Smile 發問於 科學及數學數學 · 3 月前

Maths problem, how to do, thanks?

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2 個解答

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  • SC147
    Lv 6
    3 月前
    最愛解答

    (a)

    Let height of a larger cylinder = H

    base area of a larger cylinder = πR² 

    Then, total vol. of the 2 larger cylinders = 2πR²H

    base area of a smaller cylinder = πr²

    Also, base area of a smaller cylinder = ¼ πR²

    ∴ ¼ πR² = πr²

    => R² = 4r²

    => R= 2r - - - - - - (#)

    total vol. of the 20 smaller cylinders=20πr²(6)=120πr²

    As the 2 larger cylinders are recast in 20 smaller cylinders, 

    2πR²H = 120πr²

    R²H = 60r²

    ∴ H = 60(r/R)²

      = 60(1/2)²  [ From (#) ]

      = 15 cm

    (b) 

    ratio of curved surface area of a larger cylinder to

    curved surface area of a smaller cylinder

    = 2πR(15) : 2πr(6)

    = R(15) : r(6)

    = 5/2(R/r)

    = (5/2)*(2)  [ From (#) ]

    = 5:1

  • 3 月前

    2πR^2H = 20(πr^2h) = 120πr^2 (∵ h = 6cm)

    πR^2 = 4πr^2 (∴ R = 2r)∴ 8πr^2 H = 120πr^2∴ H = 15(cm)我不確定 curved surface 是指側面積還是包含上下底.若只計側面積(個人猜測是問這個), 則 (2πRH)/(2πrh) = (R/r)(H/h) = 2(15/6) = 5若含上下底, 則表面積比值與r有關:(2πR^2+2πRH)/(2πr^2+2πrh)   = (8πr^2+60πr)/(2πr^2+12πr)   = (4r+30)/(r+6)

    是底面積倍數與側表面積倍數之加權平均.

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