**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 : 7Let the common ratio between

*y*and*z*be*a*.∴

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

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

*y*= 180Âº [from equation (1)]

or, 7

*a*+ 3*a*= 180Âºor, 10

*a*= 180Âº*or, a*= 18Âº

so,

*x*= 7*a*= 7 × 18Âº = 126ÂºQ3: 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)

(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)

∠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*= 77ÂºQ6: 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

what are laws of reflection

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