Question

The radius of the inner circle of triangle is $4cm$ and the segment into which one side is divided by the point of contact are $6cm$ and $8cm$ determine the other two sides of the triangle.

Answer 1

$BF=BD=6cm$
[length of tangent from external point are equal]
$CE=CD=8cm$
[length of tangent from external point are equal]
Let $AF=AE=x$
[length of tangent from external point are equal ]
Now,
$Areaof\Delta ABC=Areaof\Delta AOB+Areaof\Delta BOC+Areaof\Delta COA$
$=\frac{1}{2}\times 4(6+x)+\frac{1}{2}\times 4\left(14\right)+\frac{1}{2}\times 4(8+x)$
$=\frac{1}{2}\times 4(28+2x)$
$=4(14+x)$
Also, area of $\Delta ABC$ by Heron's formula
$S=\frac{14+6+x+8+x}{2}=14+x$
Area of $\Delta ABC=\sqrt{(14+x)\left(8\right)\left(6\right)x}$
So, $4(14+x)=\sqrt{48x(14+x)}$
$16(14+x{)}^2=48x(14+x)$
$14+x=3x$
$2x=14\Rightarrow x=7$
So, $AB=6+x=6+7=13cm$
$AC=8+x=8+7=15cm$

Sum = $13+15=28cm$