A wire when bent in the form of a square encloses an area of 484 m². Find the largest area enclosed by the same wire when bent to form :

(i) an equilateral triangle.

(ii) a rectangle of breadth 16 m.

(i) Area of the square = 484 m²

Let a be the length of side of the square.

⇒ a² = 484

⇒ a = $\sqrt{484}$

⇒ a = 22 m

Total length of the wire = Perimeter of the square = 4 x 22m = 88 m

Perimeter of the square = Perimeter of equilateral triangle.

⇒ 3 x side = 88 m

⇒ side = $\dfrac{88}{3}$ m

⇒ side = 29.3 m

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

= $\dfrac{\sqrt{3}}{4}$ x (29.3)² m²

= $\dfrac{\sqrt{3}}{4}$ x 858.49 m²

= 372.57 m²

Hence, the area of the equilateral triangle is 372.57 sq. m.

(ii) Given:

Length of the rectangle = 16 m

Let b be the breadth of the rectangle.

Perimeter of the rectangle = Perimeter of the square

⇒ 2(l + b) = 88 m

⇒ 2(16 + b) = 88 m

⇒ 16 + b = $\dfrac{88}{2}$ m

⇒ 16 + b = 44 m

⇒ b = 44 - 16 m

⇒ b = 28 m

Area of the rectangle = l x b

= 16 x 28 m²

= 448 m²

Hence, the area of the rectangle is 448 sq. m.