Question

$Inthegivenfigure,\angle ABC=\angle AED=90°,AB=12cm,AC=15cmandDE=3cm.Findtheareaof△ABCand△AED.$

Answer 1

In ΔABC,

$A{C}^2=A{B}^2+B{C}^2\left(byPythagorastheorem\right)$

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

$\Rightarrow B{C}^2=225-144=81$

$\Rightarrow BC=9cm$

$Areaof\Delta ABC=\frac{1}{2}\times BC\times AB$

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

$In\Delta AEDand\Delta ABC,$

$\angle ABC=\angle AED=90°\left(given\right)$

$\angle BAC=\angle DAE$

(common angle in both the triangles)

$\therefore \Delta AED\sim \Delta ABC\left(byAAsimilaritycriterion\right)$

$\Rightarrow \frac{Ar\left(\Delta ADE\right)}{Ar\left(\Delta ABC\right)}={\left(\frac{ED}{BC}\right)}^2$

$\Rightarrow \frac{Ar\left(\Delta ADE\right)}{54}{\left(\frac{3}{9}\right)}^2[\because Ar(\Delta ABC=54c{m}^2\left)\right]$

$\therefore Ar\left(\Delta ADE\right)=\frac{9}{81}\times 54=6c{m}^2$