2018 AMC 10B Problems/Problem 10
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);[/asy]](http://latex.artofproblemsolving.com/d/c/6/dc6049797801ea50f14860fa92c1464b9ae29320.png)
Contents
[hide]Solution 1
Consider the cross-sectional plane. Note that 
and we want 
, so the answer is 
. (AOPS12142015)
Solution 2
We start by finding side 
of base by using the Pythagorean theorem on 
. Doing this, we get
![\[\overline{BE}^2 = \overline{AB}^2 + \overline{AE}^2 = 3^2 + 2^2 = 11.\]](http://latex.artofproblemsolving.com/3/8/e/38e42de96f70ecaa1f0c136afe05bca91724c040.png)
Taking the square root of both sides of the equation, we get 
. We can then find the area of rectangle , noting that
![\[[BCHE] = \overline{BE} \cdot \overline{BC} = \sqrt {11} \cdot 1 = \sqrt {11}.\]](http://latex.artofproblemsolving.com/7/e/b/7eb9cc1d099728fa593bb541040ebe6aa3dd904a.png)
Taking the vertical cross-sectional plane of the rectangular prism, we see that the distance from point to base is the same as the distance from point 
to side . Calling the point where the altitude from vertex touches side as point 
, we can easily find this altitude using the area of right 
, as
![\[\frac{\overline{BF} \cdot \overline{FE}}{2} = \frac{\overline{BE} \cdot FK}{2}.\]](http://latex.artofproblemsolving.com/8/d/b/8dbc52d620bb75132c56db895c4965f1421c3d96.png)
Multiplying both sides of the equation by 2 and substituting in known values, we get
\[2 \cdot 3 = FK \cdot \sqrt {11} \Rightarrow FK = \frac{6\sqrt {11}{11}.\] (Error compiling LaTeX. Unknown error_msg)
Deducing that the altitude from vertex to base is 
and calling the point of intersection between the altitude and the base as point 
, we get the area of the rectangular pyramid to be
\[\frac{1}{3}([BCHE] \cdot MN) = \frac{1}{3}\left(\sqrt {11} \cdot \frac{6\sqrt {11}{11}\right) = \frac{66}{33} = \boxed{2}.\] (Error compiling LaTeX. Unknown error_msg)
Written by: Adharshk
Solution 3
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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
