Question

PQRS is a trapezium. OR = 9 cm, $\angle RQS={90}^o$, $PQ=5cm$ and $area\left(△QRS\right)=\frac{81}{2\sqrt{3}}c{m}^2$. Find the area of the trapezium

Answer 1

$QR=9cm,PQ=5cm$

$\Rightarrow ar\left(△QRS\right)=\frac{81}{2\sqrt{3}}c{m}^2$

$\Rightarrow \frac{1}{2}\times QR\times QS=\frac{81}{2\sqrt{3}}$

$\Rightarrow \frac{1}{2}\times 9\times x=\frac{81}{2\sqrt{3}}$

$\Rightarrow x=\frac{9}{\sqrt{3}}=3\sqrt{3}cm$

$\Rightarrow SR=\sqrt{Q{R}^2+Q{R}^2}=\sqrt{27+81}=\sqrt{108}=6\sqrt{3}cm$

Also $ar\left(△QRS\right)=\frac{1}{2}\times QT\times SR=\frac{81}{2\sqrt{3}}$

$\Rightarrow \frac{1}{2}\times y\times 6\sqrt{3}=\frac{81}{2\sqrt{3}}$

$\Rightarrow y=\frac{81}{\sqrt{3}\times 6\sqrt{3}}cm$

$y=\frac{81}{6\times 3}cm$

$y=\frac{9}{2}cm=QT$

$\Rightarrow area\left(trap\right)=\frac{1}{2}\times (PQ+SR)\times QT$

$=\frac{1}{2}\times 4.5(5+6\sqrt{3})$

$=\frac{4.5}{2}\times 15.38$

$=4.5\times 7.19$

$=32.355c{m}^2$

Hence, the answer is $32.355c{m}^2.$