2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 10 Easy

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? (2018 AMC 10B Problem, Question#10)

  • A.

    1

  • B.

    \frac{4}{3}

  • C.

    \frac{3}{2}

  • D.

    \frac{5}{3}

  • E.

    2

Answer:E

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 2.

We can start by finding the total volume of the parallelepiped. It is 2\cdot3\cdot2=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 theparallelepiped. 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 \rm E.

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.

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

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=\frac{2*\text{Area of}~FBE}{BE} (since Area=\frac{1}{2}bh).

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

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

Volume of pyramid =\frac{1}{3}*\sqrt{13}*\frac{6}{\sqrt{13}}=2, Answer is 2.

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