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,

50º + x = 180º (Linear pair)

x = 130º … (1)

Also, y = 130º (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 = 180º (Co-interior angles on the same side of the transversal)

or, z + y = 180º [from equation (1)]

or, 7a + 3a = 180º

or, 10a = 180º

or, a = 18º

so, x = 7a = 7 × 18º = 126º

Q3: In the given figure, If AB || CD, EF ⊥ CD and ∠GED = 126º, find ∠AGE, ∠GEF and ∠FGE.

Ans:

It is given that, AB || CD, EF ⊥ CD and ∠GED = 126º

From the figure, ∠GED = ∠GEF + ∠FED = 126º

or, ∠GEF + 90º = 126º

or, ∠GEF = 36º

(∠AGE and ∠GED are alternate interior angles.)

or, ∠AGE = ∠GED = 126º

However, ∠AGE + ∠FGE = 180º (Linear pair)

or, 126º + ∠FGE = 180º

or, ∠FGE = 180º − 126º = 54º

Therefore, ∠AGE = 126º, ∠GEF = 36º, ∠FGE = 54º

Q4: In the given figure, if PQ || ST, ∠PQR = 110º and ∠RST = 130º, 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 = 180º 
(Co-interior angles on the same side of transversal QR)

or, 110º + ∠QRX = 180º

or, ∠QRX = 70º

Also, ∠RST + ∠SRY = 180º (Co-interior angles on the same side of transversal SR)

or, 130º + ∠SRY = 180º

or, ∠SRY = 50º

XY is a straight line and RQ and RS intersect XY at the point R.

∴ ∠QRX + ∠QRS + ∠SRY = 180º

or, 70º + ∠QRS + 50º = 180º

or, ∠QRS = 180º − 120º = 60º

Q5: In the given figure, if AB || CD, ∠APQ = 50º and ∠PRD = 127º, find x and y.

Ans:

In the above figure, 
∠APR = ∠PRD (Alternate interior angles)

or, 50º + y = 127º

or, y = 127º − 50º

or, y = 77º

Also, ∠APQ = ∠PQR (Alternate interior angles)

∴ x = 50º and y = 77º

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