The area of a rectangle is 640 m². Taking its length as x m; find, in terms of x, the width of the rectangle. If the perimeter of the rectangle is 104 m; find its dimensions.

Given:

Area of the rectangle = 640 m²

Length = x meters

Let b be the width of the rectangle.

Area of the rectangle = l x b

⇒ x × b = 640

⇒ b = $\dfrac{640}{x}$ m

By formula,

Perimeter of the rectangle = 2(l + b)

⇒ $2\Big(x + \dfrac{640}{x}\Big)$ = 104

⇒ $\Big(x + \dfrac{640}{x}\Big) = \dfrac{104}{2}$

⇒ $\Big(x + \dfrac{640}{x}\Big)$ = 52

⇒ x² + 640 = 52x

⇒ x² - 52x + 640 = 0

⇒ x² - 32x - 20x + 640 = 0

⇒ x(x - 32) - 20(x - 32) = 0

⇒ (x - 20)(x - 32) = 0

⇒ x - 20 = 0 or x - 32 = 0

⇒ x = 20 m or x = 32 m.

Now,

⇒ b = $\dfrac{640}{x}$

Case 1 : x = 20 m

⇒ b = $\dfrac{640}{20}$ = 32 m.

Case 2 : x = 32 m

⇒ b = $\dfrac{640}{32}$ = 20 m.

Hence, the width of rectangle is $\dfrac{640}{x}$ and the dimensions of the rectangle are 20 m and 32 m.