Question

In the given figure, a ∆ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC, when area of ∆ABC is 84 cm². [CBSE 2015]

Answer 1

Here, D, E and F are the points of contact of the circle with the sides BC, AB and AC, respectively.

OD = OE = OF = 4 cm (Radii of the circle)

We know that the lengths of tangents drawn from an external point to a circle are equal.

∴ BD = BE = 8 cm

CD = CF = 6 cm

AE = AF = x cm (say)

So, BC = BD + CD = 8 cm + 6 cm = 14 cm

AB = AE + BE = x cm + 8 cm = (x + 8) cm

AC = AF + FC = x cm + 6 cm = (x + 6) cm

Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OD ⊥ BC, OE ⊥ AB and OF ⊥ AC

Now,

ar(∆OBC) + ar(∆OAB) + ar(∆OCA) = ar(∆ABC)

$$\begin{gathered} \therefore \frac{1}{2}\times \mathrm{BC}\times \mathrm{OD}+\frac{1}{2}\times \mathrm{AB}\times \mathrm{OE}+\frac{1}{2}\times \mathrm{AC}\times \mathrm{OF}=84{\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times 14\times 4+\frac{1}{2}\times \left(x+8\right)\times 4+\frac{1}{2}\times \left(x+6\right)\times 4=84 \\ \Rightarrow 28+2x+16+2x+12=84 \\ \Rightarrow 4x+56=84 \end{gathered}$$
$$\begin{gathered} \Rightarrow 4x=84-56=28 \\ \Rightarrow x=7 \end{gathered}$$

∴ AB = (x + 8) cm = (7 + 8) cm = 15 cm

AC = (x + 6) cm = (7 + 6) cm = 13 cm

Hence, the lengths of sides AB and AC are 15 cm and 13 cm, respectively.