NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Lines and Angles |

Exercise |
Ex 4.1 |

Number of Questions Solved |
6 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.1

**Question 1.**

**In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.**

**Solution:**

Here, ∠AOC and ∠BOD are vertically opposite angles.

∴ ∠AOC = ∠BOD

⇒ ∠AOC = 40° [∵ ∠BOD = 40°(Given)] …(i)

We have, ∠AOC + ∠ BOE = 70° (Given)

40°+ ∠BOE = 70° [From Eq. (i)]

⇒ ∠BOE = 30°

Also, ∠AOC + ∠COE + ∠BOE = 180° (Linear pair axiom)

⇒ 40° + ∠COE + 30° = 180°

⇒ ∠COE = 110°

Now, ∠COE + reflex ∠COE = 360° (Angles at a point)

110°+reflex ∠COE = 360°

⇒ Reflex ∠COE = 250°

**Question 2.**

**In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.**

**Solution:**

We have, ∠POY = 90°

⇒ ∠POY + ∠POX = 180° (Linear pair axiom)

⇒ ∠POX = 90°

⇒ a+b = 90°

Also, a : b = 2 : 3 (Given)

⇒ Let a = 2k,b = 3k

Now, from Eq. (j), we get

2k + 3k = 90°

⇒ 5k = 90°

⇒ k = 18°

∴ a = 2 x 18°=36°

and b=3 x 18°=54°

Now, ∠MOX + ∠XON = 1800 (Linear pair axiom)

b+ c = 180°

⇒ 540 + c= 180°

⇒ c = 126°

**Question 3.**

**In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.**

**Solution:**

∵ ∠PQS+ ∠PQR = 180° (Linear pair axiom) ,..(i)

and ∠PRT + ∠PRQ = 180° (Linear pair axiom).. .(ii)

From Eqs. (i) and (ii), we get

∠PQS + ∠PQR =∠PRT + ∠PRQ

∠PQS + ∠PRQ =∠PRT + ∠PRQ

[Given, ∠PQR = ∠PRQ]

⇒ ∠PQS = ∠PRT

**Question 4.**

**In figure, if x + y = w + z, then prove that AOB is a line.**

**Solution:**

∵ x+ y+w+ z = 360° (Angle at a point)

x + y = w + z (Given)…(i)

∴ x+ y+ x+ y = 360° [From Eq. (i)]

2(x + y) = 360°

⇒ x + y = 180° (Linear pair axiom)

Hence, AOB is a straight line.

**Question 5.**

**In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that**

**Solution:**

We have,

∠POR = ∠ROQ = 90° (∵ Given that, OR is perpendicular to PQ)

∴ ∠POS + ∠ROS = 90°

⇒ ∠ROS = 90° – ∠POS

On adding ∠ROS both sides, we get

2 ∠ROS = 90° – ∠POS + ∠ROS

⇒ 2 ∠ROS = (90° + ∠ROS) – ∠POS

⇒ 2∠ROS = ∠QOS – ∠POS (∵ ∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS)

⇒ ∠ROS = \(\frac { 1 }{ 2 }\) (∠QOS – ∠POS)

Hence proved.

**Question 6.**

**It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.**

**Solution:**

Here, YQ bisects ∠ZYP.

Hence, ∠ZYQ = ∠QYP = \(\frac { 1 }{ 2 }\) ∠ZYP ……..(i)

Given, ∠XYZ = 64° ….(ii)

∵ ∠XYZ + ∠ZYQ + ZQYP = 180° (Linear pair axiom)

⇒ 64° + ∠ZYQ + ∠ZYQ = 180° [From Eqs. (i) and (ii)]

⇒ 2 ∠ZYQ = 180° – 64°

⇒ ∠ZYQ = \(\frac { 1 }{ 2 }\) x 116°

⇒ ∠ZYQ = 58°

∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122°

Now, ∠QYP + reflex ∠QYP = 360°

58° + reflex ∠QYP = 360°

⇒ reflex ∠ QYP = 302°

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