Smile 發問於 科學及數學數學 · 5 月前

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

2 個解答

• SC147
Lv 6
5 月前
最愛解答

(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

_______________________________________________________

## :

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 !

• 5 月前

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)