Question

A wire when bent in the form of a square encloses an area = $576c{m}^2$. Find the largest area enclosed by the same wire when bent to form an equilateral triangles.

• A. $256\sqrt{2}c{m}^2$
• B. $256\sqrt{3}c{m}^2$
• C. $256\sqrt{5}c{m}^2$
• D. $256\sqrt{7}c{m}^2$

Answer 1

The correct option is B $256\sqrt{3}c{m}^2$

Given, a wire when bent in the form of a square encloses an area of $576$ sq. m .
Area of Square$={\left(\text{side}\right)}^2$
$576={\left(\text{side}\right)}^2$
$\text{side}=\sqrt{576}=24cm$
Perimeter of square $=4\times \text{side}$
$=4\times 24$
$=96cm$.
When same wire is formed into equilateral triangle, then perimeter of square is equals to perimeter of triangle.
$\therefore$ Perimeter of equilateral triangle $=96cm$
or, $3\times \text{side}=96cm$
$\text{side}=\frac{96}{3}=32cm.$
Area of equilateral triangle thus formed $=\frac{\sqrt{3}\times {\left(\text{side}\right)}^2}{4}$
$=\frac{\sqrt{3}\times {32}^2}{4}$
$=256\sqrt{3}{cm}^2$