Question

A wire when bent in the form of an equilateral triangle encloses an area of $64\sqrt{3}c{m}^2.$ The area enclosed by the same wire when bent to form a square is ___ $c{m}^2$

• A. 16
• B. 81
• C. 100
• D. 144

Answer 1

The correct option is D

144

Let the side of the equilateral $\Delta le$ 'a' and the side of the square be 'x' we know, area of equilateral $\Delta =\frac{\sqrt{3}}{4}{a}^2$

$64\sqrt{3}=\frac{\sqrt{3}}{4}{a}^2$

$\therefore a^2=8^2\times 2^2$

$a=16cm$

Perimeter of equilateral $\Delta =3\left(16\right)$

= 48 cm

Hence, perimeter of square = 48 cm

i.e., $4\times =48\Rightarrow x=12cm$

Area of square $=x^2$

$=(12{)}^2$

$=144$

Hence (D)