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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.
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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