Question

In the following figure seg $AB\perp$ seg $BC$, seg $DC\perp$ seg $BC$. If $AB=2$ and $DC=3$, find $\frac{A\left(△ABC\right)}{A\left(△DCB\right)}$.

Answer 1

In the given figure seg $AB\perp$ seg $BC$ and seg $DC\perp$ seg $BC$
$AB=2$ and $DC=3$ (Given)
To find: $\frac{A\left(△ABC\right)}{A\left(△DCB\right)}$
$△ABC$ and $△DCB$ have equal height.
$\therefore \frac{A\left(△ABC\right)}{A\left(△DCB\right)}=\frac{\frac{1}{2}\times AB\times BC}{\frac{1}{2}\times DC\times BC}$
$=\frac{AB}{DC}=\frac{2}{3}$