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Question
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.
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Solution
Let the sides of two squares by a and b respectively.
Then, area of one square, S1 = a2
And, area of second square, S2 = b2
Given, S1 + S2 = 400 cm2
⇒ a2 + b2 = 400 cm2 …..(1)
Also, difference in perimeter = 16 cm
⇒ 4a - 4b = 16 cm
⇒ a - b = 4
⇒ a = (4 + b)
Substituting the value of 'a' in (1), we get
(4 + b)2 + b2 = 400
⇒ 16 + 8b + b2 + b2 = 400
⇒ 2b2 + 8b - 384 = 0
⇒ b2 + 4b - 192 = 0
⇒ b2 + 16b - 12b - 192 = 0
⇒ b(b + 16) - 12(b + 16) = 0
⇒ (b +16)(b - 12) = 0
⇒ b + 16 = 0 or b - 12 = 0
⇒ b = - 16 or b = 12
Since, the side of a square cannot be negative, we reject - 16.
Thus, b = 12
⇒ a = 4 + b = 4 + 12 = 16
Hence, the sides of a square are 16 cm and 12 cm respectively.
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