2018 AMC 10B Problems/Problem 10


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 $\boxed{(E) 2}$.(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 $\overline{BC}$ is given to be $1$, we have that $\overline{FM}$ 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 $FBE$ drawn from $F$ to $BE$.

$BE=\sqrt{BF^2 + EF^2}=\sqrt{13}$ Area of $BCHE = BC * BE = \sqrt{13}$

$h = 2 *$ Area of $FBE / BE$ (since Area $= \frac{1}{2}bh$).

Area of $FBE = \frac{1}{2} * FB * FE = 3$

$h = 2 * 3 / \sqrt{13} = \frac{6}{\sqrt{13}}$

Volume of pyramid $=\frac{1}{3} * \sqrt{13} * \frac{6}{\sqrt{13}} = 2$

Answer is $\boxed{\textbf{E } 2}$


Solution 5

We can start by identifying the information we need. We need to find the area of rectangle $EHCB$ and the height of rectangular prism $EHCBM$.

In order to find the area of $EHCB,$ we can use the Pythagorean Theorem. We find that $EB = \sqrt{13}$, so the area of rectangle $EHCB = \sqrt{13}$. We shall refer to this as $x$.

In order to find the height of rectangular prism $EHCBM$, we can examine triangle $EFB$. We can use the Geometric Mean Theorem to find that when an altitude is dropped from point $F,$ $\overline{EB}$ is split into segments of length $\dfrac{4 \cdot \sqrt{13}}{13}$ and $\dfrac{9 \cdot \sqrt{13}}{13}$. Taking the geometric mean of these numbers, we find that the altitude has length $\dfrac{6 \cdot \sqrt{13}}{13}$. This is also the height of the rectangular prism, which we shall refer to as $y$.

Plugging $x$ and $y$ into the formula $V = \dfrac{b \cdot h}{3},$ we find that the volume is $\boxed{2}$. The answer is $\boxed{E}$.

Solution 6

We start by setting the formula for the volume of a rectangular pyramid: $\frac{1}{3}Bh$. By the Pythagorean Theorem, we know that $BE = \sqrt{BF^2 + EF^2} = \sqrt{13}$. Therefore, the area of the base is $1 \times \sqrt{13} = \sqrt{13}$. Next, we would like to know the height of the pyramid. We can observe that the altitude from point $F$ in $\triangle EFB$ is parallel to the height of the pyramid and therefore congruent because those two altitudes are on the same plane of base $EBCH$. From this, we only need to find the altitude from point $F$ in $\triangle EFB$ and plug it into our formula for the volume of a rectangular pyramid. This is easy because we already know the area of $\triangle EFB$ and the base from point $F$, so all we need to do is divide: $\frac{2 \times 3}{\sqrt{13}} = \frac{6}{\sqrt{13}} = \frac{6\sqrt{13}}{13}$. Now all we need to do is plug in all our known values into the volume formula: $\frac{1}{3}Bh = \frac{\sqrt{13} \times \frac{6\sqrt{13}}{13}}{3} = \boxed{(E) 2}$


Solution 7

AMC 10B 10 2018.jpg

Using the Pythagorean Theorem, we can easily find that $EB = \sqrt{2^2 + 3^2} = \sqrt{13}$. Quickly computing, we find the area of the base, $BCHE = \sqrt{13} \cdot 1 = \sqrt{13}$. Now we can make the following adjustments to our 3d shape as shown in the diagram. All we need now is to solve for the height, or $XM$. We can set up to following equation due to our knowledge of altitudes(of the hypotenuse)in right triangles. We can set up the following equations: \begin{align*} b(a+b) &= (MH_1)^2 \\ a(a+b) &= (MH_2)^2 \\ b\sqrt{13} &= 3^2 \\ a\sqrt{13} &= 2^2 \\ b &= \dfrac{9}{\sqrt{13}} \\ a &= \dfrac{4}{\sqrt{13}} \\ (MX)^2 &= ab \\ (MX)^2 &= \dfrac{9 \cdot 4}{13} \\ MX &= \dfrac {3 \cdot 2}{\sqrt{13}} \\ \end{align*} Thus $\triangle V_{BCHEM} = \dfrac{\text{(height)}\cdot \text{(base)}}{3} = \dfrac{MX \cdot BCHE}{3} = \dfrac {\sqrt{13} \cdot \dfrac{3 \cdot 2}{\sqrt{13}}}{3}$ $= \boxed{\textbf{(E) } 2}$

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See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AMC 10 Problems and Solutions

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