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.
Given : △ABC that is drawn to circumscribe a circle with radius r = 3 cm and BD = 6 cm DC = 9cm
Also, area(△ABC) = 54 cm2
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)
x2 + 15x - 54 = 0
x2 + 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