Question

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

• A. $2:1$
• B. $1:1$
• C. $1:2$
• D. none

Answer 1

The correct option is C $1:2$

In $△ABC$

$⟹{AB}^2+{BC}^2={AC}^2$

$⟹{AB}^2+{AB}^2={AC}^2$

$⟹{AC}^2={2AB}^2-\left(1\right)$

Ratio of areas of similar triangle is equal to ratio of squares of their corresponding sides.

$⟹\frac{Area\left(△ABE\right)}{Area\left(△ACD\right)}=\frac{{AB}^2}{{AC}^2}$

using(1)

$⟹\frac{Area\left(△ABE\right)}{Area\left(△ACD\right)}=\frac{{AB}^2}{{2AB}^2}=\frac{1}{2}$