To solve this problem, we can use some important properties of a triangle that circumscribes a circle.

Key Properties:

1. In a triangle $ABC$ circumscribing a circle, the sum of the lengths of the two tangents drawn from an external point to the points of contact with the circle are equal.

   Let:

   • $AB = c$, $BC = a$, and $CA = b$.
   • The point of tangency on $BC$ is $D$, on $CA$ is $E$, and on $AB$ is $F$.
   • The tangents from $B$ and $C$ to the circle are such that $BD = 6 \, \text{cm}$ and $DC = 9 \, \text{cm}$, so: $a = BD + DC = 6 + 9 = 15 \, \text{cm}.$

2. If the circle has a radius $r$ and the area of $\triangle ABC$ is $A$, then the semi-perimeter $s$ of $\triangle ABC$ is related to $A$ and $r$ by: $A = r \times s.$

Step-by-Step Solution:

1. Calculate the Semi-perimeter $s$:

Given:

• Radius $r = 3 \, \text{cm}$
• Area $A = 54 \, \text{cm}^2$

Using $A = r \times s$: $54 = 3 \times s$ $s = \frac{54}{3} = 18 \, \text{cm}.$

2. Set Up Equations for $AB$ and $AC$:

Since $s = \frac{a + b + c}{2}$, we have: $18 = \frac{15 + b + c}{2}$ Multiplying both sides by 2: $36 = 15 + b + c$ $b + c = 21.$

Let:

• $AB = c$
• $AC = b$

Also, from the tangents' property:

• $AF = AE$
• $BF = BD = 6 \, \text{cm}$
• $CE = DC = 9 \, \text{cm}$

Thus: $c = AF + FB = AE + 6$ $b = AE + CE = AE + 9$

Since $b + c = 21$, and substituting the values: