÷ÈÓ°Ö±²¥

The opposite sides of a parallelogram are congruent.

In the given figure, $\stackrel{¯}{PS}â‰\dots \stackrel{¯}{QR}$implies that $a=8$and $\stackrel{¯}{PQ}â‰\dots \stackrel{¯}{SR}$implies that $b=15$.

Thus,

$\begin{array}{c}PQ=SR\\ PQ=SO+SR\\ a=6+4\\ =10\end{array}$

And$PS=QR$

Both pairs of opposite sides of a parallelogram are parallel, that is, $\stackrel{¯}{PS}âˆ\yen \stackrel{¯}{QR}$and $\stackrel{¯}{SR}$is a transversal then same side interior angles are supplementary.

From the figure, it can be observed that the transversal $\stackrel{¯}{SR}$is perpendicular to the line $\stackrel{¯}{PS}$, then $\stackrel{¯}{SR}$ is perpendicular to the line $\stackrel{¯}{QR}$. That is,

$\begin{array}{c}m∠SRQ=90\\ x=90°\end{array}$

Consider $\mathrm{Δ}QOR$, the sum of the measures of all the angles of a triangle is 180.

Therefore,

$\begin{array}{c}m∠QRO+m∠ROQ+m∠OQR=180\\ 90+45+y=180\\ 135+y=180\\ y=45\end{array}$

Consider $\mathrm{Δ}QOR$, here $∠OQRâ‰\dots ∠ROQ$.

Therefore, by converse isosceles triangle theorem, $\stackrel{¯}{OR}â‰\dots \stackrel{¯}{QR}$, that is,

$\begin{array}{c}QR=OR\\ =4\end{array}$

As, $PS=QR$,

$\begin{array}{c}PS=4\\ b=4\end{array}$

Therefore, the values of a, b, x and y are $10,4,90°$ and $45°$ respectively.