Haryana State Board HBSE 9th Class Maths Solutions Chapter 6 Lines and Angles Ex 6.2 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 6 Lines and Angles Exercise 6.2

Question 1.

In the figure 6-49, find the values of x and y and then show that AB || CD.

Solution :

x + 50Â° = 180Â°,

(Linear pair axiom)

â‡’ x = 180Â° – 50Â°

â‡’ x = 130Â° …(i)

y = 130Â° …(ii)

(Vertically opposite angles)

From (i) and (ii), we get

x = y

Thus, a pair of alternate interior angles x and y are equal. Therefore, by theorem 6.3, we have

AB || CD. Hence proved.

Question 2.

In figure 6.50, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Solution:

Since, CD || EF and transversal line l intersects CD at Q and EF at R.

âˆ CQR = âˆ QRF,

(Alternate interior angles)

â‡’ âˆ CQR = z

âˆ PQC + âˆ CQR = 180Â°,

(Linear pair ax,iom)

â‡’ y + z = 180Â°

Now, y : z = 3 : 7

Sum of ratios = 3 + 7 = 10

âˆ´ y = \(\frac {3}{10}\) Ã— 180Â°

â‡’ y = 54Â°

and z = \(\frac {7}{10}\) Ã— 180Â°

â‡’ z = 126Â°

Since AB || CD and transversal line l intersects them at P and Q respectively. Therefore,

âˆ APQ + âˆ PQC = 180Â°,

(Co-interior angles are supplementary)

â‡’ x + y = 180Â°

â‡’ x + 54Â° = 180Â° [âˆµ y = 54Â°]

â‡’ x = 180Â° – 54Â°

â‡’ x = 126Â°

Hence, x = 126Â°.

Question 3.

In figure 6.51, if AB || CD, EF âŠ¥ CD and âˆ GED = 126Â°, find âˆ AGE, âˆ GEF and âˆ FGE.

Solution :

Since, AB || CD and transversal GE cuts them at G and E respectively.

âˆ AGE = âˆ GED,

(Alternate interior angles)

â‡’ âˆ AGE = 126Â°

[âˆµ âˆ GED = 126Â° given]

âˆ GED = 126Â°

and âˆ FED = 90Â° (EF âŠ¥ CD)

Now, âˆ GEF = âˆ GED – âˆ FED

â‡’ âˆ GEF = 126Â° – 90Â°

â‡’ âˆ GEF = 36Â°

âˆ FGE + âˆ GED = 180Â°

(Co-interior angles are supplementary)

â‡’ âˆ FGE + 126Â° = 180Â°

â‡’ âˆ FGE = 180Â° – 126Â°

â‡’ âˆ FGE = 54Â°

Hence, âˆ AGE = 126Â°, âˆ GEF = 36Â° and âˆ FGE = 54Â°

Question 4.

In figure 6.52, if PQ || ST, âˆ PQR = 110Â° and âˆ RST = 130Â°, find âˆ QRS.

(Hint: Draw a line parallel to ST through point R.)

Solution :

Draw a line EF || ST through point R.

âˆ TSR + âˆ FRS = 180Â°

(Co-interior angles are supplementary)

â‡’ 130Â° + âˆ FRS = 180Â°,(âˆµ âˆ TSR = 130Â°)

â‡’ âˆ FRS = 180Â° – 130Â°

â‡’ âˆ FRS = 50Â°

PQ || ST, (Given)

EF || ST, (By construction)

PQ || EF, (By theorem 6.6)

âˆ PQR + âˆ ERQ = 180Â°

(Co-interior angles are supplementary)

â‡’ 110Â° + âˆ ERQ = 180Â° (âˆµ âˆ PQR = 110Â°)

â‡’ âˆ ERQ = 180Â° – 110Â°

â‡’ âˆ ERQ = 70Â°

Now, âˆ ERQ + âˆ QRF = 180Â°,

(Linear pair axiom)

â‡’ âˆ ERQ + âˆ QRS + âˆ FRS = 180Â°

â‡’ 70Â° + âˆ QRS + 50Â° = 180Â°

(âˆµ âˆ ERQ = 70Â° and âˆ FRS = 50Â°)

â‡’ 120Â° + âˆ QRS = 180Â°

â‡’ âˆ QRS = 180Â° – 120Â°

â‡’ âˆ QRS = 60Â°

Question 5.

In figure 6.54, if AB || CD, âˆ APQ = 50Â° and âˆ PRD = 127Â°, find x and y.

Solution :

AB || CD and transversal PQ intersects them at P and Q respectively.

âˆ´ âˆ PQR = âˆ APQ

(Alternate interior angles)

â‡’ x = 50Â°, (âˆµ âˆ APQ = 50Â°)

Also, AB || CD and transversal PR intersects them at P and R respectively.

âˆ´ âˆ APR = âˆ PRD

â‡’ âˆ APQ + âˆ QPR = 127Â°,(âˆµ âˆ PRD = 127Â°)

â‡’ 50Â° + y = 127Â°, (âˆµ âˆ APQ = 50Â°, âˆ QPR = y)

â‡’ y = 127Â° – 50Â°

â‡’ y = 77Â°

Hence, x = 50Â° and y = 77Â°.

Question 6.

In figure 6.55, 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.

Solution:

Given: Two mirrors PQ and RS are placed such that PQ || RS. 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.

To prove: AB || CD.

Construction: Draw normals BM on PQ and CN on RS.

Proof: BM âŠ¥ PQ and CN âŠ¥ RS.

â‡’ BM || CN

Now, BM || CN and BC is the transversal.

âˆ´ âˆ 2 = âˆ 3, (Alternate interior angles) …(i)

âˆ 1 = âˆ 2 and âˆ 3 = âˆ 4,

(By laws of reflection) …(ii)

From (i), we have

2âˆ 2 = 2âˆ 3

â‡’ âˆ 2 + âˆ 2 = âˆ 3 + âˆ 3

â‡’ âˆ 1 + âˆ 2 = âˆ 3 + âˆ 4,

[From (ii), âˆ 1 = âˆ 2, âˆ 3 = âˆ 4]

â‡’ âˆ ABC = âˆ BCD

Thus, a pair of alternate interior angles âˆ ABC and âˆ BCD are equal. Therefore, by theorem 6.3, we have AB || CD. Hence proved