Question

In the given figure, $DE\parallel BC$, if $AD=1.5cm,BD=2AD$ then $\frac{ar\left(\Delta ADE\right)}{ar\left(trapeziumBCED\right)}$

Answer 1

Given AD= 1.5 cm

$\therefore BD$ = 3 cm and AB= AD+BD

= 1.5+3

= 4.5 cm

proof : Given , in triangle ADE and ABC

$\angle A$ is common and DE||BC.

$\angle ADE=\angle ABC$

$\Rightarrow \angle AED=\angle ACB$ (corresponding angles)

$\therefore \Delta ADE\cong \Delta ABC$ (AA similarity)

$\frac{ar\left(\Delta ADE\right)}{ar\left(\Delta ABC\right)}=\frac{A{D}^2}{A{B}^2}=\frac{(1.5{)}^2}{(4.5{)}^2}=\frac{1}{9}$

$\frac{ar\left(\Delta ADE\right)}{ar\left(\Delta ABC\right)-ar\left(\Delta ADE\right)}=\frac{1}{9-1}$

$\therefore \frac{ar\left(\Delta ADE\right)}{ar\left(trapeziumBCED\right)}=\frac{1}{8}$