Question

In the adjoining figure, for the right angled triangles ABC and BCD, $\frac{AB}{CD}=\frac{1}{2}$. Find $\frac{A\left(\Delta ABC\right)}{A\left(\Delta BCD\right)}$.

Answer 1

$\frac{AB}{CD}=\frac{1}{2}$ find $\frac{A\left(\Delta ABC\right)}{A\left(\Delta BCD\right)}$

Area of triangle $=\frac{1}{2}\times$ base$\times$ height

So $\frac{A\left(\Delta ABC\right)}{A\left(\Delta BCD\right)}=\frac{\frac{1}{2}\times BC\times AB}{\frac{1}{2}\times BC\times CD}=\frac{AB}{CD}=\frac{1}{2}$.