2008 AMC 12B Problems/Problem 18: Difference between revisions

Revision as of 09:50, 7 April 2013

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 $\frac{bh}{3}=\frac{12\cdot 196}{3}=\boxed{784 \Rightarrow E}$ .

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

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 ==See also==
 {{AMC12 box|year=2008|ab=B|num-b=17|num-a=19}}
 [[Category:Introductory Geometry Problems]]
+[[Category:3D Geometry Problems]]

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2008 AMC 12B Problems/Problem 18: Difference between revisions

Revision as of 09:50, 7 April 2013

Problem

Solution

See also