Results 1 to 4 of 4

Math Help - Cone with water in it

  1. #1
    Member
    Joined
    May 2007
    Posts
    150

    Cone with water in it

    Water is being poured into a conical cup that is 12 cm tall.
    a. When the water in the cup is 9 cm deep, what percentage of the cup is filled?
    b. When the cup is 75% filled, how deep is the water?

    I think it has to do with ratio, since it only gives you the height?
    Along with the answer, can someone give me like a cheat sheet way of doing this, with either percentage,/ height,/ radius, etc.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by stones44 View Post
    Water is being poured into a conical cup that is 12 cm tall.
    a. When the water in the cup is 9 cm deep, what percentage of the cup is filled?
    b. When the cup is 75% filled, how deep is the water?

    I think it has to do with ratio, since it only gives you the height?
    Along with the answer, can someone give me like a cheat sheet way of doing this, with either percentage,/ height,/ radius, etc.
    you may use similar triangles there..
    suppose that the radius of the cone is R and the radius of the cone at the hight of 9 ft is r, then the ratio is given by
    \frac{R}{12}=\frac{r}{9}
    which implies that
    r= \frac {9R}{12}

     V_{cone} = \frac{1}{3}\pi R^2 H
    given th situation, the volume of the cone when full is
    V_{cone, full}=4\pi R^2

    while when if 9 ft full,
    V_{cone,9 ft}=3\pi r^2

    then, the percentage is
    \frac{V_{cone,9 ft}}{V_{cone, full}} = \frac{3\pi r^2}{4\pi R^2}
    = \frac{3\pi (\frac {9R}{12})^2}{4\pi R^2}
    = \frac{3\frac{9^2 \pi R^2}{12^2}}{4\pi R^2}
    = \frac{(3)(9^2)}{(4)(12^2)}
    then multiply by 100%..

    for b)
    you are looking for h. so, using similar triangles again, you have
    \frac{r}{h}=\frac{R}{12}
    so that
    r=\frac{hR}{12}
    so, 75% full,
    \frac{V_{cone,h}}{V_{cone,full}}=\frac{3}{4}
    so,
    \frac{\frac{1}{3}h\pi r^2}{4\pi R^2} = \frac{\frac{1}{3}h\pi (\frac{hR}{12})^2}{4\pi R^2} = \frac{3}{4}
    can you continue? you will notice that R^2 will cancel and you can now solve for h.
    Attached Thumbnails Attached Thumbnails Cone with water in it-sketch2.jpg  
    Last edited by kalagota; November 1st 2007 at 11:50 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2007
    Posts
    150
    dont you have to 2/3 to the volume equation?

    im confused how your doing your algebra.....:-/
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by stones44 View Post
    dont you have to 2/3 to the volume equation?

    im confused how your doing your algebra.....:-/
    ahh, yeah.. i was thinking it was a cylinder.. wait, i'll edit it..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Pumping water out of a cone.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 8th 2010, 08:47 PM
  2. Replies: 3
    Last Post: March 20th 2010, 04:18 PM
  3. Emptying a water tank (Inverted circular cone)
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 20th 2009, 05:05 PM
  4. Replies: 0
    Last Post: February 10th 2009, 01:25 PM
  5. Replies: 1
    Last Post: October 22nd 2007, 06:47 PM

Search Tags


/mathhelpforum @mathhelpforum