2018 AMC 10B Problems/Problem 10
Problem
In the rectangular parallelepiped shown, =
,
=
, and
=
. Point
is the midpoint of
. What is the volume of the rectangular pyramid with base
and apex
?
Solution 1
Consider the cross-sectional plane and label its area . Note that the volume of the triangular prism that encloses the pyramid is
, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is
, so the answer is
. (AOPS12142015)
Solution 2
We can start by finding the total volume of the parallelepiped. It is , because a rectangular parallelepiped is a rectangular prism.
Next, we can consider the wedge-shaped section made when the plane cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is
. Since BC is given to be
, we have that FM is
. Using the formula for the volume of a triangular pyramid, we have
. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume
as well.
The original wedge we considered in the last step has volume , because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have
. Thus, the volume of the figure we are trying to find is
. This means that the correct answer choice is
.
Written by: Archimedes15
NOTE: For those who think that it isn't a rectangular prism, please read the problem. It says "rectangular parallelepiped." If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.
Solution 3
If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is , with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is
. We can obtain the answer by subtracting twice this value from the diagonal half prism, or
Solution 4
You can calculate the volume of the rectangular pyramid by using the formula, .
is the area of the base,
, and is equal to
. The height,
, is equal to the height of triangle
drawn from
to
.
Area of
(since
).
Volume of pyramid $=\frac{\sqrt{13} * \frac{6}{\sqrt{13}}{3} = 2$ (Error compiling LaTeX. Unknown error_msg)
Answer is
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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All AMC 10 Problems and Solutions |
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