#### Question

Is it possible to design a rectangular park of perimeter 80 and area 400 m^{2}? If so find its length and breadth.

#### Solution 1

Let the breadth of the rectangle be = x meters. Then

Perimeter = 80 meters

2(length + breadth) = 80

(length + x) = 40

length = 40 - x

And area of the rectangle

length x breadth = 400

(40 - x)x = 400

40x - x^{2} = 400

x^{2} - 40x + 400 = 0

x^{2} - 20x - 20x + 400 = 0

x(x - 20) - 20(x - 20) = 0

(x - 20)(x - 20) = 0

Yes, it is possible.

Hence, breadth of the rectangular park be 20 meters and length be 20 meters.

#### Solution 2

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

Perimeter = 2 (l + b) = 80

l + b = 40

Or, b = 40 - l

Area = l×b = l(40 - l) = 40l - l^{2}40l - l^{2} = 400

l^{2} - 40l + 400 = 0

Comparing this equation with al^{2} + bl + c = 0, we get

a = 1, b = -40, c = 400

Discriminant = b^{2} - 4ac

(-40)^{2} - 4 × 400

= 1600 - 1600 = 0

b^{2} - 4ac = 0

Therefore, this equation has equal real roots. And hence, this situation is possible.

Root of this equation,l = `-b/(2a)`

`l = (40)/(2(1)) = 40/2 = 20`

Therefore, length of park, l = 20 m

And breadth of park, b = 40 - l = 40 - 20 = 20 m.