Question

In the given figure, a triangel ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point o f contant D, are of lengths 6 cm and 9 cm respectively. If the area of $\Delta$ = 54 $c{m}^2$ then find the lengths of sides AB and AC.

Answer 1

Consider the above figure. We assume that the circle touches AB in F, BC in D and AC in E. Also given is BD = 9cm and DC = 3cm.Let AF = x.

For ΔABC, AF = AE = x (∵ tangents drawn from an external point to a circle are congruent i.e. AE and AF are tangent drawn from external point A.)

Similarly we have, BE = BD = 3cm (∵ congruent tangents from point B)

And CF = CD = 9cm (∵ congruent tangents from point C)

Now, AB = AE + EB = x + 3

BC = BD + DC = 12

AC = AF + FC = x + 9

Then,

2s = AB + BC +CA = x + 3 + 12 + 1 + x + 9 = 2x + 24

∴ s = x + 12

Using Heron’s formula,

Area of $\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{(12+x)(12+x-12)(12+x-x-9)(12+x-x-3)}$

$=\sqrt{(12+x)\left(x\right)\left(3\right)\left(9\right)}$

$=\sqrt{27(12x+x^2)}$

$=3\sqrt{3(12x+x^2)}$ -----(1)

Area of $\Delta OBC=\frac{1}{2}\times 12\times 3=18$

Area of $\Delta OAB=\frac{1}{2}\times (3+x)\times 3=\frac{9+3x}{2}$

Area of $\Delta OAC=\frac{1}{2}\times (9+x)\times 3=\frac{27+3x}{2}$

Total Area = Area of $\Delta OBC$ + Area of $\Delta OAB$ + Area of $\Delta OAC$

$=8+\frac{9+3x}{2}+\frac{27+3x}{2}$

$\Rightarrow 3\sqrt{3(12x+x^2)}=8+\frac{9+3x}{2}+\frac{27+3x}{2}$

$\Rightarrow 3\sqrt{3(12x+x^2)}=\frac{36+9+3x+27+3x}{2}$

$\Rightarrow 3\sqrt{3(12x+x^2)}=\frac{72+6x}{2}=12+x$

Squaring both sides,

$\Rightarrow [\sqrt{3(12x+x^2)}{]}^2=(12+x{)}^2$

$\Rightarrow 3(12x+x^2)=(12+x{)}^2$

$\Rightarrow 36x+3x^2=144+24x+x^2$

$\Rightarrow 2x^2+12x-144=0$

$\Rightarrow x^2+6x-72=0$

$\Rightarrow (x+12)(x-6)=0$

$\Rightarrow (x=6$

Hence AB = x + 3 = 9 + 3 = 12 cm

BC = 12 cm

AC = x + 9 = 15 cm