Is it possible to design a rectangular park of perimeter 80 m and area

1.	Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? If so, find its length and breadth.
Answer» Given: Perimeter of a rectangle 2(1 + b) = 80 ⇒ 6 + b = 80/2 = 40 Area of the rectangle, l × b = 400 If possible, let us suppose that length of the rectangle = x m say Then its breadth by equation (1) = 40 – x By problem area = x . (40 – x) = 400 ⇒ 40x – x² = 400 ⇒ x² – 40x + 400 = 0 Here a = 1; b = -40; c = +400 b² – 4ac = (-40)² – 4(1)(+400) = 1600 – 1600 = 0 ∴ The roots are real and equal. They are \(\frac{-b}{2a}\), \(\frac{-b}{2a}\) i.e., \(\frac{-(-40)}{2\times1}\) = \(\frac{40}{2a}\) = 20 ∴ The dimensions are 20 m, 20 m. (∴ The park is in square shape)

Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? If so, find its length and breadth.

Answer»

Given: Perimeter of a rectangle 2(1 + b) = 80

⇒ 6 + b = 80/2 = 40

Area of the rectangle, l × b = 400

If possible, let us suppose that length of the rectangle = x m say

Then its breadth by equation (1) = 40 – x

By problem area = x . (40 – x) = 400

⇒ 40x – x² = 400

⇒ x² – 40x + 400 = 0

Here a = 1; b = -40; c = +400

b² – 4ac = (-40)² – 4(1)(+400)

= 1600 – 1600 = 0

∴ The roots are real and equal.

They are \(\frac{-b}{2a}\), \(\frac{-b}{2a}\)

i.e., \(\frac{-(-40)}{2\times1}\) = \(\frac{40}{2a}\) = 20

∴ The dimensions are 20 m, 20 m.

(∴ The park is in square shape)