Question

Sum of the areas of two squares is 468m square. if the difference of there perimeters is 24 m, find the sides of the two squares?

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