# Chapter 6 Lines and Angles | Class 9 NCERT CBSE Mathematics Solution | NCERT Maths Textbook Exercise 6.2

Class IX NCERT CBSE Mathematics Solutions
Lines and Angles
NCERT Mathematics Textbook Exercise 6.1 Solved
(Page 103)
Q1: In the given figure, find the values of x and y and then show that AB || CD.
Ans:
In the above figure,
50Âº + x = 180Âº (Linear pair)
x = 130Âº … (1)
Also, y = 130Âº (Vertically opposite angles)
As x and y are alternate interior angles for lines AB and CD and also measures of these angles are equal to each other, therefore, line AB || CD.
Q2: In the given figure, if AB || CD, CD || EF and y: z = 3: 7, find x.
Ans:
Given that AB || CD and CD || EF
or, AB || CD || EF (Lines parallel to the same line are parallel to each other)
Also, x = z (Alternate interior angles) . . . . . . . (1)
It is given that y : z = 3 : 7
Let the common ratio between y and z be a.
y = 3a and z = 7a
Also, x + y = 180Âº (Co-interior angles on the same side of the transversal)
or, z + y = 180Âº [from equation (1)]
or, 7a + 3a = 180Âº
or, 10a = 180Âº
or, a = 18Âº
so, x = 7a = 7 × 18Âº = 126Âº
Q3: In the given figure, If AB || CD, EF CD and GED = 126Âº, find AGE, GEF and FGE.
Ans:
It is given that, AB || CD, EF ⊥ CD and ∠GED = 126Âº
From the figure, ∠GED = ∠GEF + ∠FED = 126Âº
or, ∠GEF + 90Âº = 126Âº
or, ∠GEF = 36Âº
(∠AGE and ∠GED are alternate interior angles.)
or, ∠AGE = ∠GED = 126Âº
However, ∠AGE + ∠FGE = 180Âº (Linear pair)
or, 126Âº + ∠FGE = 180Âº
or, ∠FGE = 180Âº − 126Âº = 54Âº
Therefore, ∠AGE = 126Âº, ∠GEF = 36Âº, ∠FGE = 54Âº
Q4: In the given figure, if PQ || ST, PQR = 110Âº and RST = 130Âº, find QRS.
[Hint: Draw a line parallel to ST through point R.]

Let us draw a line XY || ST and passing through point R.
we can write, ∠PQR + ∠QRX = 180Âº
(Co-interior angles on the same side of transversal QR)
or, 110Âº + ∠QRX = 180Âº
or, ∠QRX = 70Âº
Also, ∠RST + ∠SRY = 180Âº (Co-interior angles on the same side of transversal SR)
or, 130Âº + ∠SRY = 180Âº
or, ∠SRY = 50Âº
XY is a straight line and RQ and RS intersect XY at the point R.
∴ ∠QRX + ∠QRS + ∠SRY = 180Âº
or, 70Âº + ∠QRS + 50Âº = 180Âº
or, ∠QRS = 180Âº − 120Âº = 60Âº
Q5: In the given figure, if AB || CD, APQ = 50Âº and PRD = 127Âº, find x and y.
Ans:
In the above figure,
∠APR = ∠PRD (Alternate interior angles)
or, 50Âº + y = 127Âº
or, y = 127Âº − 50Âº
or, y = 77Âº
Also, ∠APQ = ∠PQR (Alternate interior angles)
x = 50Âº and y = 77Âº
Q6: In the given figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Let us draw BM ⊥ PQ and CN ⊥ RS.
Since, PQ || RS
Therefore, BM || CN
Thus, BM and CN are two parallel lines and a transversal line BC intersects these lines at B and C respectively.
So, ∠2 = ∠3 (Alternate interior angles)
However, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)
So, ∠1 = ∠2 = ∠3 = ∠4
Also, ∠1 + ∠2 = ∠3 + ∠4
or, ∠ABC = ∠DCB (but, these are alternate interior angles)
∴ AB || CD