Question

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

• A. $366\sqrt{3}$ cm${}^2$
• B. $326\sqrt{3}$ cm${}^2$
• C. $256\sqrt{3}$ cm${}^2$
• D. $297\sqrt{3}$ cm${}^2$

Answer 1

The correct option is C $256\sqrt{3}$ cm${}^2$

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