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Difference between revisions of "2018 AMC 10B Problems/Problem 10"

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==Solution 4==
 
==Solution 4==
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You can calculate the volume of the rectangular pyramid by using the formula, <math>\frac{Ah}{3}</math>. <math>A</math> is the area of the base, <math>BCHE</math>, and is equal to <math>BC * BE</math>. The height, <math>h</math>, is equal to the height of triangle <math></math>.
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You can calculate the volume of the rectangular pyramid by using the formula, <math>\frac{Ah}{3}</math>. <math>A</math> is the area of the base, <math>BCHE</math>, and is equal to <math>BC * BE</math>. The height, <math>h</math>, is equal to the height of triangle <math>FBE</math> .
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$BE=\sqrt{BF^2 + EF^2}=\sqrt{13}
  
 
==See Also==
 
==See Also==

Revision as of 10:17, 3 April 2019

Problem

In the rectangular parallelepiped shown, $AB$ = $3$, $BC$ = $1$, and $CG$ = $2$. Point $M$ is the midpoint of $\overline{FG}$. What is the volume of the rectangular pyramid with base $BCHE$ and apex $M$?


[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]

$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2$

Solution 1

Consider the cross-sectional plane and label its area $b$. Note that the volume of the triangular prism that encloses the pyramid is $\frac{bh}{2}=3$, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is $\frac{bh}{3}$, so the answer is $\boxed{2}$. (AOPS12142015)

Solution 2

We can start by finding the total volume of the parallelepiped. It is $2 \cdot 3 \cdot 1 = 6$, because a rectangular parallelepiped is a rectangular prism.

Next, we can consider the wedge-shaped section made when the plane $BCHE$ cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is $\frac{1}{2} \cdot 2 \cdot 3 = 3$. Since BC is given to be $1$, we have that FM is $\frac{1}{2}$. Using the formula for the volume of a triangular pyramid, we have $V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}$. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume $\frac{1}{2}$ as well.

The original wedge we considered in the last step has volume $3$, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have $3 - \frac{1}{2} \cdot 2 = 2$. Thus, the volume of the figure we are trying to find is $2$. This means that the correct answer choice is $\boxed{E}$.

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 $\frac{1}{3}Bh$, 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 $\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}$. We can obtain the answer by subtracting twice this value from the diagonal half prism, or $(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})=$ $\boxed{2}$

Solution 4

You can calculate the volume of the rectangular pyramid by using the formula, $\frac{Ah}{3}$. $A$ is the area of the base, $BCHE$, and is equal to $BC * BE$. The height, $h$, is equal to the height of triangle $FBE$ . $BE=\sqrt{BF^2 + EF^2}=\sqrt{13}

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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