Question

The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.

Answer 1

Let the shorter side $=xm$

diagonal of the rectangle $=x+16m$

Longer side $=x+14m$

$\therefore (AB{)}^2+(BC{)}^2=(AC{)}^2$

$\therefore (x+14{)}^2+x^2=(x+16{)}^2$

$\Rightarrow x^2+28x+196+x^2=x^2+32x+256$

$\Rightarrow x^2+x^2-x^2+28x-32x+196-256=0$

$\Rightarrow x^2-4x+60=0$

$\Rightarrow x^2-10x+6x-60=0$

$\Rightarrow x(x-10)+6(x-10)=0$

$\Rightarrow (x-10)(x+6)=0$

Hence, $x=10$ or $-6$

$x=-6$ is not admissible

So, $x=10$

Breadth of rectangle $=10m$

Length of rectangle $=10+14=24m$