Question

In the given figure, AB || CD. Find the measure of x, y and z.

• A. $x={50}^{\circ }$, $y={90}^{\circ }$ and $z={40}^{\circ }$
• B. $x={40}^{\circ }$, $y={90}^{\circ }$ and $z={50}^{\circ }$
• C. $x={40}^{\circ }$, $y={50}^{\circ }$ and $z={90}^{\circ }$
• D. $x={90}^{\circ }$, $y={40}^{\circ }$ and $z={50}^{\circ }$

Answer 1

The correct option is B

$x={40}^{\circ }$, $y={90}^{\circ }$ and $z={50}^{\circ }$

Here, AB || CD, and PQ and PR are the two transversals.

We know that co-interior angles are supplementary.

$\Rightarrow \angle APQ+\angle CQP={180}^{\circ }$

$\Rightarrow x+\angle CQP={180}^{\circ }$

$\Rightarrow x+{140}^{\circ }={180}^{\circ }$

$\Rightarrow x={40}^{\circ }$

Similarly,

$\angle BPR+\angle PRD={180}^{\circ }$ (Co-interior angles)

$\Rightarrow z+{130}^{\circ }={180}^{\circ }$

$\Rightarrow z={50}^{\circ }$

Now, angles x, y and z form a line, hence they sum up to ${180}^{\circ }$.

$\Rightarrow x+y+z={180}^{\circ }$ (Linear pair)

$\Rightarrow {40}^{\circ }+y+{50}^{\circ }={180}^{\circ }$

$\Rightarrow y+{90}^{\circ }={180}^{\circ }$

$\Rightarrow y={90}^{\circ }$

$\therefore x={40}^{\circ },y={90}^{\circ }andz={50}^{\circ }$