Q.1 In figure, find the values of x and y and then show that AB || CD.
Since AB || CD and transversal PQ intersects them at R and S respectively.
Since [Alternate angles]
But [Vertically opp. angles]
[Since ]
Hence , x = y = 130º
Q.2 In figure , if AB || CD, CD|| EF and y : z = 3 : 7 , find x.
Since CD|| EF and transversal PQ intersects them at S and T respectively.
Since [Alternate Angles ]
[Since being linear pair]
Given y : z = 3 : 7
So, the sum of ratios = 3 + 7 = 10
Therefore
and
Since AB || CD and transversal PQ intersects them at R and S respectively.
Since [Consecutive interior angles are supplementary]
[Since y = 54º]
Hence, x = 126º
Q.3 In figure , if AB || CD, EF CD and = 126º, find
Since AB || CD and transversal GE cuts them at G and E respectively.
Since [Alternate angles]
[Since (given)]
and , [Alternate angles]
Hence,
Q.4 In figure , if PQ || ST, and , find .
Sol. Produce PQ to intersect SR at point M.
Now, PM || ST and transversal SM intersects them at M and R respectively.
Since [Alternate angles]
130º [Since 130º(given)]
180º – 130º = 50º [Since 180º( linear pairs)]
Since ray RQ stands at Q on PM.
Since 180º (Linear pair)
110º + = 180º
70º
Since = 180º – (70º + 50º) = 60º [Since sum of the angles of a triangle is 180º]
Q.5 In figure , if AB || CD, 50º and 127º , find x and y.
Therefore AB || CD and transversal PQ intersects them at P and Q respectively.
Therefore [Alternate angles]
x = 50º [Since 50º (given)]
Since AB || CD and transversal PR intersects them at P and R respectively.
Therefore [Alternate angles]
= 127º [Since 127º]
50º + y = 127º [Since 50º]
y = 127º – 50º = 77º
Hence , x = 50º and y = 77º
Q.6 In 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.
Two plane mirrors PQ and RS are placed parallel to each other i.e., PQ || RS. An incident ray AB after reflections takes the path BC and CD.
BN and CM are the normals to the plane mirrors PQ and RS respectively.
Since BN PQ, CM RS and PQ || RS
Therefore BN RS
BN || CM
Thus BN and CM are two parallel lines and a transversal BC cuts them at B and C respectively.
Therefore [Alternative interior angles]
But , [By laws of reflection]
Therefore,
[Since ]
Thus, lines AB and CD are intersected by transversal BC such that -
i.e., alternate interior angles are equal.
Therefore, AB || CD.
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