Question

In triangle ABC ; Angle $A={90}^0$ , side AB = x cm, AC = (x+5) cm and area $=150c{m}^2$. Find the sides of triangle.

Answer 1

R.E.F. Image.

Area $=150c{m}^2$

Area $=\frac{1}{2}\times base\times height=\frac{1}{2}\times AC\times AB$

$150=\frac{1}{2}\times (x+5)\times x$

$\Rightarrow 300=x^2+5x$

$\Rightarrow x^2+5x-300=0$

$\Rightarrow x^2+20x-15x-300=0$

$\Rightarrow x^2(x+20)-15(x+20)=0$

$\Rightarrow (x+20)(x-15)=0$

$\Rightarrow x=15$

$\therefore AB=15cm,AC=x+5=15+5=20cm,$

$BC=\sqrt{{15}^2+{20}^2}=25cm$