Math Class 9 Chapter 6 Exercise 6.2 Questions Solutions Questions 1 and 2.
Exercise 6.2
Question 1.
In figure Given below, Find the values of x and y and Then show that AB || CD.
Sol. y = 130° [Vertically opposite angles]
Further, 50° + x = 180° [Linear pair]
⇒ x = 130°
Hence, x = y = 130°
Transversal intersects lines AB and CD.
Such that x = y. [Alternate interior angles]
Hence, AB || CD
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Question 2.In Figure, If AB || CD, CD || EF And y : z = 3 : 7, Find x.
AB || CD and CD || EF ...(i) [Given]
∠1 = y [Vertically opposite angles]
∠1 + z = 180° [CD EF and ∠1, ∠z are on the
same side of the transversal]
⇒ y + z = 180° ...(ii)
Given: y : z = 3 : 7 ⇒ y÷z = 3
÷7 ⇒ y = 3z÷7
∴
3z÷7+ z = 180° ⇒
10z÷7 = 180° [From (ii)]
⇒ z = 126°
∴ y = 180° – 126° = 54° [From (i)]
Now, AB CD and transversal intersects these lines.
∴ x + y = 180° ⇒ x + 54° = 180° ⇒ x = 180° – 54°
= 126°.
