Question

The sum of the area of the two square fields is 468m2.If the difference of their perimeter is 24m.find the sides of the two square.

Answer 1

Let us say that the sides of the two squares are 'a' and 'b'

Sum of their areas = a^2 + b^2 = 468

Difference of their perimeters = 4a - 4b = 24

=> a - b = 6

=> a = b + 6

So, we get the equation

(b + 6)^2 + b^2 = 468

=> 2b^2 + 12b + 36 = 468

=> b^2 + 6b - 216 = 0

=> b = 12

=> a = 18

The sides of the two squares are 12 and 18.