Question

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

Answer 1

In right-angled $△ABC$
$A{C}^2=A{B}^2+B{C}^2$
$\Rightarrow (x+6{)}^2=(x-3{)}^2+(x+4{)}^2$
$\Rightarrow (x^2+12x+36)=(x^2-6x+9)+(x^2+8x+16)$
$\Rightarrow x^2-10x-11=0$
$\Rightarrow (x-11)(x+1)=0$
But length of the side of a triangle can not be negative.
$\Rightarrow x=11cm$
$\therefore AB=(x-3)=(11-3)=8cm$
$BC=(x+4)=(11+4)=15cm$
$AC=(x+6)=(11+6)=17cm$