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== | ||
+ | [[Category:Introductory Geometry Problems]] |
Revision as of 13:39, 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
.