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

Revision as of 23:07, 28 January 2020

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 $\boxed{(E) 2}$ . (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 $\boxed{\textbf{E } 2}$

~OlutosinNGA

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 just use the Pythagorean Theorem. We find that $EB = \sqrt{13}$ , so $[EHCB] = \sqrt{13}$ .

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 $\dfrac{9 \cdot \sqrt{13}}{13}$ . Thus, the altitude has length $\dfrac{6 \cdot \sqrt{13}}{13}$ . This is also the height of the rectangular prism.

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

@@ Line 87: / Line 87: @@
 In order to find the height of rectangular prism <math>EHCBM</math>, we can examine triangle <math>EFB</math>. We can use the Geometric Mean Theorem to find that when an altitude is dropped from point <math>F,</math>  <math>\overline{EB}</math> is split into segments of length <math>\dfrac{4 \cdot \sqrt{13}}{13}</math> and <math>\dfrac{9 \cdot \sqrt{13}}{13}</math>. Thus, the altitude has length <math>\dfrac{6 \cdot \sqrt{13}}{13}</math>. This is also the height of the rectangular prism.
-Plugging these two numbers into the formula <math>V = \dfrac{b \cdot h}{3},</math> we find that the area is <math>\boxed{2}</math>. The answer is <math>\boxed{E}</math>.
+Plugging these two numbers into the formula <math>V = \dfrac{b \cdot h}{3},</math> we find that the volume is <math>\boxed{2}</math>. The answer is <math>\boxed{E}</math>.
 ==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
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