Question

In the figure, PB and QA are perpendicular to segment AB. If OA = 5 cm, PO = 7cm and area ($\Delta$QOA) = 150 $c{m}^2$, find the area of $\Delta$POB.

• A. $233c{m}^2$
• B. $420c{m}^2$
• C. $300c{m}^2$
• D. $294c{m}^2$

Answer 1

The correct option is D

$294c{m}^2$

Consider $\Delta$ QOA and $\Delta$ POB

We have QA || PB,

$\therefore$ $\angle$ AQO = $\angle$ PBO [Alternate angles]

$\angle$ QAO = $\angle$ BPO [Alternate angles]

and

$\angle$ QOA = $\angle$ BOP [Vertically opposite angles]

$\Delta$ QOA ~ $\Delta$BOP [AA similarity]

$\therefore$ $\frac{OQ}{OB}$ = $\frac{OA}{OP}$

The ratio of the area of two similar triangles are equal to the ratio of the squares of their corresponding sides
So,
$\frac{area\left(POB\right)}{area\left(QOA\right)}=\frac{(OP{)}^2}{(OA{)}^2}=\frac{7^2}{5^2}Given,area\left(QOA\right)=150c{m}^2\Rightarrow area\left(POB\right)=\frac{49}{25}\left(150\right)=294c{m}^2$