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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 sides of the squares are x m and = y m.Then
According to question,
Sum of the difference of their perimeter=64 m
4x - 4y = 64
x - y = 16
y = x - 16 .................... (1)
And sum of the areas of square = 640 m2
x2 + y2 = 640 ............ (2)
Putting the value of x in equation (2) from equation (1)
x2 + (x - 16)2 = 640
x2 + x2 - 32x + 256 = 640
2x2 - 32x + 256 - 640 = 0
2x2 - 32x - 384 = 0
2(x2 - 16x - 192) = 0
x2 - 16x - 192 = 0
x2 - 24x + 8x - 192 = 0
x(x - 24) + 8(x - 24) = 0
(x - 24)(x + 8) = 0
x - 24 = 0
x = 24
or
x + 8 = 0
x = -8
Sides of the square never are negative.
Therefore, putting the value of x in equation (1)
y = x - 16 = 24 - 16 = 8
Hence, sides of the square be 24m and 8m respectively.
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