Question

ABC is an isosceles triangle right angled at B. Equilateral ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of $\Delta ABE$ and $\Delta ACD$ .

Answer 1

Given that:- $△ABC$ is an isosceles triangle and $\angle ABC=90°$

$\therefore AB=BC$

$△ABE\sim △ACD(\because \text{All equilateral triangles are similar})$

To find:- $\frac{ar\left(△ABE\right)}{ar\left(△ACD\right)}=?$

Solution:-

In $△ABC$,

Using pythagoras theorem,

${AC}^2={AB}^2+{BC}^2$

${AC}^2={AB}^2+{AB}^2[\because AB=AC]$

${AC}^2=2{AB}^2.....\left(i\right)$

Now In $△ABE$ and $△ACD$

$△ABE\sim △ACD\left(\text{Given}\right)$,

As we know that ratio of area of similar triangles is equal to the ratio of squares of their corresponding sides.

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

$\Rightarrow \frac{ar\left(△ABE\right)}{ar\left(△ACD\right)}=\frac{{AB}^2}{2{AB}^2}\left[\text{From}\left(i\right)\right]$

$\Rightarrow \frac{ar\left(△ABE\right)}{ar\left(△ACD\right)}=\frac{1}{2}$

$\Rightarrow ar\left(△ABE\right):ar\left(△ACD\right)=1:2$

Hence the ratio between the area of $△ABE$ to the area of $△ACD$ is $1:2$.