Answer:
The perimeter of Δ ABC is 40 cm ⇒ 2nd answer
Step-by-step explanation:
* Lets explain how to solve the problem
- Circle D is inscribed in triangle ABC
- The circle touches the side AB at H , side BC at F , side CA at G
- BF and BH are tangents to circle D from point B
∴ BF = BH ⇒ tangents drawn from a point outside the circle
- CF and CG are tangents to circle D from point C
∴ CF = CG ⇒ tangents drawn from a point outside the circle
- AG and AH are tangents to circle D from point A
∴ AG = AH ⇒ tangents drawn from a point outside the circle
∵ CG = 6 cm ⇒ given
∴ CF = 6 cm
∵ CB = 11 cm ⇒ given
∵ CB = CF + FB
∴ 11 = 6 + FB ⇒ subtract 6 from both sides
∴ FB = 5 cm
∵ FB = BH
∴ BH = 5 cm
∵ AH = 9 cm ⇒ given
∵ AH = AG
∴ AG = 9 cm
∵ AB = AH + HB
∴ AB = 9 + 5 = 14 cm
∵ AC = AG + GC
∴ AC = 9 + 6 = 15 cm
∵ BC = 11 cm ⇒ given
∵ The perimeter of Δ ABC = AB + BC + CA
∴ The perimeter of Δ ABC = 14 + 11 + 15 = 40 cm
* The perimeter of Δ ABC is 40 cm
. Using this, we can find that the height is 10. Take the average of the lengths to get l = 27.