Question

An isosceles right triangle has its area $20x^2$. Find the area of the equilateral triangle constructed on one of its legs?

Answer 1

In the isosceles right triangle $ABC$, $AB=BC=a$ cm.

We know that the area of triangle $ABC$ is $A=\frac{1}{2}AB\times BC$.

Here, it is given that the area of triangle is $20x^2$ and $AB=BC=a$ cm, therefore,

$20x^2=\frac{1}{2}\times a\times a\Rightarrow 2\times 20x^2=a^2\Rightarrow a^2=40x^2.......\left(1\right)$

We know that the area of an equilateral triangle $ABC$ is $A=\frac{\sqrt{3}}{4}AB\times BC$.

Here, it is given that$AB=BC=a$ cm, therefore,

$A=\frac{\sqrt{3}}{4}\times a\times a\Rightarrow A=\frac{\sqrt{3}{a}^2}{4}\Rightarrow A=\frac{\sqrt{3}\times 40x^2}{4}\left(from\right(1\left)\right)\Rightarrow A=10\sqrt{3}{x}^2$

Hence, area of equilateral triangle is $10\sqrt{3}{x}^2$.