.

∴ ∠x + ∠y + ∠p + ∠q = 360°

or ∠x + ∠y + ∠x + ∠y = 360°

(∵ ∠x + ∠y = ∠p + ∠q)

2 (∠x + ∠y) = 360°

⇒ ∠x + ∠y = 180°

⇒ AOB is a straight line.

We hope the given RBSE Solutions for Class 9 Maths Chapter 5 Plane Geometry and Line and Angle Ex 5.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 9 Maths Solutions Chapter 5 Plane Geometry and Line and Angle Exercise 5.1, drop a comment below and we will get back to you at the earliest.

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