Question

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 cm² then find the lengths of sides AB and AC.

Answer 1

In the given figure, ∆ABC circumscribed the circle with centre O.

Radius OD = 3 cm

BD = 6 cm, DC = 9 cm

Area of ∆ABC = 54 cm²

To find : Lengths of AB and AC.

AF and EA are tangents to the circle at point A.

Let AF = EA = x

BD and BF are tangents to the circle at point B.

BD = BF = 6 cm

CD and CE are tangents to the circle at point C.

CD = CE = 9 cm

Now, new sides of the triangle are:

AB = AF + FB = x + 6 cm

AC = AE + EC = x + 9 cm

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

Squaring both sides, we have

54² = 54x(x + 15)

x² + 15x – 54 = 0

Solve this quadratic equation and find the value of x.

x² + 18x – 3x – 54 = 0

x(x + 18) – 3(x + 18) = 0

(x – 3)(x + 18) = 0

Either x = 3 or x = – 18

But x cannot be negative.

So, x = 3

Answer:

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

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