Results 1 to 2 of 2
Like Tree1Thanks
  • 1 Post By Soroban

Math Help - solid mensuration

  1. #1
    Newbie REDnight's Avatar
    Joined
    Jan 2014
    From
    Paris
    Posts
    10
    Thanks
    2

    Post solid mensuration

    hello everyone !!
    ..I'm new here and i wanna ask some help...if its ok with you..

    I'm having an assignment today and it really gives me a hard time answering this problem,...

    "A cylindrical tin can holding 2 gal. has its height equal to the diameter of its base. Another cylindrical tin can with the same capacity has its height equal to twice the diameter of its base. Find the ratio required for making the two can with covers."

    ..well my book got the answer but doesn't have any solutions...

    please help me with this...
    .
    ..btw this question comes from the "SOLID MENSURATION BY KERN AND BLAND" and i tried to look for answers in the internet...all i got is the "SOLID MENSURATION BY ASIN" but the sad thing is, he's writings were too small and i cant even see it clearly...and because this copy that i have is just a scan from another scan book...

    hope you guys can help me
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,686
    Thanks
    617

    Re: solid mensuration

    Hello, REDnight!

    Welcome aboard!



    A cylindrical tin can holding 2 gal. has its height equal to the diameter of its base.
    Another cylindrical tin can with the same capacity has its height equal to twice the diameter of its base.
    Find the ratio required for making the two can with covers.

    The shorter can has radius R and height H = 2R.
    Its volume is: . V_1 \:=\:\pi R^2H \:=\:\pi R^2(2R) \:=\:2\pi R^3

    The taller can has radius r and height h = 4r.
    Its volume is: . V_2 \:=\:\pi r^2h \:=\:\pi r^2(4r) \:=\:4\pi r^3

    The volumes are equal: . 2\pi R^3 \:=\:4\pi r^3 \quad\Rightarrow\quad R^3 \:=\:2r^3 \quad\Rightarrow\quad \frac{R}{r} \:=\:\sqrt[3]{2} .[1]


    The short can has surface area: . A_1 \:=\:2\pi R^2 + 2\pi RH \:=\:2\pi R^2 + 2\pi R(2R) \:=\:6\pi R^2

    The tall can has surface area: . A_2 \:=\:2\pi r^2 + 2\pi rh \:=\:2\pi r^2 + 2\pi r(4r) \:=\:10\pi r^2


    Ratio: . \frac{A_1}{A_2} \:=\:\frac{6\pi R^2}{10\pi r^2} \:=\:\frac{3R^2}{5r^2} \:=\:\frac{3}{5}\left(\frac{R}{r}\right)^2

    Substitute [1]: . \text{Ratio} \:=\:\frac{3}{5}\left(\sqrt[3]{2}\right)^2 \:=\:\frac{3\sqrt[3]{4}}{5}
    Thanks from REDnight
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with my Solid Mensuration Problem
    Posted in the Geometry Forum
    Replies: 3
    Last Post: August 21st 2013, 07:07 PM
  2. Solid mensuration -
    Posted in the Geometry Forum
    Replies: 1
    Last Post: August 26th 2012, 10:07 PM
  3. Problems on Solid Mensuration
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: December 1st 2010, 11:02 PM
  4. Solid Mensuration
    Posted in the Geometry Forum
    Replies: 0
    Last Post: February 16th 2010, 01:00 PM
  5. Plane and Solid Mensuration
    Posted in the Geometry Forum
    Replies: 0
    Last Post: August 11th 2009, 05:15 AM

Search Tags


/mathhelpforum @mathhelpforum