Maths problem, how to do, thanks?

(a) Let mid-pt. of AB = M
sin 60⁰ = AM/9
AM = 9 sin 60⁰ = 9√3/2
∴ AB = 2(9√3/2) = 9√3 = 15.6 cm  (to 3 sig.fig.)

(b) Arc AB = 2π(9)(120/360) ---------→ 《arc length = 2πr(θ/360)》
　　　　　= 6π 
∴ perimeter of shaded region =  6π + 9√3 = 34.4 cm (to 3 sig.fig.)

(c) area of sector = π(9²)(120/360) -→ 《area of sector = πr²(θ/360)》
 　　　　　　　　= 27π
area of △OAB = ½(9)(9)sin120⁰  --→《area of △ = ½ absinθ, where θ- included ∠》
 　　　　　　 = (81/2)(√3/2)
 　　　　　　 = 81√3/4
∴ area of shaded region = 27π -  81√3/4 = 49.7 cm² (to 3 sig.fig.) 

SC老師謝謝您,這題我看明白了,之前我想不通,原來AB線只要按你说加中點,就可以解決了,谢谢你.