Compute the area of trapezium PQRS in figure

We have,

ar(trap. PQRS) = ar(rect. PSRT) + ar(ΔQRT)

⇒ ar(trap. PQRS) = PT × RT + 1/2(QT × RT)

= 8 × RT + 1/2(8 × RT) = 12 × RT

In ΔQRT, we have

$\mathrm{QR}^{2}=\mathrm{QT}^{2}+\mathrm{RT}^{2}$

$\Rightarrow \mathrm{RT}^{2}=\mathrm{QR}^{2}-\mathrm{QT}^{2}$

$\Rightarrow R T^{2}=17^{2}-8^{2}=225$

⇒ RT = 15

Hence, Area of trapezium $=12 \times 15 \mathrm{~cm}^{2}=180 \mathrm{~cm}^{2}$