Question

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

(i) an equilateral triangle.

(ii) a rectangle whose adjacent sides differ by 4 cm.

Answer 1

Area of a square$={side}^2$

$\Rightarrow 576={side}^2$

$\Rightarrow$ Side$=24$m

Length of wire$=$ Perimeter of square$=4\times$Side

$=4\times 24=96$m

$\left(i\right)$Perimeter of equilateral triangle$=$Length of wire

$3\times$Side of equilateral triangle$=96$

Side$=\frac{96}{3}=32$m

Area of an equilateral triangle$\frac{\sqrt{3}}{4}{\left(side\right)}^2=\frac{\sqrt{3}}{4}{\left(32\right)}^2=\frac{\sqrt{3}\times 1024}{4}=256\sqrt{3}{m}^2$

$\left(ii\right)$Perimeter of rectangle$=$Length of wire

$2(l+b)=96$

$\Rightarrow 2(36+b)=96$

$\Rightarrow 36+b=48$

$\Rightarrow b=48-36=12$m

Area of rectangle$l\times b=36\times 12=432m^2$

The equilateral triangle enclosed the more area as compare to a rectangle.