Question

$ABC$ is an isosceles triangle right angled at $B.$ Similar triangles $ACD$ and $ABE$ are constructed on sides $AC$ and $AB.$ The ratio between the areas of $△ABE$ and $△ACD$ is:

• A. $\sqrt{2}:1$
• B. $1:2$
• C. $2:1$
• D. $1:\sqrt{2}$

Answer 1

Answer: (B)

$1:2$

$ABC$ is the right-angled isosceles triangle.
Hence, $AB=BC$
and, using pythagoras theorem,
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=2A{B}^2$
$AC=\sqrt{2}AB$

Given, $△ABE\sim △ACD,$

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

(By relation between sides and area of similar triangle)

$\frac{A\left(ABE\right)}{A\left(ACD\right)}=\frac{1}{2}$.