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.
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Q1. In Fig. 6.13, lines AB and CD intersect at O. If AOC +BOE = 70° and BOD = 40°, find BOE and reflex COE.
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.
We know that the sum of linear pair are always equal to 180°
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.
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
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).
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.
Here , XP is a straight line
So , XYZ + ZYP = 1800
Putting the value of XYZ = 640 we get ,
640 + ZYP = 1800
ZYP = 1160
YQ bisects ZYP ,
ZYQ = QYP
ZYP = 2ZYQ
ZYQ = QYP = 580
XYQ = XYZ + ZYQ
By putting the value of XYZ = 640 and ZYQ = 580 we get
XYQ = 640 + 580
XYQ = 1220
QYP = 1800 + XYQ
XYQ = 1220
QYP = 1800 + 1220
QYP = 3020