Question

A piece of wire $132$ dm long is bent successively in the shape of an equilateral triangle, a square, a regular hexagon and a circle. Then the area included is largest when the shape is

• A. Triangle
• B. Square
• C. Hexagon
• D. Circle

Answer 1

The correct option is D Circle

Let each side of equilateral triangle be $a$ a square be $b$, a regular hexagon be $c$.

Let the radius of circle be $r$

$3a=132$

$a=\frac{132}{3}$

$4b=132$

$\Rightarrow b=\frac{132}{4}$

$6c=132$

$\Rightarrow c=\frac{132}{6}$

$2\pi r=132$

$\Rightarrow r=\frac{132}{2\pi }$

Areas

Equilateral triangle: $\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}(\frac{132}{3}{)}^2$

$=\frac{\sqrt{3}}{4}\times (44{)}^2=484\sqrt{3}d{m}^2$

$=838.29d{m}^2$

Squar: $b^2=(\frac{132}{4}{)}^2$

$={33}^2$

$=1089d{m}^2$

Hexagon : $6\times \frac{\sqrt{3}}{4}{c}^2=\frac{6\sqrt{3}}{4}\times \frac{132}{6}\times \frac{132}{6}$

$=726\sqrt{3}d{m}^2$

$=1257.43d{m}^2$

Circle : $\pi r^2=\pi (\frac{132}{2\pi }{)}^2=\pi \times \frac{(132{)}^2}{4{\pi }^2}$

$=\frac{(132{)}^2}{4\pi }=1387.26d{m}^2$

Here , area is largest for circle.