Smile 發問於 科學及數學數學 · 3 月前

# Maths problem, how to do, thanks?

### 2 個解答

• 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.