Difference between revisions of "2018 AMC 10B Problems/Problem 10"
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<math>BE=\sqrt{BF^2 + EF^2}=\sqrt{13}</math> Area of <math>BCHE = BC * BE = \sqrt{13}</math> | <math>BE=\sqrt{BF^2 + EF^2}=\sqrt{13}</math> Area of <math>BCHE = BC * BE = \sqrt{13}</math> | ||
− | <math>h = 2 * Area of FBE / BE</math> (since <math> | + | <math>h = 2 *</math> Area of <math>FBE / BE</math> (since Area <math>= \frac{1}{2}bh</math>). |
− | <math> | + | Area of <math>FBE = \frac{1}{2} * FB * FE = 3</math> |
<math>h = 2 * 3 / \sqrt{13} = \frac{6}{\sqrt{13}}</math> | <math>h = 2 * 3 / \sqrt{13} = \frac{6}{\sqrt{13}}</math> |
Revision as of 11:51, 3 April 2019
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
Area of (since Area ).
Area of
Volume of pyramid
Answer is
~OlutosinNGA
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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