A wire when bent in the form of a square encloses an area = 576 cm². 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.

(i) Area of the square = 576 cm²

Let a be the length of side of the square.

⇒ a² = 576

⇒ a = $\sqrt{576}$

⇒ a = 24 cm

Total length of the wire = Perimeter of the square = 4 x 24 cm = 96 cm

Perimeter of the square = Perimeter of equilateral triangle.

⇒ 3 x side = 96 cm

⇒ side = $\dfrac{96}{3}$ cm

⇒ side = 32 cm

Area of equilateral triangle = $\dfrac{\sqrt{3}}{4}$ x side²

= $\dfrac{\sqrt{3}}{4}$ x (32)² cm²

= $\dfrac{\sqrt{3}}{4}$ x 1,024 cm²

= 256 $\sqrt{3}$ cm²

Hence, the area of equilateral triangle is 256 $\sqrt{3}$ cm².

(ii) Given:

Let l be the length and b be the breadth of the rectangle.

l - b = 4 ……………(1)

Perimeter of rectangle = Perimeter of square

⇒ 2(l + b) = 96 cm

⇒ 2(l + b) = 96 cm

⇒ l + b = $\dfrac{96}{2}$ cm

⇒ l + b = 48 cm ……………(2)

Add equation (1) and (2), we get

⇒ (l - b) + (l + b) = 4 + 48

⇒ l - b + l + b = 52

⇒ 2l = 52

⇒ l = $\dfrac{52}{2}$

⇒ l = 26 cm

So, b = l - 4 = 26 - 4 = 22 cm

Area of rectangle = l x b

= 26 x 22 cm²

= 572 cm²

Hence, the area of rectangle is 572 cm².