Question

In $\Delta ABC$, $\angle B={90}^{\circ }$, Find the sides of the triangle, if $AB=(x-3)cm,BC=(x+4)cm$ and $AC=(x+6)cm$.

Answer 1

In a right angled triangle with right angle at $B$,

$AB$ and $BC$ for the sides and $AC$ is the hypotenuse.

So, as per the pythagoras theorem, ${AC}^2={AB}^2+{BC}^2$

$\Rightarrow {(x+6)}^2={(x-3)}^2+{(x+4)}^2$

$\Rightarrow x^2+12x+36=x^2-6x+9+x^2+8x+16$

$\Rightarrow x^2-10x-11=0$

$\Rightarrow x^2-11x+x-11=0$

$\Rightarrow x(x-11)+1(x-11)=0$

$\Rightarrow x=-1,x=11$

When $x=-1,$ side $AB=-4$ which is not possible.

Hence, $x=11$ and sides are $AC=x+6=11;AB=x-3=8;BC=x+4=15$