Question

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

• A. $12\sqrt{2}c{m}^2$
• 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 square has side $a$ and triangle $t$

Given, $4a=3t\Rightarrow a=\frac{3}{4}t$ Diagonial of square $=\sqrt{2}a$

$\Rightarrow \sqrt{2}a=12\sqrt{2}\Rightarrow a=12cm\Rightarrow t=\frac{4}{3}\times 12=16cm$

Area of equilateral triangle $=\frac{\sqrt{3}}{4}{t}^2=\frac{\sqrt{3}}{4}\left(256\right)=64\sqrt{3}$

$\Rightarrow \left(D\right)$