The document Ex 6.2 NCERT Solutions- Lines and Angles Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.

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**Question 1. In the following figure, find the values of x and y and then show that AB || CD.**

**Solution:** In the figure, we have CD and PQ intersect at Y.

âˆ´ y = 130Âº [Vertically opposite angles]

Again, PQ is a straight line and EA stands on it.

âˆ´ âˆ AEP + âˆ AEQ = 180Âº [Linear pair]

or 50Âº + x = 180Âº

â‡’ x = 180Âº - 50Âº = 130Âº â€¦(2)

From (1) and (2), x = y

But they are the angles of a pair of interior alternate angles.

âˆ´ AB || CD.

**Question 2. In the following Figure, if AB || CD, CD ||EF and y : z = 3 : 7, find x.**

**Solution: **

âˆ´ AB || EF and PQ is a transversal.

âˆ´ Interior alternate angles are equal.

âˆ´ âˆ x= âˆ z â€¦(1)

Again, AB || CD,

âˆ´ Interior opposite angles are supplementary,

â‡’ y + z = 180Âº

But y : z = 3 : 7

From (1) and (2), we have x = 126Âº

**Question 3. In the following figure, if AB || CD, EF âŠ¥ CD and âˆ GED = 126Âº, find âˆ AGE, âˆ GEF and âˆ FGE. Solution:** AB || CD and GE is a transversal.

âˆ´ Interior alternate angles are equal.

âˆ´ âˆ AGE = âˆ GED

But âˆ GED = 126Âº [Given]

âˆ´ âˆ AGE = 126Âº

Since âˆ GED = 126Âº

âˆ´ âˆ GEF + âˆ FED = âˆ GED

or âˆ GEF + 90Âº = 126Âº

or âˆ GEF = 126 âˆ 90Â° = 36Âº

Next, AB || CD and GE is a transversal.

âˆ´ âˆ FGE + âˆ GED = 180Âº

or âˆ FGE + 126Âº = 180Âº

or âˆ FGE = 180Âº - 126Âº = 54Âº

Thus, âˆ AGE = 126Âº, âˆ GEF = 36Âº and âˆ FGE = 54Âº

**Question 4. In the following figure, if PQ || ST, âˆ PQR = 110Âº and âˆ RST = 130Âº, find âˆ QRS. Hint: Draw a line parallel to ST through point R.**

**Solution:** âˆµ PQ || ST [Given] and EF || ST [Construction] âˆ´ PQ || EF

and QR is a transversal,

âˆ´ Interior alternate angles are equal i.e âˆ PQR = âˆ QRF

But âˆ PQR = 110Âº [Given]

âˆ´ âˆ QRF = âˆ QRS + âˆ SRF = 110Âº â€¦(1)

Again ST || EF [Construction] and RS is a transversal.

âˆ´ âˆ RST + âˆ SRF = 180Âº

or 130Âº + âˆ SRF = 180Âº

â‡’ âˆ SRF = 180Âº - 130Âº = 50Âº

Now, from (1), we have

âˆ QRS + 50Âº = 110Âº

â‡’ âˆ QRS = 110Âº - 50Âº = 60Âº

Thus, âˆ QRS = 60Âº.

**Question 5. In the following figure, if AB || CD, âˆ APQ = 50Âº and âˆ PRD = 127Âº, find x and y. Solution: **We have AB || CD [Given] and PQ is a transversal.

âˆ´ Interior alternate angles are equal.

âˆ´ âˆ APQ = âˆ PQR

or 50Âº = x [âˆµ APQ = 50Âº (Given)] â€¦.(1)

Again, AB || CD and PR is a transversal.

âˆ´ âˆ APR = âˆ PRD [Interior alternate angles]

â‡’ âˆ APR = 127Âº [âˆµ It is given that âˆ PRD = 127Âº]

But âˆ APR = âˆ APQ + âˆ QPR

âˆ´ âˆ APQ + âˆ QPR = 127Âº

â‡’ 50Âº + y = 127Âº [âˆµ It is given that âˆ APQ = 50Âº]

â‡’ y = 127Âº - 50Âº = 77Âº

Thus, x = 50Âº and y = 77Âº

**Question 6. In the following 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.**

**Solution: REMEMBER**

(i) Perpendiculars to the parallel lines are parallel.

(ii) According to the laws of reflection, angle of incidence = angle of reflection

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