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# Ex 6.2 NCERT Solutions- Lines and Angles Class 9 Notes | EduRev

## Mathematics (Maths) Class 9

Created by: Full Circle

## Class 9 : Ex 6.2 NCERT Solutions- Lines and Angles Class 9 Notes | EduRev

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.
All you need of Class 9 at this link: Class 9

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. ∴ y = 130º                       [Vertically opposite angles]
Again, PQ is a straight line and EA stands on it.
∴ ∠ AEP + ∠ AEQ = 180º            [Linear pair]
or 50º + x = 180º
⇒ x = 180º -  50º = 130º                   …(2)
From (1) and (2), x = y
But they are the angles of a pair of interior alternate angles.
∴ AB || CD.

Question 2. In the following Figure, if AB || CD, CD ||EF and y : z = 3 : 7, find x. Solution:  ∴ AB || EF and PQ is a transversal.
∴ Interior alternate angles are equal.
∴ ∠ x= ∠ z                           …(1)
Again, AB || CD,
∴ Interior opposite angles are supplementary,
⇒ y + z = 180º
But y : z = 3 : 7 From (1) and (2), we have x = 126º

Question 3. In the following figure, if AB || CD, EF ⊥ CD and ∠ GED = 126º, find ∠ AGE, ∠ GEF and ∠ FGE.
Solution:
AB || CD and GE is a transversal. ∴ Interior alternate angles are equal.
∴ ∠ AGE = ∠ GED
But ∠ GED = 126º                [Given]
∴ ∠ AGE = 126º
Since ∠ GED = 126º
∴ ∠ GEF + ∠ FED = ∠ GED
or ∠ GEF + 90º = 126º
or ∠ GEF = 126 ∠ 90° = 36º
Next, AB || CD and GE is a transversal.
∴ ∠ FGE + ∠ GED = 180º
or ∠ FGE + 126º = 180º
or ∠ FGE = 180º - 126º = 54º
Thus, ∠ AGE = 126º, ∠ GEF = 36º and ∠ FGE = 54º

Question 4. In the following figure, if PQ || ST, ∠ PQR = 110º and ∠ RST = 130º, find ∠ QRS.
Hint: Draw a line parallel to ST through point R. Solution: ∵ PQ || ST [Given] and EF || ST [Construction] ∴ PQ || EF
and QR is a transversal, ∴ Interior alternate angles are equal i.e ∠ PQR = ∠ QRF
But ∠ PQR = 110º                    [Given]
∴ ∠ QRF = ∠ QRS + ∠ SRF = 110º              …(1)
Again ST || EF [Construction] and RS is a transversal.
∴ ∠ RST + ∠ SRF = 180º
or 130º + ∠ SRF = 180º
⇒ ∠ SRF = 180º - 130º = 50º
Now, from (1), we have
∠ QRS + 50º = 110º
⇒ ∠ QRS = 110º - 50º = 60º
Thus, ∠ QRS = 60º.

Question 5. In the following figure, if AB || CD, ∠ APQ = 50º and ∠ PRD = 127º, find x and y.
Solution:
We have AB || CD [Given] and PQ is a transversal. ∴ Interior alternate angles are equal.
∴ ∠ APQ = ∠ PQR
or 50º = x [∵ APQ = 50º                    (Given)] ….(1)
Again, AB || CD and PR is a transversal.
∴ ∠ APR = ∠ PRD                       [Interior alternate angles]
⇒ ∠ APR = 127º                         [∵ It is given that ∠ PRD = 127º]
But ∠ APR = ∠ APQ + ∠ QPR
∴ ∠ APQ + ∠ QPR = 127º
⇒ 50º + y = 127º                       [∵ It is given that ∠ APQ = 50º]
⇒ y = 127º - 50º = 77º
Thus, x = 50º and y = 77º

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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