In δbca, cb = 11 cm, cg = 6 cm, ah = 9 cm. find the perimeter of δbca. triangle bca with inscribed circle d. segments bf and bh, cf and cg, and ag and gh are tangent to circle d. 32 cm 40 cm 35 cm 42 cm

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

i believe so, according to my "strange" way of figuring out weird math problems like this.

it's possible to build a triangle with side lengths of 3, 3, and 9.- true ( b.)