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 $△ABE$ and $△ACD$.

• A. $1:2$
• B. $3:1$
• C. $1:3$
• D. $4:1$

Answer 1

The correct option is A $1:2$

Given $ABC$ is an isosceles right angled triangle with $\angle B={90}^{\circ }$

$⟹AB=BC$

By Pythagoras theorem, $A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=2A{B}^2$ --------(i) (as $AB=BC$)

Given that, $ABE\cong ADC$

We know, Ratios of areas of similar triangles is equal to ratio of squares of their corresponding sides.

Hence,

$\frac{Areaof△ABE}{Areaof△ACD}=\frac{A{B}^2}{A{C}^2}$

$\frac{Areaof△ABE}{Areaof△ACD}=\frac{A{B}^2}{2A{B}^2}$ -------from(i)

$\frac{Areaof△ABE}{Areaof△ACD}=\frac{1}{2}$