Difference between revisions of "2008 AMC 12B Problems/Problem 18"
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Therefore, <math>h=\sqrt{15^2-9^2}=12</math>, and the volume of the pyramid is <math>\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784 \Rightarrow E}</math>. | Therefore, <math>h=\sqrt{15^2-9^2}=12</math>, and the volume of the pyramid is <math>\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784 \Rightarrow E}</math>. | ||
==See also== | ==See also== | ||
| + | {{AMC12 box|year=2008|ab=B|num-b=17|num-a=19}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
Revision as of 12:40, 1 December 2008
Problem
A pyramid has a square base 
and vertex 
. The area of square is 
, and the areas of 
and 
are 
and 
, respectively. What is the volume of the pyramid?
Solution
Let 
be the height of the pyramid and 
be the distance from to 
. The side length of the base is 14. The side lengths of and are 
and 
, respectively. We have a systems of equations through the Pythagorean Theorem:
Setting them equal to each other and simplifying gives
.
Therefore, 
, and the volume of the pyramid is 
.
See also
| 2008 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 17 |
Followed by Problem 19 |
| 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 12 Problems and Solutions | |
