Question

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is $12\sqrt{2}cm$, then the area of the triangle is:

• A. $24\sqrt{2}c{m}^2$
• B. $24\sqrt{3}c{m}^2$
• C. $48\sqrt{3}c{m}^2$
• D. $64\sqrt{3}c{m}^2$

Answer 1

The correct option is D $64\sqrt{3}c{m}^2$

Let ABCD be a square.
Given, diagonal of the square $12\sqrt{2}\text{cm}$.
Then, BD = $12\sqrt{2}\text{cm}$


$\angle BCD={90}^{\circ }$ ($\because$ ABCD is a square)
Then, by Pythagoras' theorem, we have
$2(DC{)}^2=(12\sqrt{2}{)}^2$
$\Rightarrow D{C}^2=\frac{144\times 2}{2}\Rightarrow DC=12\text{cm}$

$\therefore \text{Perimeter of square}=4\left(12\right)=48cm$
($\because AB=BC=CD=AD$)

Given, the square and equilateral triangle have equal perimeters.
Perimeter of an equilateral triangle with side 'a' is 3a.
So, $3a=48\Rightarrow a=\frac{48}{3}=16\text{cm}$

We know, area of an equilateral triangle with side 'a' is given by $\frac{\sqrt{3}}{4}{a}^2$.

$\therefore \text{Area}=\frac{\sqrt{3}}{4}(16{)}^2=\frac{\sqrt{3}}{4}\times 16\times 16=64\sqrt{3}{\text{cm}}^2$