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Q6. The diagonal of a rectangular field is 60 metres more than the shorter side. If, the longer side is 30 metres more than the shorter side, find the sides of the field.
Answer :

Let shorter side of rectangle = x metres
Let diagonal of rectangle = (x + 60) metres
Let longer side of rectangle = (x + 30) metres
According to pythagoras theorem,
\((x + 60)^2 = (x + 30)^2 + x^2\)
=>\( x^2 + 3600 + 120x = x^2 + 900 + 60x + x^2\)
=>\(x^2 - 60x - 2700 = 0\)
Comparing equation with \(x^2 - 60x - 2700 = 0\) standard form \(ax^2 + bx + c = 0\),
We get a = 1, b = - 60 and c = - 2700
Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{60 ± \sqrt{(60)^2 - 4(1)(-2700)}} \over {(2)(1)}}\)
=>\(x = {{60 ± \sqrt{3600 + 10800}} \over {2}} = {{60 ± \sqrt{14400}} \over {2}}\)
=>\(x = {{60 + 120} \over {2}} , {{60 - 120} \over {2}}\)
=>\(x = 90,-30\)
We ignore –30. Since length cannot be in negative.
Therefore, x = 90 which means length of shorter side = 90 metres
And length of longer side = x + 30 = 90 + 30 = 120 metres
Therefore, length of sides are 90 and 120 in metres.