Question

A triangle $ABC$ is drawn to circumscribe a circle of radius $4cm$ such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of the lengths $8cm$ and $6cm$ respectively. Find the sides $AB$ and $AC.$

Answer 1

It is known that two tangents from same point on the circle are equal

Hence $BD=BF=8$cm

$CD=CE=6$cm

$AE=AF=$(say $x$ cm)

Now as it is a incircle so it would have the formula

r$=\frac{area}{s}$ where s is the semiperimetre

s$=\frac{6+8+6+8+x+x}{2}$

s$=14+x$

area$=\sqrt{s(s-a)(s-b)(s-c)}$ (by heron's formula

$r=4$ cm (given)

So$\Rightarrow$ $4=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$

$\Rightarrow$ $4=\frac{\sqrt{(14+x)(14+x-6-x)(14+x-8-x)(14+x-14)}}{14+x}$

$\Rightarrow$$4=\frac{\sqrt{(14+x)\left(8\right)\left(6\right)\left(x\right)}}{14+x}$

Squaring both sides

$\Rightarrow$$16=\frac{8\times 6\times x}{14+x}$

$\Rightarrow$$14+x=3x$

$\Rightarrow$$14=2x$

$\Rightarrow$$x=7$

hence $AB=AF+BF$

$=8+x=15cm$

$AC=AE+CE$

$=6+x=13cm$