Question

For the parallelogram given above, find the values of x and y.

Answer 1

Given, $ABCD$ is a parallelogram.

We know that in parallelogram adjacent angles are supplementary.

$\therefore \angle A+\angle B={180}^o$

$\Rightarrow$ $4x+{20}^o+7y={180}^o$

$\Rightarrow$ $4x+7y={160}^o$ ------ ( 1 )

Also, $\angle A+\angle D={180}^o$

$\Rightarrow$ $4x+{20}^o+6x+3y-8={180}^o$

$\Rightarrow$ $10x+3y={168}^o$ ----- ( 2 )

Now, multiplying equation ( 1 ) by $10$ and equation ( 2 ) by $4$ we get,

$40x+70y={1600}^o$ ----- ( 3 )

$40x+12y={672}^o$ ----- ( 4 )

Now subtracting equation ( 4 ) from equation ( 3 ) we get,

$58y={928}^o$

$\Rightarrow$ $y={16}^o$

Substituting value of $y$ in equation ( 1 )

$\Rightarrow$ $4x+7\left({16}^o\right)={160}^o$

$\Rightarrow$ $4x={160}^o-{112}^o$

$\Rightarrow$ $4x={48}^o$

$\therefore$ $x={12}^o$

Hence, the values of $x$ and $y$ are ${12}^o$ and ${16}^o$ respectively.