In the given figure, a triangle 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 of contact D, are of lengths 6 cm and 9 cm respectively. If the area of ABC = 54 cm^{2} then find the lengths of sides AB and AC.

Given : △ABC that is drawn to circumscribe a circle with radius r = 3 cm and BD = 6 cm DC = 9cm

Also, area(△ABC) = 54 cm^{2}

To Find : AB and AC

Now,

As we know tangents drawn from an external point to a circle are equal.

Then,

FB = BD = 6 cm [Tangents from same external point B]

DC = EC = 9 cm [Tangents from same external point C]

AF = EA = x (let) [Tangents from same external point A]

Using the above data, we get

AB = AF + FB = x + 6 cm

AC = AE + EC = x + 9 cm

BC = BD + DC = 6 + 9 = 15 cm

Now we have heron's formula for area of triangles if its three sides a, b and c are given

Where,

⇒

So for △ABC

a = AB = x + 6

b = AC = x + 9

c = BC = 15 cm

⇒

And

Squaring both sides, we get,

54(54) = 54x(x + 15)

x^{2} + 15x - 54 = 0

x^{2} + 18x - 3x - 54 = 0

x(x + 18) - 3(x + 18) = 0

(x - 3)(x + 18) = 0

x = 3 or - 18

but x = - 18 is not possible as length can't be negative.

So

AB = x + 6 = 3 + 6 = 9 cm

AC = x + 9 = 3 + 9 = 12 cm

12