Question

The radius of the incircle of a triangle is $4cm$ and the segments into which one side divided by the point of contact are $6cm$ and $8cm$. Determine the value of x.

Answer 1

We know length of the tangents drawn from the external point are equal.

$BF=BD=6cm$

$CE=CD=8cm$

Let AF=AE=x cm

$AreaoftriangleABC=AreaoftriangleAOB+AreaoftriangleBOC+Areaof△COA$

$AreaoftriangleABC=\frac{1}{2}\times 4(6+x)+\frac{1}{2}\times 4\left(14\right)+\frac{1}{2}\times 4(8+x)$

$=4(14+x)$

Area of triangle ABC by Heron's formula

$S=\frac{14+6+x+8+x}{2}=14+x$

Area of triangle $=\sqrt{(14+x)\left(8\right)\left(6\right)\left(x\right)}$

$4(14+x)=\sqrt{48x(14+x)}$

$14+x=3x$

$2x=14$

$x=7$