Question

In the given figure, 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 the area of ∆ABC = 21 cm² then find the lengths of sides AB and AC.

Answer 1

Given:

∆ABC is circumscribed a circle with centre O and radius 2 cm.

Point D divides BC as

BD = 4 cm, DC = 3 cm, OD = 2 cm

Area of ∆ABC = 21 cm²

Join OA, OB, OC, OE and OF.

From figure:

BF and BD are tangents to the circle.

So, BF = BD = 4 cm

CD and CE are tangents to the circle.

So, CE = CD = 3 cm

AF and AE are tangents to the circle.

Let say, AE = AF = x cm

Now,

Area of ∆ABC = 1/2 x Perimeter of ∆ABC x Radius

21 = 1/2[AB + BC + CA] OD

21 = 1/2 [4 + 3 + 3 + x + x + 4] x 2

21 = 14 + 2x

x = 3.5

Therefore,

AB = AF + FB = 3.5 + 4 = 7.5 cm

AC = AE + CE = 3.5 + 3 = 6.5 cm