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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>.
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==See also==
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[[Category:Introductory Geometry Problems]]

Revision as of 12:39, 1 December 2008

Problem

A pyramid has a square base $ABCD$ and vertex $E$. The area of square $ABCD$ is $196$, and the areas of $\triangle ABE$ and $\triangle CDE$ are $105$ and $91$, respectively. What is the volume of the pyramid?

$\textbf{(A)}\ 392 \qquad \textbf{(B)}\ 196\sqrt {6} \qquad \textbf{(C)}\ 392\sqrt {2} \qquad \textbf{(D)}\ 392\sqrt {3} \qquad \textbf{(E)}\ 784$

Solution

Let $h$ be the height of the pyramid and $a$ be the distance from $h$ to $CD$. The side length of the base is 14. The side lengths of $\triangle ABE$ and $\triangle CDE$ are $2\cdot105\div14=15$ and $2\cdot91\div14=13$, respectively. We have a systems of equations through the Pythagorean Theorem:

$13^2-(14-a)^2=h^2 \\ 15^2-a^2=h^2$

Setting them equal to each other and simplifying gives $-27+28a=225 \implies a=9$.

Therefore, $h=\sqrt{15^2-9^2}=12$, and the volume of the pyramid is $\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784 \Rightarrow E}$.

See also

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