Question

In the following figure, seg BE $\perp$ seg AB and seg BA $\perp$ seg AD. If BE$=6$ and AD$=9$, find $\frac{A\left(\Delta ABE\right)}{A\left(\Delta BAD\right)}$.

Answer 1

In the figure above , given that $BE\perp AB$ and $BA\perp AD$

Given that $BE=6$ and $AD=9$

The area of triangle $ABE$ is $\frac{1}{2}\times AB\times BE$

The area of triangle $BAD$ is $\frac{1}{2}\times BA\times AD$

The ratio of areas of triangle $ABE$ and $BAD$ is $\frac{BE}{AD}=\frac{6}{9}=\frac{2}{3}$