NCERT Solutions For Class 9 Math Chapter – 6 Exercise – 6.1

NCERT Solutions For Class 9 Math Chapter – 6 Exercise – 6.1

NCERT Solutions For Class 9 Math Chapter – 6 Lines and Angles helps students in learning basic concepts of probability which is include in the second term’s CBSE syllabus 2021-2022 NCERT Solutions For Class 9 Maths provide answers to all the questions in exercise present at the end of the chapter These solutions are prepared by our mathematics experts who are highy experienced in the field of education.

Experts at Sunstarup have created the NCERT Solutions after extensive search on each topic Students can refer to this study study material to boost their confidence and attempt the second term exam smartly. The concepts are explained with steps, shortcuts to remember formula , tips and tricks to solve the numerical problems wisely and quickly.

Q1. In Fig. 6.13, lines AB and CD intersect at O. If AOC +BOE = 70° and BOD = 40°, find BOE and reflex COE.

Ncert solutions class 9 chapter 6-1

Solution:

From the diagram, we have

(∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) forms a straight line.

So, ∠AOC+∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180°

Now, by putting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get

∠COE = 110° and ∠BOE = 30°

So, reflex ∠COE = 360o – 110o = 250o

In Fig. 6.14, lines XY and MN intersect at O. If POY = 90° and a : b = 2 : 3, find c.

Ncert solutions class 9 chapter 6-2

Solution:

We know that the sum of linear pair are always equal to 180°

So,

POY +a +b = 180°

Putting the value of POY = 90° (as given in the question) we get,

a+b = 90°

Now, it is given that a : b = 2 : 3 so,

Let a be 2x and b be 3x

∴ 2x+3x = 90°

Solving this we get

5x = 90°

So, x = 18°

∴ a = 2×18° = 36°

Similarly, b can be calculated and the value will be

b = 3×18° = 54°

From the diagram, b+c also forms a straight angle so,

b+c = 180°

c+54° = 180°

∴ c = 126°

In Fig. 6.15, PQR = PRQ, then prove that PQS = PRT.

Ncert solutions class 9 chapter 6-3

Solution:

Since ST is a straight line so,

∠PQS+∠PQR = 180° (linear pair) and

∠PRT+∠PRQ = 180° (linear pair)

Now, ∠PQS + ∠PQR = ∠PRT+∠PRQ = 180°

Since ∠PQR =∠PRQ (as given in the question)

∠PQS = ∠PRT. (Hence proved)

Q4. In fig 6.16 If x + y = w + z then prove that AOB is a line

Solution:-

X + y = 180⁰

X + y + w + z = 360⁰

X+y = w + z

So , (x+y) + (x+y) = 360⁰

2(x+y) = 360⁰

(x+y) = 180⁰ (Hence proved)

Q5. In Fig 6.17 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).

Solution:-

POQ = 180⁰

POS + ROS + ROQ = 180⁰ – 90⁰ (Since POR = ROQ = 90⁰)

POS + ROS = 90⁰

QOS = ROQ + ROS

QOS = 90⁰ + ROS

QOS – ROS = 90⁰

POS + ROS = 90⁰ and QOS – ROS = 90⁰ we get

POS + ROS = QOS – ROS

2ROS + POS = QOS

Or , ROS = 1/2 (QOS – POS) (Hence proved).

Q6. It is given that XYZ = 64⁰ 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:-