Difference between revisions of "2018 AMC 10B Problems/Problem 10"
(→Solution 2) |
(→Solution 2) |
||
Line 64: | Line 64: | ||
Deducing that the altitude from vertex <math>M</math> to base <math>BCHE</math> is <math>\frac{6\sqrt {13}}{13}</math> and calling the point of intersection between the altitude and the base as point <math>N</math>, we get the area of the rectangular pyramid to be | Deducing that the altitude from vertex <math>M</math> to base <math>BCHE</math> is <math>\frac{6\sqrt {13}}{13}</math> and calling the point of intersection between the altitude and the base as point <math>N</math>, we get the area of the rectangular pyramid to be | ||
− | <cmath>\frac{1}{3}([BCHE] \cdot MN) = \frac{1}{3}\left(\sqrt {13} \cdot \frac{6\sqrt {13}}{13}\right) = \frac{ | + | <cmath>\frac{1}{3}([BCHE] \cdot MN) = \frac{1}{3}\left(\sqrt {13} \cdot \frac{6\sqrt {13}}{13}\right) = \frac{6}{3} = \boxed{2}.</cmath> |
Written by: Adharshk | Written by: Adharshk |
Revision as of 20:26, 16 February 2018
In the rectangular parallelpiped shown, =
,
=
, and
=
. Point
is the midpoint of
. What is the volume of the rectangular pyramid with base
and apex
?
Contents
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
Taking the square root of both sides of the equation, we get . We can then find the area of rectangle
, noting that
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
Multiplying both sides of the equation by 2 and substituting in known values, we get
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
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.