#### Question

the below given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the

area of ΔABC is 54 cm^{2}, then find the lengths of sides AB and AC.

#### Solution

Let the given circle touch the sides AB and AC of the triangle at points F and E respectively and let the length of line segment AF be x.

Now, it can be observed that:

BF = BD = 6 cm (tangents from point B)

CE = CD = 9 cm (tangents from point C)

AE = AF = x (tangents from point A)

AB = AF + FB = x + 6

BC = BD + DC = 6 + 9 = 15

CA = CE + EA = 9 + x

2s = AB + BC + CA = x + 6 + 15 + 9 + x = 30 + 2x

s = 15 + x

s – a = 15 + x – 15 = x

s – b = 15 + x – (x + 9) = 6

s – c = 15 + x – (6 + x) = 9

`Area of triangle ABC =sqrt(s(s-a)(s-b)(s-c))`

`54=sqrt((15+x)(x)(6)(9))`

`54=3sqrt(6(15x+x^2))`

`18=sqrt(6(15x+x^2))`

`324=6(15x+x^2)`

`54=15x+x^2`

`15x+x^2-54=0`

`x^2+ 18x -3x -54 0=0`

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

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

x = 18 and x = 3

As distance cannot be negative, x = 3

AC = 3 + 9 = 12

AB = AF + FB = 6 + x = 6 + 3 = 9