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Question

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.

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Solution

Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and 4y respectively and their areas will be x2 and y2 respectively. It is given that
4x - 4y = 24
x - y = 6
x = y + 6
Also,x2+y2=468
(6+y)2+y2=468
36+y2+12y+y2=468
2y2+12y+432=0
y2+6y216=0
y2+18y12y216=0
y(y+18)12(y+18)=0
(y+18)(y12)=0
y=18,12
However, side of a square cannot be negative. Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

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