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

Solution :

Let x and y be the side length of squares.

x2 + y2 = 468 (1)

4x - 4y = 24

x - y = 6

x = 6 + y (2)

(6+y)2 + y2 = 468

36+y2+12y+y2-468 = 0

2y2 + 12y - 432 = 0

y2 + 6y - 216 = 0

(y - 12)(y + 18) = 0

y = 12 and y = -18 (not admissible)

If y = 12, then x = 18

So, the side length of required squares are 12 and 18 respectively.

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