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Q11. Sum of areas of two squares is 468 \(m^2\). If, the difference of their perimeters is 24 metres, find the sides of the two squares.
Answer :

Let perimeter of first square = x metres
Let perimeter of second square = (x + 24) metres
Length of side of first square = \({{x} \over {4}}\) metres {Perimeter of square = 4 × length of side}
Length of side of second square = \({{x + 24} \over {4}}\) metres
Area of first square = side × side = \({{x} \over {4}} × {{x} \over {4}} = {{x^2} \over {16}} m^2\)
Area of second square = \(({{x + 24} \over {4}})^2\)
According to given condition:
\({{x^2} \over {16}} + ({{x + 24} \over {4}})^2 = 468\)
=>\({{x^2} \over {16}} + {{x^2 + 576 + 48x} \over {16}} = 468\)
=>\({{x^2 + x^2 + 576 + 48x} \over {16}} = 468\)
=> \(2x^2 + 576 + 48x = (468)(16)\)
=> \(2x^2 + 576 + 48x = 7488\)
=>\(2x^2 + 48x - 6912 = 0\)
=>\(x^2 + 24x - 3456 = 0\)
Comparing equation \(x^2 + 24x - 3456 = 0\) with standard form \(ax^2 + bx + c = 0\),
We get a = 1, b =24 and c = -3456
Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-24 ± \sqrt{(24)^2 - 4(1)(-3456)}} \over {(2)(1)}}\)
=>\(x = {{-24 ± \sqrt{576 + 13824}} \over {2}} = {{-24 ± \sqrt{14400}} \over {2}}\)
=>\(x = {{-24 + 120} \over {2}} , {{-24 - 120} \over {2}}\)
=>\(x = 48, -72\)
Perimeter of square cannot be in negative. Therefore, we discard x=-72.
Therefore, perimeter of first square = 48 metres
And, Perimeter of second square = x + 24 = 48 + 24 = 72 metres
=> Side of First square = \({{Perimeter} \over {4}} = {{48} \over {4}} = 12\) m
And, Side of second Square = \({{Perimeter} \over {4}} = {{72} \over {4}} = 18\) m