Maths problem: I don't know how to do (b), thanks.?

(a) 
2πr(60⁰/360⁰)+2r = 5π+30
πr/3+2r = 5π+30
r(π+6)/3 = 5(π+6)
∴ r = 15

(b)
area of △AOB = (1/2) r*r sin60⁰ - - - -(#)
　　　　　　　= (1/2)15² (√3/2)
　　　　　　　= 225√3/4
area of sector = πr²(60⁰/360⁰) = π*15²/6 = 225π/6
∴ area of shaded part = 225π/6 - 225√3/4
　　　　　　　　　　= 225(π/6 -√3/4)
So, area of shaded part / area of sector
= 225(π/6 -√3/4) / 225π/6
= (π/6 -√3/4) / π/6
= 1 - (3√3 / 2π)
≒ 0.17
< 1/5

I agree with Emily's claim that: area of shaded part is < 1/5 area of sector 
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答 "Explain your answer" 這類題目 :
很多時都要有 「計算」; 而 其中入「文字」解釋, 則視乎需要而定 !

如上題 :-
要知道 是否 "area of shaded part  < 1/5 area of sector", 
一定要靠 「計算」!
最後, 再加上「文字」 去做 CONCLUSION : whether you AGREE or not !

扇形面積 = πr^2 × (60/360)
         = πr^2/6
三角形面積 = (1/2)(底r)(高 √3r/2)
           = √3 r^2/4
陰影部分面積 = πr^2/6 - √3 r^2/4
             = r^2(π/6-√3/4)
Ratio = [r^2(π/6-√3/4)]/(πr^2/6)
      = (π/6-√3/4)/(π/6)
      = 1-3√3/(2π) ≒ 0.173 < 1/5

又:
扇形周長 = 2πr×(60/360) + 2r 
               = r(π+6)/3 = 5π+30
∴ r = 15(cm)