Question

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the tangents 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

Length of two tangents drawn from the same point to the circle are equal.
$\therefore$ CF = CD = 6 cm
$\therefore$ BE = BD = 8 cm
$\therefore$ AE = AF = x
We observed that,
AB = AE + EB = x + 8
BC = BD + DC = 8 + 6 = 14
CA = CF + FA = 6 + x
Let perimeter of triangle ABC be 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 $\Delta AOF+\Delta COD+\Delta DOB)$
$=2\times [(\frac{1}{2}\times OF\times AF)+(\frac{1}{2}\times CD\times OD)+(\frac{1}{2}\times DB\times OD)]$
$=2\times \frac{1}{2}(4x+24+32)=56+4x...\left(ii\right)$
Equating equations (i) and (ii) we get,
$\sqrt{(14+x)48x}=56+4x$
Squaring both sides,
$48x(14+x)=(56+4x{)}^2$
$\Rightarrow 48x=\frac{\left[4\right(14+x){]}^2}{(14+x)}$
$\Rightarrow 48x=16(14+x)$
$\Rightarrow 48x=224+16x$
$\Rightarrow 32x=224$
$\Rightarrow$ x = 7 cm
So, AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm.

Hence, AB=15 cm , CA = 13 cm