. Sum of the areas of two squares is


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

Best Answer

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 or  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 + 6y – 216  = 0

= y2 +18y – 12y – 216 = 0

= y(y+18)(y – 12) = 0

= y = - 18  or 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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