Question

In the figure $DE\parallel BC$ , $DE=4cm$, $BC=8cm$, $A\left(△ADE\right)=25{cm}^2$, find $A\left(△ABC\right)$

Answer 1

In $△$s, ABC and ADE,

$\angle BAC=\angle DAE$(Common)

$\angle ADE=\angle ABC$(Corresponding angles of parallel lines)

$\angle AED=\angle ACB$(corresponding angles of parallel lines)

Therefore, $△ABC\sim △ADE$

For similar triangles,

ratio of area of triangles = ratio of square of corresponding sides

Hence, $\frac{Ar.△ABC}{Ar.△ADE}=\frac{B{C}^2}{D{E}^2}$

$\frac{Ar.△ABC}{25}=\frac{8\times 8}{4\times 4}$

$Ar.△ABC=25\times 4$

$Ar.△ABC=100c{m}^2$