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Quadratic Equations
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30/05/2021 11:58 am
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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.
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30/05/2021 11:59 am
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.
4x – 4y = 24
x – y = 6
x = y + 6
Also, x2 + 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(y +18) -12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
⇒ y = -18, 12
As we know, the side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.