In Figure 5, 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 triangle ABC is 54 cm², then find the lengths of sides AB and AC.

Given OD = 3 cm
BD = BE = 6 cm (equal tangent)
DC = CF = 9 cm (equal tangent)
Area of triangle ABC = 54 cm² (Given)
Area of triangle OBC = 1/2 × 15 × 3 = 45/2 cm²
Area of triangle OAC = 1/2(x + 9) × 3 = 3(x + 9)/2 cm²
Area of triangle OAB = 1/2(x + 6) × 3 = 3(x + 6)/2 cm²
54 = 3/2[(x + 9) + (x + 6) + 15]
→ 54 = 3/2(2x + 30)
→ 54 = 3(x + 15)
→ 18 = x + 15
→ x = 3
Then sides are x + 9 = 3 + 9 = 12, x + 6 = 3 + 6 = 9
So sides of triangle ABC are 12, 9 and 15.