$XYZW$ is a square. If $XY = 17 \text{ cm}$,then find the area of $XYZW$ in $\text{cm}^2$.

In $\Delta PQR$,$M$ and $N$ are the midpoints of $PQ$ and $PR$ respectively. $X$ is any point on $QR$. Prove that,$ar(MXN) = \frac{1}{4} ar(PQR)$.

In $\Delta ABC$,$\angle B = 90^{\circ}$,$AB = 14 \, cm$ and $AC = 50 \, cm$,then find the area of $\Delta ABC$ in $cm^2$.

Write True or False and justify your answer:
$PQRS$ is a parallelogram whose area is $180 \, cm^{2}$ and $A$ is any point on the diagonal $QS$. The area of $\triangle ASR = 90 \, cm^{2}$.

In $\Delta PQR$,$\angle Q = 90^{\circ}$,$QR = 21 \text{ cm}$ and $PR = 29 \text{ cm}$,then find the area of $\Delta PQR$ in $\text{cm}^2$.

In the figure,$PSDA$ is a parallelogram. Points $Q$ and $R$ are taken on $PS$ such that $PQ = QR = RS$ and $PA \parallel QB \parallel RC$. Prove that $\operatorname{ar}(PQE) = \operatorname{ar}(CFD)$.