Question

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

Answer 1

Area of $△ABC=\frac{1}{2}\times b\times h$

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

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

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

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

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

$\Rightarrow x\ne -20$ and $x=15$

$AB=x=15$cm

$AC=x+15=15+5=20$cm

$BC=\sqrt{{AB}^2+{BC}^2}=\sqrt{{20}^2+{15}^2}=\sqrt{400+225}=\sqrt{625}=25$cms

Hence the sides of the triangle are $15$cm,$20$cm and $25$cm.