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 $$\Delta ABC = 54 cm^2$$ then find the lengths of sides AB and AC.
In the given figure, $$\Delta ABC$$ circumscribed the circle with centre O.
Radius $$OD = 3 cm$$
$$BD = 6 cm, DC = 9 cm$$
Area of $$\Delta ABC = 54 cm^2$$
To find : Length 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:
$$AB = AF + FB = x + 6 cm$$$$AC = AE + EC = x + 9 cm$$$$BC = BD + DC = 6 + 9 = 15 cm$$
Now, using Heron's formula:
Area of triangle $$ABC = \sqrt{s(s - a) (s - b) (s-c)}$$
Where $$S = \dfrac{a + b + c}{2}$$
$$S = 1/2(x + 6 + x + 9+15) = x + 15$$
Area of $$ABC = \sqrt{(x + 15) (x + 15 - (x + 6)) (x + 15 - (x - 9)) (x + 15 - 15)}$$
Or
$$54 = \sqrt{(x + 15) (9)(6) (x)}$$
Squaring both sides, we have
$$54^2 = 54 x (x + 15)$$$$x^2 + 15x - 54 = 0$$
Solve this quadratic equation and find the value of x.
$$x^2 + 18x - 3x - 54 = 0$$$$x(x + 18) - 3(x + 18) = 0$$$$(x - 3) (x + 18) = 0$$
Either $$x = 3$$ or $$x = - 18$$
But x cannot be negative.
So, $$x = 3$$
Answer :-
$$AB = x + 6 = 3 + 6 = 9 cm$$
$$AC = x + 9 = 3 + 9 = 12 cm$$
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 Δ ABC = 21 cm2 then find the lengths of sides AB and AC.
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 of AB and AC. 
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see given figure). Find the sides AB and AC.
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 ΔABC = 21 cm2, then find the lengths of sides AB and AC.
