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Revision as of 18:24, 4 July 2013

Problem

Faces $ABC^{}_{}$ and $BCD^{}_{}$ of tetrahedron $ABCD^{}_{}$ meet at an angle of $30^\circ$. The area of face $ABC^{}_{}$ is $120^{}_{}$, the area of face $BCD^{}_{}$ is $80^{}_{}$, and $BC=10^{}_{}$. Find the volume of the tetrahedron.

Solution

Since the area $BCD=80=\frac{1}{2}\cdot10\cdot16$, the perpendicular from D to BC has length 16.

The perpendicular from D to ABC is $16 \cdot \sin 30^\circ=8$. Therefore, the volume is $\frac{8\cdot120}{3}=\boxed{320}$.

See also

1992 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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