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प्रश्न
Sum of the areas of two squares is 640 m2. If the difference of their perimeters is 64 m. Find the sides of the two squares.
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उत्तर
Let the side of the larger square be x and the side of the smaller square be y.
The sum of the areas is 640 m2.
x2 + y2 = 640 ...(1)
The difference of their perimeters (4 × side) is 64 m.
4x − 4y = 64
4(x − y) = 64
x − y = `64/4`
x − y = 16
x = y + 16 ...(2)
Putting the value of x in equation (2) from equation (1).
(y + 16)2 + y2 = 640
(y2 + 32y + 256) + y2 = 640
2y2 + 32y + 256 − 640 = 0
2y2 + 32y − 384 = 0
Divide the entire equation by 2 to simplify:
y2 + 16y − 192 = 0
y2 + 24y − 8y − 192 = 0
y2 + 24y − 8y − 192 = 0
y(y + 24) − 8(y + 24) = 0
(y + 24) (y − 8) = 0
y + 24 = 0 or y − 8 = 0
y = −24 or y = 8
Sides of the square are never negative.
∴ y = 8
∴ Side of the smaller square y = 8 m
Side of the larger square y + 16
= 8 + 16
= 24 m
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