Question

ABC is an isosceles triangle, right-angled at B. Similar triangles, $△ACD$ and $△ABE$ are constructed on sides AC and AB. Find the ratio of the areas of $\Delta ABE$ and $\Delta ACD$.

• A. $1$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{2}$

Let's assume,
AB = BC = x
It is given that, $\Delta ABC$ is right-angled at B

$\therefore A{C}^2=A{B}^2+B{C}^2$
$\Rightarrow A{C}^2=x^2+x^2$
$\Rightarrow AC=\sqrt{2}x$

It is given that,

$\Delta ABE\sim \Delta ACD$

$\Rightarrow \frac{Area\left(\Delta ABE\right)}{Area\left(\Delta ACD\right)}=\frac{A{B}^2}{A{C}^2}$

$\Rightarrow \frac{Area\left(\Delta ABE\right)}{Area\left(\Delta ACD\right)}=\frac{x^2}{(\sqrt{2}x{)}^2}$

$\Rightarrow \frac{Area\left(\Delta ABE\right)}{Area\left(\Delta ACD\right)}=\frac{1}{2}$

$\therefore Area\left(\Delta ABE\right):Area\left(\Delta ACD\right)=1:2$