A wire when bent in the form of an equilateral triangle encloses an area of $36\sqrt{3}$ cm². Find the area enclosed by the same wire when bent to form:

(i) a square, and

(ii) a rectangle whose length is 2 cm more than its width.

Given,

Area of equilateral triangle = $36\sqrt{3}$ cm²

Consider x cm as the side of equilateral triangle

We know that,

Area of an equilateral triangle = $\dfrac{\sqrt{3}}{4}(\text{ side})^2$

Substituting the values we get,

$\Rightarrow \dfrac{\sqrt{3}}{4}(\text{ side})^2 = 36\sqrt{3} \\[1em] \Rightarrow \text{side}^2 = \dfrac{36 \sqrt{3} \times 4}{\sqrt{3}} \\[1em] \Rightarrow \text{side}^2 = 144 \\[1em] \Rightarrow \text{side} = \sqrt{144} = 12 \text{ cm}.$

By formula,

Perimeter of equilateral triangle = 3 × side = 3 × 12 = 36 cm.

(i) As the same wire is now bent to form a square.

∴ Perimeter of equilateral triangle = Perimeter of square

36 = 4 × side

Side = $\dfrac{36}{4}$ = 9 cm.

Area of square = side × side = 9 × 9 = 81 cm².

Hence, area enclosed by wire in form of square = 81 cm².

(ii) As the same wire is now bent to form a rectangle.

∴ Perimeter of triangle = Perimeter of rectangle ……..(1)

According to the condition given for rectangle,

Length is 2 cm more than its width

Let width of rectangle = x cm

∴ Length of rectangle = (x + 2) cm

Perimeter of rectangle = 2(l + b)

Substituting the values in equation 1 we get,

⇒ 36 = 2[(x + 2) + x]

⇒ 36 = 2[2x + 2]

⇒ 4x + 4 = 36

⇒ 4x = 32

⇒ x = 8 cm.

∴ Length of rectangle = x + 2 = 8 + 2 = 10 cm and Breadth of rectangle = x = 8 cm.

By formula,

Area of rectangle = length × breadth

= 10 × 8

= 80 cm².

Hence, area enclosed by wire in form of square = 80 cm².