1983 AIME Problems/Problem 11

Revision as of 11:14, 25 April 2008 by I like pie (talk | contribs) (Undo revision by I like pie (Talk))


The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All edges have length $s$. Given that $s=6\sqrt{2}$, what is the volume of the solid? 1983Number11.JPG


First, we find the height of the figure by drawing a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle is the median of equilateral triangle $ADE$ one of the legs is $3\sqrt{2}$. We apply the Pythagorean Theorem to find that the height is equal to $6$.

Next, we complete the figure into a triangular prism, and find the area, which is $\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432$.

Now, we subtract off the two extra pyramids that we included, whose combined area is $2\cdot \left( \frac{12\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144$.

Thus, our answer is $432-144=\boxed{288}$.

See also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions
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