Question

$ABCD$ is a square. Equilateral triangles $ACF$ and $ABE$ are drawn on the the diagonal $AC$ and side $AB$ respectively. Find area of $△ACF$ : area of $△ABE$.

• A. $8:1$
• B. $\sqrt{2}:1$
• C. $4:1$
• D. $2:1$

Answer 1

The correct option is D

$2:1$

Let the length of AB be $a\text{units.}.....\left(i\right)$

We know that diagonal of a square $=\sqrt{2}\times \text{side length}$
$\Rightarrow AC=\sqrt{2}\times AB\Rightarrow AC=\sqrt{2}a....\left(ii\right)$

Now, $△$ ABE and $△$ ACF are equilateral triangles.

Each angle of both $△$ ABE and $△$ ACF is ${60}^{\circ }$.

$\therefore △ABE\sim △$ ACF (by AAA similarity criterion)

So, from (i) and (ii), we can conclude that,
$\frac{ar\left(ACF\right)}{ar\left(ABE\right)}=\frac{A{C}^2}{A{B}^2}=\frac{2a^2}{a^2}=\frac{2}{1}$