Sum of the areas of two squares is 468m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Let the side of the first square be 'a'm and that of the second be 'A'm.
Area of the first square=a²sq m.
Area of the second square=A²sq m.
Their perimeters would be 4a and 4A respectively.
Given 4A−4a=24
A−a=6–(1)
A²+a²=468–(2)
From (1), A=a+6
Substituting for A in (2), we get
(a+6)²+a²=468
a²+12a+36+a²=468
2a²+12a+36=468
a²+6a+18=234
a²+6a−216=0
a²+18a−12a−216=0
a(a+18)−12(a+18)=0
(a−12)(a+18)=0
a=12, −18
So, the side of the first square is 12m. and the side of the second square is 18m.