**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 : 7Let the common ratio between

*y*and*z*be*a*.∴

*y*= 3*a*and*z*= 7*a*Also,

*x*+*y*= 180º (Co-interior angles on the same side of the transversal)*or, z*+

*y*= 180º [from equation (1)]

or, 7

*a*+ 3*a*= 180ºor, 10

*a*= 180º*or, a*= 18º

so,

*x*= 7*a*= 7 × 18º = 126ºQ3: 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)

(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)

∠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*= 77ºQ6: 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

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