**NCERT Solutions Class 9 Maths Chapter 6 Lines and Angles** – Here are all the NCERT solutions for Class 9 Maths Chapter 6. This solution contains questions, answers, images, explanations of the complete Chapter 6 titled Lines and Angles of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, then you must come across Chapter 6 Lines and Angles. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles in one place.

## NCERT Solutions Class 9 Maths Chapter 6 LINES AND ANGLES

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Maths for Class 9 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 6 LINES AND ANGLES , Maths, Class 9.

Class | 9 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 6 |

Chapter Name |
LINES AND ANGLES |

### NCERT Solutions Class 9 Maths chapter 6 LINES AND ANGLES

Class 9, Maths chapter 6, LINES AND ANGLES solutions are given below in PDF format. You can view them online or download PDF file for future use.

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### Question & Answer

Q.1:In Fig. lines AB and CD intersect at O. If ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°, find ∠ BOE and reflex ∠ COE.

Ans :AB is a straight line, rays OC and OE stand on it. \(\begin{array}{l}{\therefore \angle \mathrm{AOC}+\angle \mathrm{COE}+\angle \mathrm{BOE}=180^{\circ}} \\ {\Rightarrow(\angle \mathrm{AOC}+\angle \mathrm{BOE})+\angle \mathrm{COE}=180^{\circ}} \\ {\Rightarrow 70^{\circ}+\angle \mathrm{COE}=180^{\circ}} \\ {\Rightarrow \angle \mathrm{COE}=180^{\circ}-70^{\circ}=110^{\circ}} \\ {\text { Reflex } \angle \mathrm{COE}=360^{\circ}-110^{\circ}=250^{\circ}}\end{array}\) CD is a straight line. rays OE and 0B stand on it. \(\begin{array}{l}{\therefore \angle \mathrm{COE}+\angle \mathrm{BOE}+\angle \mathrm{BOD}=180^{\circ}} \\ {\Rightarrow 110^{\circ}+\angle \mathrm{BOE}+40^{\circ}=180^{\circ}} \\ {\Rightarrow \angle \mathrm{BOE}=180^{\circ}-150^{\circ}=30^{\circ}}\end{array}\)

Q.2:In Fig. lines XY and MN intersect at O. If ∠ POY = 90° and a : b = 2 : 3, find c.

Ans :Let the common ratio between a and b be x. a = 2x, and b = 3x XY is a straight line, rays OM and OP stand on it. \(\begin{array}{l}{\therefore \angle \mathrm{XOM}+\angle \mathrm{MOP}+\angle \mathrm{POY}=180^{\circ}} \\ {b+a+\angle \mathrm{POY}=180^{\circ}} \\ {3 x+2 x+90^{\circ}=180^{\circ}} \\ {5 x=90^{\circ}} \\ {x=18^{\circ}} \\ {a=2 x=2 \times 18=36^{\circ}} \\ {b=3 x=3 \times 18=54^{\circ}}\end{array}\) MN is a straight line. Ray OX stands on it. \(\begin{array}{l}{\therefore b+c=180^{\circ}(\text { Linear Pair })} \\ {54^{\circ}+c=180^{\circ}} \\ {c=180^{\circ}-54^{\circ}=126^{\circ}} \\ {\therefore c=126^{\circ}}\end{array}\)

Q.3:In Fig. ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT

Ans :In the given figure, ST isa straight line and ray QP stands on it. \(\begin{array}{l}{\therefore \angle \mathrm{PQS}+\angle \mathrm{PQR}=180^{\circ}(\text { Linear Pair })} \\ {\angle \mathrm{PQR}=180^{\circ}-\angle \mathrm{PQS}(1)} \\ {\angle \mathrm{PRT}+\angle \mathrm{PRQ}=180^{\circ}(\text { Linear Pair })} \\ {\angle \mathrm{PRQ}=180^{\circ}-\angle \mathrm{PRT}(2)} \\ {\text { It is given that } \angle \mathrm{PQR}=\angle \mathrm{PRQ} \text { . }}\end{array}\) Equating equations (1) and (2), we obtain \(\begin{array}{l}{180^{\circ}-\angle \mathrm{PQS}=180^{\circ}-\angle \mathrm{PRT}} \\ {\angle \mathrm{PQS}=\angle \mathrm{PRT}}\end{array}\)

Q.4:In Fig. if x + y = w + z, then prove that AOB is a line.

Ans :\(\begin{array}{l}{\text { It can be observed that, }} \\ {x+y+z+w=360^{\circ}(\text { Complete angle })} \\ {\text { It is given that, }} \\ {x+y=z+w} \\ {\therefore x+y+x+y=360^{\circ}} \\ {2(x+y)=360^{\circ}} \\ {x+y=180^{\circ}}\end{array}\) Since x and y form a linear pair, AOB is a line.

Q.5:In Fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ ROS = 1/2 (∠ QOS – ∠ POS).

Ans :It is given that \(\mathrm{OR} \perp \mathrm{PQ}\) \(\begin{array}{l}{\square \square \mathrm{POR}=90^{\circ}} \\ {\square \square \mathrm{POS}+\square \mathrm{SOR}=90^{\circ}}\end{array}\) \(\begin{array}{l}{\square R O S=90^{\circ}-\square P O S \ldots(1)} \\ {\square Q O R=90^{\circ}(A s O R \square P Q)}\end{array}\) On adding equations (1) and (2), we obtain \(2 \square \mathrm{ROS}=\square \mathrm{QOS}-\square \mathrm{POS}\) \(\square \mathrm{ROS}=\frac{1}{2}(\square \mathrm{QOS}-\square \mathrm{POS})\)

## NCERT / CBSE Book for Class 9 Maths

You can download the NCERT Book for Class 9 Maths in PDF format for free. Otherwise you can also buy it easily online.

- Click here for NCERT Book for Class 9 Maths
- Click here to buy NCERT Book for Class 9 Maths

### All NCERT Solutions Class 9

- NCERT Solutions for Class 9 English
- NCERT Solutions for Class 9 Hindi
- NCERT Solutions for Class 9 Maths
- NCERT Solutions for Class 9 Science
- NCERT Solutions for Class 9 Social Science
- NCERT Solutions for Class 9 Sanskrit

### All NCERT Solutions

You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

Class 1 | Class 2 | Class 3 |

Class 4 | Class 5 | Class 6 |

Class 7 | Class 8 | Class 9 |

Class 10 | Class 11 | Class 12 |

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