Question

In triangle $ABC$; angle $A={90}^{\circ }$, side $AB=x$ cm, $AC=(x+5)$ cm and area $=$ $150$ cm${}^2$. Find the $AB$

Answer 1

Since, it is a right angled triangle, $AB$ is the base and $AC$ is the height.
Area of triangle $=\frac{1}{2}\times$ base $\times$ height $=150$ sq cm
$\Rightarrow \frac{1}{2}\times x\times (x+5)=150$
$\Rightarrow x^2+5x=300$
$\Rightarrow x^2+20x-15x-300=0$
Solving it $x=15;-20$
Since side cannot be negative, $AB$ $=15$ cm

$AC$ $=x+5=20$ cm
Since the third side $BC$ is the hypotenuse, ${BC}^2={15}^2+{20}^2=625$.
Hence, $BC$ $=25$ cm