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.

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.