Question

In the figure given below, sides PB and QA are perpendiculars drawn to the line segment AB.

If PO = 6 cm, QO = 9 cm and area of $\Delta POB=120c{m}^2$, then find the area of $\Delta QOA$.

Answer 1

In $\Delta POB$ and $\Delta QOA$,

$\angle PBO=\angle QAO={90}^{\circ }$

$\angle POB=\angle QOA$
(Since vertically opposite angles are equal)$\left[1Mark\right]$

$\Delta POB\sim \Delta QOA$
(by AA similarity criterion) $\left[1Mark\right]$

$\Rightarrow \frac{ar\left(\Delta POB\right)}{ar\left(\Delta QOA\right)}=\frac{P{O}^2}{Q{O}^2}$

(The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides).

$\Rightarrow \frac{120c{m}^2}{ar\left(QOA\right)}=\frac{6^2}{9^2}$

$\Rightarrow ar\left(QOA\right)=\frac{120c{m}^2\times 81}{36}=270c{m}^2$ $\left[1Mark\right]$