Question

In fig. 4, a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 4 cm and 3 cm respectively. If area of $\Delta$ABC = 21 $c{m}^2$, then find the lengths of sides AB and AC.

Answer 1

Join O with A, O with B, O with C, O with E, and O with F.

We have, OD =OF =OE = 2cm (radii)

BD = 4 cm and DC = 3 cm

$\therefore$ BC = BD + DC = 4 cm + 3 cm = 7 cm

Now, BF = BD = 4 cm [Tangents from the same point]

CE = DC = 3 cm

Let AF = AE = x cm

Then, AB = AF + BF = (4 + x) cm and AC = AE + CE = (3 + x) cm

It is given that
$ar\left(\Delta OBC\right)+ar\left(\Delta OAB\right)+ar\left(\Delta OAC\right)=ar\left(\Delta ABC\right)$
$\Rightarrow \frac{1}{2}\times BC\times OD+\frac{1}{2}\times AB\times OF+\frac{1}{2}\times AC\times OE=21$
$\Rightarrow \frac{1}{2}\times 7\times 2+\frac{1}{2}\times (4+x)\times 2+\frac{1}{2}\times (3+x)\times 2=21$
$\Rightarrow \frac{1}{2}\times 2(7+4+x+3+x)=21$
$\Rightarrow 14+2x=21$
$\Rightarrow 2x=7$
$\Rightarrow x=3.5$

Thus, AB = (4 + 3.5) cm = 7.5 cm and AC = (3 + 3.5) cm = 6.5 cm.