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.?

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2 個解答

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  • 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 

    _______________________________________________________

    ## : 

    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 !

  • 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)

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