Class 9 NCERT Solutions Maths Chapter 6 - Lines And Angles
Ex. 6.1
Ex. 6.2
Lines And Angles Exercise Ex. 6.1
Solution 1
Solution 2
Let common ratio between a and b is x, a = 2x and b = 3x.
XY is a straight line, OM and OP rays stands on it.
XOM + MOP + POY = 180 b + a + POY = 180
3x + 2x + 90 = 180
5x = 90
x = 18
a = 2x
= 2 * 18
= 36
b = 3x
= 3 * 18
= 54
Now, MN is a straight line. OX ray stands on it.
b + c = 180
54 + c = 180
c = 180 54 = 126
c = 126
Solution 3
In the given figure, ST is a straight line and QP ray stand on it.
PQS + PQR = 180 (Linear Pair)
PQS + PQR = 180 (Linear Pair)PQR = 180 - PQS (1)PRT + PRQ = 180 (Linear Pair)PRQ = 180 - PRT (2)Given that
PQR = PRQ. Now, equating equations (1) and (2), we have180 -
PQS = 180 - PRTPQS = PRTSolution 4
We may observe that
x + y + z + w = 360 (Complete angle)
It is given that
x + y = z + w
x + y + x + y = 360
x + y + z + w = 360 (Complete angle)
It is given that
x + y = z + w
x + y + x + y = 3602(x + y) = 360
x + y = 180
Since x and y form a linear pair, thus AOB is a line.
Solution 5
Given that OR 
PQ
PQ 
POR = 90 
POS + SOR = 90ROS = 90 - POS ... (1)QOR = 90 (As OR PQ) QOS - ROS = 90ROS = QOS - 90 ... (2) On adding equations (1) and (2), we have
2 ROS = QOS - POS
ROS = QOS - POS
Solution 6
Given that line YQ bisects 
PYZ.
Hence,QYP = ZYQ
Now we may observe that PX is a line. YQ and YZ rays stand on it.
PYZ.Hence,
QYP = ZYQ Now we may observe that PX is a line. YQ and YZ rays stand on it.
XYZ + ZYQ + QYP = 180 
64 + 2QYP = 180 2QYP = 180 - 64 = 116 QYP = 58 Also, ZYQ = QYP = 58
ZYQ = QYP = 58 Reflex QYP = 360o - 58o = 302o
QYP = 360o - 58o = 302oXYQ = XYZ + ZYQ = 64o + 58o = 122o
Lines And Angles Exercise Ex. 6.2
Solution 1
Given that AB || CD and CD || EF
AB || CD || EF (Lines parallel to a same line are parallel to each other) Now we may observe that
x = z (alternate interior angles) ... (1)
Given that y: z = 3: 7
Let common ratio between y and z be a
y = 3a and z = 7a
x = z (alternate interior angles) ... (1)
Given that y: z = 3: 7
Let common ratio between y and z be a
y = 3a and z = 7aAlso x + y = 180 (co-interior angles on the same side of the transversal)
z + y = 180 [Using equation (1)]
7a + 3a = 180
10a = 180
a = 18
x = 7 a = 7 
18 = 126
x = 7 a = 7 18 = 126Solution 2
It is given that
AB || CD
AB || CD
EF 
CD
CDGED = 126
GEF + FED = 126
GEF + FED = 126 GEF + 90 = 126 GEF = 36Now,
AGE and GED are alternate interior angles AGE = GED = 126 But AGE +FGE = 180 (linear pair)
AGE +FGE = 180 (linear pair) 126 + FGE = 180 FGE = 180 - 126 = 54
AGE = 126, GEF = 36, FGE = 54Solution 3
Let us draw a line XY parallel to ST and passing through point R.
PQR + QRX = 180 (co-interior angles on the same side of transversal QR)
110 + QRX = 180
QRX = 70
Now,
RST +SRY = 180 (co-interior angles on the same side of transversal SR)
130 +SRY = 180
SRY = 50
XY is a straight line. RQ and RS stand on it.
QRX + QRS + SRY = 180
70 +QRS + 50 = 180
QRS = 180 - 120 = 60
PQR + QRX = 180 (co-interior angles on the same side of transversal QR) 110 + QRX = 180 QRX = 70Now,
RST +SRY = 180 (co-interior angles on the same side of transversal SR)130 +
SRY = 180SRY = 50XY is a straight line. RQ and RS stand on it.
QRX + QRS + SRY = 180 70 +
QRS + 50 = 180QRS = 180 - 120 = 60Solution 4
APR = PRD (alternate interior angles)50 + y = 127
y = 127 - 50
y = 77
Also
APQ = PQR (alternate interior angles)50 = x
x = 50 and y = 77Solution 5
Let us draw BM 
PQ and CN RS.
As PQ || RS
So, BM || CN
Thus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.
PQ and CN RS. As PQ || RS
So, BM || CN
Thus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.
2 = 3 (alternate interior angles)But 1 = 2 and 3 = 4 (By laws of reflection)
1 = 2 and 3 = 4 (By laws of reflection) 1 = 2 = 3 = 4Now, 1 + 2 = 3 + 4
1 + 2 = 3 + 4ABC = DCBBut, these are alternate interior angles
AB || CD