Cubic Yards of Material in Great Wall of China

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geekily

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I was doing so well on my homework - first time in a long time that everything was making sense - and then I got to this last problem:

The Great Wall of China is about 1500 miles long. The cross-section of the wall is a trapezoid 25 feet high, 25 feet wide at the bottom, and 15 feet wide at the top. How many cubic yards of material make of this wall?

I wasn't quite sure if I should be looking for surface area or volume, but since the problem was part of the volume section, I went with that. I drew a trapezoid and labeled it, then drew a rectangular prism, because I put the 2 triangles of the trapezoid together to make a second rectangle, and then just combined it with the first rectangle. Now I had a width of 20 feet on the top and bottom. Since the volume of a prism is area of the base x height, I took 20 x 25 x 25 = 12500. Then I realized that maybe the area of the base should be 20 x 20, but either way, I'm not doing it right. Then, I converted the 1500 mi to feet, which was 7920000 and divided that by 20, the width of one section. That was 396,000, which I then multiplied by 125,000, the area of the prism. Then I divided by 3 to get it into yards, arriving at an answer of 1,650,000,000.

According to the answer key, the correct answer is 1.47 x 10^8 cubic yards, so I'm obviously very, very wrong and I have no idea as to why. I've never done a problem like this before, so it's very possible that everything I've done isn't even appropriate for this kind of problem, but honestly, I don't know either way. Please help!

Thanks so much!
 
You divide by 27 to get cubic yards. 3 is square yards.

\(\displaystyle \L\\\frac{\frac{1}{2}(25)(25+15)(7920000)}{27}=1.47\cdot{10^{8}} \;\ yd^{3}\)
 
Wow, then, I did it right for the most part?

Just one more question... where does the 1/2 come from in your equation?
 
You used it in the formula for the area of a trapezoid?.

\(\displaystyle \L\\\frac{(b+B)h}{2}\)
 
Oh, I actually just put the one triangle on top of the other to change it from a trapezoid to a rectangular prism, and then found used area of the base x height for the volume of the prism. Is that wrong?
 
No, not necessarily wrong, but harder than it needed to be. There is a formula for trapezoid area. Which is what a cross-section of the wall is.
 
Ah, I see now. Thank you!

The only thing that confuses me is that it doesn't seem like this problem had any steps related to volume, but it's in the "volume" section in my book.
 
The volume is found by multiplying the area of the cross section times the length of the wall. Area (2 dimensional) times length (1 dimensional) equals volume (3 dimenstional).
 
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