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

Math Help - Solid mensuration problems

  1. #1
    Junior Member
    Joined
    Dec 2013
    From
    Philippines
    Posts
    26

    Lightbulb Solid mensuration problems

    Hello Its me again XD
    just want to give a sample problem ^^






    Solid mensuration problems-ex.png


    a railway cut 200ft long and 30ft wide at the bottom is made with sloping sides, which are 80ft . and 60ft in length . the 80-ft side is inclined 45o and the other side is inclined 30o to the horizontal. find the cost of removing the earth at $2 per load , if the trucks have a capacity of 4cu.yd.





    can't figure out the steps on solving

    thanks for the help in advance^^
    -MJ
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123

    Re: Solid mensuration problems

    Quote Originally Posted by mventurina1 View Post
    Hello Its me again XD
    just want to give a sample problem ^^

    a railway cut 200ft long and 30ft wide at the bottom is made with sloping sides, which are 80ft . and 60ft in length . the 80-ft side is inclined 45o and the other side is inclined 30o to the horizontal. find the cost of removing the earth at $2 per load , if the trucks have a capacity of 4cu.yd.

    can't figure out the steps on solving

    thanks for the help in advance^^
    -MJ
    Good morning,

    one question first: In your text you state that the inclination of the 80'-side is 45 but in your sketch the inclination is given with 40. So which value is correct?
    I'll use the 45 inclination. If this is wrong replace the 45 by 40.

    1. Draw a sketch with all measures.

    2. You've got 2 right triangles. The left one is an isosceles right triangle with d = x. Therefore

    d^2+d^2 = 80^2~\implies~d = 40\sqrt{2}=x

    3. With the right one you'll get:

    \frac t{60'}=\sin(30^\circ) = \frac12~\implies~t=30'
    and
    \frac y{60'}=\cos(30^\circ) = \frac12 \cdot \sqrt{3}~\implies~y=30 \cdot \sqrt{3}

    4. The volume of the complete ditch is a prism with the orange area as base area and the height of 200'. So you have first to determine the value of the orange area which is

    Area_{orange} = trapezium - \Delta_{green} - \Delta_{blue}

    The area A_T of the trapezium is calculated by:

    A_T=\frac{d+t}2 \cdot (x+10+y)

    A_T=\frac{40\sqrt{2}+30}2 \cdot (40\sqrt{2}+10+30 \cdot \sqrt{3}) = 50\left((4\sqrt{2}+3)(4\sqrt{2}+1+3 \cdot \sqrt{3})  \right)\approx 5130.488 \ (ft)^2

    Consequently
    A_{orange} = A_T - \frac12 x^2 - \frac12 \cdot y \cdot t

    A_{orange} = 50\left((4\sqrt{2}+3)(4\sqrt{2}+1+3 \cdot \sqrt{3})  \right) - 1600 - 450 \sqrt{3} \approx 2751.065\ (ft)^2

    5. The volume of the complete ditch is now:

    V = A_{orange} \cdot 200' \approx 550212.94\ (ft)^3

    6. Now you have to transform the number of (ft)^3 into (yd)^3 because I'm not familiar with imperial measures. Sorry! (Maybe this is correct: 1(yd)^3 = 27 (ft)^3)

    7. Afterwards calculate the number of loads and consequently the costs for the transport.
    Attached Thumbnails Attached Thumbnails Solid mensuration problems-eisenbahneinschnitt.png  
    Thanks from mventurina1
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. solid mensuration
    Posted in the New Users Forum
    Replies: 1
    Last Post: January 15th 2014, 06:20 AM
  2. Solid mensuration -
    Posted in the Geometry Forum
    Replies: 1
    Last Post: August 26th 2012, 11:07 PM
  3. Problems on Solid Mensuration
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: December 2nd 2010, 12:02 AM
  4. Solid Mensuration
    Posted in the Geometry Forum
    Replies: 0
    Last Post: February 16th 2010, 02:00 PM
  5. Plane and Solid Mensuration
    Posted in the Geometry Forum
    Replies: 0
    Last Post: August 11th 2009, 06:15 AM

Search Tags


/mathhelpforum @mathhelpforum