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

1984 AIME-9.png

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 (ProblemsAnswer KeyResources)
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