Difference between revisions of "2018 AMC 10B Problems/Problem 10"
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==Solution 1== | ==Solution 1== | ||
| − | Consider the cross-sectional plane and label it b. Note that the volume of the triangular prism that encloses the pyramid is <math>bh/2=3</math>, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is <math>bh/3</math>, so the answer is <math>\boxed{2}</math>. (AOPS12142015) | + | Consider the cross-sectional plane and label it <math>b</math>. Note that the volume of the triangular prism that encloses the pyramid is <math>bh/2=3</math>, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is <math>bh/3</math>, so the answer is <math>\boxed{2}</math>. (AOPS12142015) |
==Solution 2== | ==Solution 2== | ||
Revision as of 12:50, 19 February 2018
Contents
Problem
In the rectangular parallelpiped shown, 
= 
, 
= 
, and 
= 
. Point 
is the midpoint of 
. What is the volume of the rectangular pyramid with base 
and apex ?
![[asy] size(250); defaultpen(fontsize(10pt)); pair A =origin; pair B = (4.75,0); pair E1=(0,3); pair F = (4.75,3); pair G = (5.95,4.2); pair C = (5.95,1.2); pair D = (1.2,1.2); pair H= (1.2,4.2); pair M = ((4.75+5.95)/2,3.6); draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H); draw(B--C); draw(F--G); draw(A--D--H--C--D,dashed); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,E); label("$D$",D,W); label("$E$",E1,W); label("$F$",F,SW); label("$G$",G,NE); label("$H$",H,NW); label("$M$",M,N); dot(A); dot(B); dot(E1); dot(F); dot(G); dot(C); dot(D); dot(H); dot(M); label("3",A/2+B/2,S); label("2",C/2+G/2,E); label("1",C/2+B/2,SE); [/asy]](http://latex.artofproblemsolving.com/8/8/5/885c4e50ebdac89dd6aba698666a62eaa1b5eba6.png)
Solution 1
Consider the cross-sectional plane and label it 
. 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.
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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