Question

In the given figure, $\angle ABC=\angle AED={90}^{\circ }$, AB = 12 cm, AC = 15 cm and DE = 3 cm. The area of $\Delta ABC$ and $\Delta AED$ are ___ and ___ respectively.

• A. $54c{m}^2,6c{m}^2$
• B. $52c{m}^2,8c{m}^2$
• C. $56c{m}^2,6c{m}^2$
• D. $52c{m}^2,6c{m}^2$

Answer 1

The correct option is A

$54c{m}^2,6c{m}^2$

In $\Delta ABC$,

$A{C}^2=A{B}^2+B{C}^2$ (by Pythagoras theorem)

$\Rightarrow B{C}^2=(15{)}^2-(12{)}^2$

$\Rightarrow BC=225-144=81$

$\Rightarrow$ BC = 9 cm

Area of $△ABC=\frac{1}{2}\times BC\times AB$

$=\frac{1}{2}\times 9\times 12=54c{m}^2$

In $\Delta AED$ and $\Delta ABC$,

$\angle ABC=\angle AED={90}^{\circ }$ (given)

$\angle BAC=\angle DAE$
(common angle in both the triangles)


$\therefore \Delta AED\sim \Delta ABC$ (by AA similarity criterion)

$\Rightarrow \frac{ar\left(\Delta ADE\right)}{ar\left(\Delta ABC\right)}=\frac{E{D}^2}{B{C}^2}$

$\Rightarrow \frac{ar\left(\Delta ADE\right)}{54}=\frac{3^2}{9^2}[\because ar(\Delta ADE=54c{m}^2\left)\right]\Rightarrow ar\left(\Delta ADE\right)=\frac{9}{81}\times 54=6c{m}^2$