 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.

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