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Difference between revisions of "1984 AIME Problems/Problem 9"

(there's clearly a couple of typos in the prob, + solution, box)
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== See also ==
 
== See also ==
 
{{AIME box|year=1984|num-b=8|num-a=10}}
 
{{AIME box|year=1984|num-b=8|num-a=10}}
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* [[AIME Problems and Solutions]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]

Revision as of 13:25, 6 May 2007

Problem

In tetrahedron $\displaystyle ABCD$, edge $\displaystyle AB$ has length 3 cm. The area of face $\displaystyle ABC$ is $\displaystyle 15\mbox{cm}^2$ and the area of face $\displaystyle ABD$ is $\displaystyle 12 \mbox { cm}^2$. These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\displaystyle \mbox{cm}^3$.

Solution

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Position face $ABC$ on the bottom. Since $[\triangle ABD] = 12 = \frac{1}{2} AB \cdot h_{ABD}$, we find that $h_{ABD} = 8$. The height of $ABD$ forms a $30-60-90$ with the height of the tetrahedron, so $h = \frac{1}{2} 8 = 4$. The volume of the tetrahedron is thus $\frac{1}{3}Bh = \frac{1}{3} 15 \cdot 4 = 020$.

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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