Question #a72dd

1 Answer
BeeFree
Oct 25, 2015

Trisha - this problem is virtually identical to the "pile of ore" problem we worked on earlier.

Explanation:

In the pile of ore problem, there were triangles at the two ends and trapezoids at the two sides. This problem is the same except we replace the two triangles with trapezoids.

Use the same formula for the volume of a prismatoid:

#V=1/6[B + (4M) + b]*h#

#B=# the area of the base #=66*210=13860 ft^2#
#b=# the area of the top #=24*286=6864 ft^2#
#M=# the area of the middle section #=(66+24)/2*(210+286)/2=11160 ft^2#
#h=# the height #=14 ft#

Now, plug the results into the volume formula:

#V=1/6[13860 + (4*11160) + 6864]*14=152516 ft^3#

Convert to yards (1 yard = 3 feet):

#V=152516/3^3=5648.7 yd^3#

Hope that helped

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