1992 AIME Problems/Problem 7

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 $\frac{8\cdot120}{3}=\boxed{320}$ .

Solution 2

The area of $ABC$ is 120 and $BC$ =10, the slant height is 24. Height from $A$ to $BCD$ is $24 \cdot \sin 30^\circ=12$ . Since area of $BCD$ is 80, the volume of tetrahedron $ABCD$ = $\frac{80\cdot12}{3}=\boxed{320}$ .

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
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1992 AIME Problems/Problem 7

Contents

Problem

Solution

Solution 2

See also