1. ## Solid Geometry Problem

An irregular pile of earth is 15ft. high and covers 500sq ft. Its Mid-sections and level top are estimated to contain 400 and 200 sq. ft., respectively. Find the cost of removing it at 60 cents per load. If the tower measures 3 by 4 by 9 ft.

2. ## Re: Solid Geometry Problem

To find the volume of dirt, use the formula:

$V=\frac{h}{6}(B+4M+T)$

where $h$ is the height of the pile, $B$ is the area of the bottom, $M$ is the area of the mid-section and $T$ is the area of the top.

Then divide this volume by the volume of the "tower" to find how many loads there are, and multiply this by 60 cents to find the total cost of removing the dirt.

3. ## Re: Solid Geometry Problem

Perhaps a better formula for the volume of dirt, since the pile dies not seem to have a constant angle of repose, is:

$V=\frac{h}{6}\left(B+\sqrt{BM}+2M+\sqrt{MT}+T \right)$

4. ## Re: Solid Geometry Problem

Granular materials form conical piles according to their angles of repose.Earth 30=40deg,sand 34 deg.Standard formula for frustum volume
V=1/3pih(r1^2+r2^2+r1*r2)