1984 AIME Problems/Problem 9

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

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $\frac{1}{3}Bh = \frac{1}{3} 15 \cdot 4 = 020$ .

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

Page

Toolbox

Search

1984 AIME Problems/Problem 9

Problem

Solution

See also