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

C

$64\sqrt{3}c{m}^2$

Diagonal of a square $12\sqrt{2}cm$

Since, square of Diagonal$=2(DC{)}^2=(12\sqrt{2}{)}^2$
$D{C}^2=\frac{144\times 2}{2}=144\Rightarrow DC=12cm$
$\therefore$ Perimeter of square$=4\left(12\right)cm$
$=48cm$
Perimeter of equilateral triangle = 3a where a is side of the triangle
$a=\frac{48}{3}=16cm$
Area of equilateral triangle $=\frac{\sqrt{3}}{a}{a}^2$
$=\frac{\sqrt{3}}{4}(16{)}^2=\frac{\sqrt{3}}{4}\times 16\times 16c{m}^2$
$=64\sqrt{3}c{m}^2$