Question

Find the dimensions of a rectangular park whose perimeter is 60 m and area 200 m².

Answer 1

Let the length of the rectangle be x
And the breadth of the rectangle be y
xy = 200  (∵ area = length × breadth)

And 2(x + y) = 60     [∵ perimeter = 2(length + breadth)]

substitute `y = 200/x` in 2(x + y) = 60

Equation becomes:

`2(x + 200/x) = 60`

`2((x^2 + 200)/x) = 60`

`2x^2 - 60x + 400 = 0`

`x^2 - 30x + 200 = 0`

Using

`x = (-b±sqrt(b^2 - 4ac))/(2a)`

Plugging the values we get:

`x = (30±sqrt((-30)^2 - 4(1)(200)))/2`

`x = (30±10)/2`

`x = 40/2, 20/2`

x = 20,10

when x = 20 then y = 10
And when x = 10 then y = 20.