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Question -

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.



Answer -

Let the sides of the two squares be x m and y m.
Therefore, their perimeter will be 4x and 4y respectively
And area of the squares will be x2 and y2 respectively.
Given,
4x – 4y = 24
x – y = 6
x = y + 6

Also, x+ y2 =468

(6 + y2) + y2 =468

36 + y2 + 12y + y2 =468

2y2 + 12y + 432 = 0

 y2 + 6y – 216 = 0

 y2 + 18y – 12y –216 = 0

 y(+18) -12(y + 18) = 0

(y + 18)(y – 12) = 0

 y = -18, 12

As we know, the side of a square cannot benegative.

Hence, the sides of the squares are 12 m and(12 + 6) m = 18 m.


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