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1984 AIME Problems/Problem 9

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Problem

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

Solution

For non-asymptote version of image, see Image:1984_AIME-9.png.
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Position face ABC on the bottom. Since [\triangle ABD] = 12 = \frac{1}{2} \cdot 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 = \boxed{020}.

See also

1984 AIME (ProblemsResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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