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11. 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.

11. 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.

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 60782 Points
one year ago
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, 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.

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