1. ## Solid mensuration problems

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 2. ## Re: Solid mensuration problems Originally Posted by mventurina1 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.