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Sum of the areas of two squares is 400 cm2. If the difference of their perimeters is 16 cm, find the sides of the two squares.
Asked by Topperlearning User | 26th Jul, 2017, 12:36: PM
Expert Answer:
Let the sides of the two squares be x cm and y cm where x > y.
Then, their areas are x2 and y2 and their perimeters are 4x and 4y.
By the given condition:
x2 + y2 = 400...(1)
and 4x - 4y = 16
4(x - y) = 16 x - y = 4
x = y + 4... (2)
Substituting the value of x from (2) in (1), we get:
(y + 4)2 + y2 = 400
y2 + 16 + 8y + y2 = 400
2y2 + 16 + 8y = 400
y2 + 4y - 192 = 0
y2 + 16y - 12y - 192 = 0
y(y + 16) - 12 (y + 16) = 0
(y + 16) (y - 12) = 0
y = -16 or y = 12
Since, y cannot be negative, y = 12.
So, x = y + 4 = 12 + 4 = 16
Thus, the sides of the two squares are 16 cm and 12 cm.