Question

In the figure $segBE\perp segABandsegBA\perp segAD$.
If $BE=6$ and $AD=9$, If $\frac{A\left(△ABE\right)}{A\left(△BAD\right)}=\frac{m}{3}$.Find $m$

Answer 1

$\frac{A\left(△ABE\right)}{A\left(△BAD\right)}=\frac{\frac{1}{2}AB\times BE}{\frac{1}{2}AB\times AD}$ (Area of triangle = $\frac{1}{2}$ base $\times$ height)

$\frac{A\left(△ABE\right)}{A\left(△BAD\right)}=\frac{BE}{AD}$

$\frac{A\left(△ABE\right)}{A\left(△BAD\right)}=\frac{6}{9}$

$\frac{A\left(△ABE\right)}{A\left(△BAD\right)}=\frac{2}{3}$