Question

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

Answer 1

In $\Delta ABC,$
Length of two tangents drawn from the same point to the circle are equal,
$\therefore CF=CD=6cm\therefore BE=BD=8cm\therefore AE=AF=x$
We observed taht
AB +AE +EB =x +8
BC =BD +DC =8+6 =14
CA =CF +FA =6+x
Now semiperime of circle s,
$\Rightarrow 2s=AB+BC+CA$
=x+8+14+6+x
=28 +2x
$\Rightarrow s=14+x$
Area of $\Delta ABC=\sqrt{s}(s-a)(s-b)(s-c)$
$=\sqrt{(14+x)(14+x-14)(14+x-x-6)(14+x-x-8)}=\sqrt{(14+x)\left(x\right)\left(8\right)\left(6\right)}=\sqrt{(14+x)48x}...\left(i\right)$
Also, Area of $\Delta ABC=2\times$ area of $(\Delta AOF+\Delta COD+\Delta DOB)$
$=2\times \left[\right(\frac{1}{2}\times OF\times AF)+(\frac{1}{2}\times CD\times OD)+(\frac{1}{2}\times DB\times OD\left)\right]=2\times \frac{1}{2}(4x+24+32)=56+4x....\left(ii\right)$
Equating equation (i) and (ii) we get,
$\sqrt{(14+x)}48x=56+4x$
Squaring both sides,
$48x(14+x=(56+4x{)}^2$

$\Rightarrow 48x=\frac{{\left[\right(14+x\left)\right]}^2}{14+x}\Rightarrow 48x=16(14+x)\Rightarrow 48x=224+16x\Rightarrow 32x=224\Rightarrow x=7cm$
Hence, AB =x+8 =7 +8=15 cm
CA =6 +x =6 +7 =13 cm