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 between the areas of $\Delta ABE$ and $\Delta ACD$. [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark
Calculation : 2 Marks

Let 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,

$\Delta ABE\sim \Delta ACD$

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

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

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