Question

In the figure, $\angle POQ={30}^{\circ }$ and radius OP =12 cm, then find the given area of $△$ OPQ (Given $\pi$ = 3.14).

• A. $36c{m}^2$
• B. $30c{m}^2$
• C. $32c{m}^2$
• D. $34c{m}^2$

Answer 1

The correct option is A $36c{m}^2$

From figure,
$\sin {30}^{\theta }=\frac{1}{2}=\frac{PM}{OP}$

$\therefore PM=6$

$P{M}^2+O{M}^2=O{P}^2$

$\therefore 36+O{M}^2={12}^2$

$O{M}^2=108,OM=10.4$

$QM=1.6,PM=6$

$P{M}^2+M{P}^2={1.6}^2+6^2=38.56=P{Q}^2$

$PQ=8.63$

$\therefore a=12,h=6,Area=\frac{12\times 6}{2}=36c{m}^2$