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प्रश्न
The sum of the areas of two squares is 400 sq.m. If the difference between their perimeters is 16 m, find the sides of two squares.
उत्तर
Let the sides of the two squares be x cm and y cm(x > y).
Then, their areas are x2 and y2 and their perimeters are 4x and 4y.
According to the first condition,
Sum of the areas of two squares is 400 sq.m
∴ x2 + y2 = 400 ........(i)
According to the second condition,
difference between the perimeters is 16 m
∴ 4x − 4y = 16
∴ 4(x − y) = 16
∴ x − y = 4
∴ x = y + 4
Substituting the value of x in equation (i), we get
(y + 4)2 + y2 = 400
∴ y2 + 8y + 16 + y2 = 400
∴ 2y2 + 8y – 384 = 0
∴ y2 + 4y – 192 = 0
∴ y2 + 16y – 12y – 192 = 0
∴ y(y + 16) – 12(y + 16) = 0
∴ (y + 16) (y – 12) = 0
∴ y + 16 = 0 or y – 12 = 0
∴ y = – 16 or y = 12
But, y ≠ – 16 as the side of a square cannot be negative.
∴ y = 12
∴ x = y + 4 = 12 + 4 = 16
∴ The sides of the two squares are 16 cm and 12 cm.
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