Question

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.

Answer

Answer: 320

Concepts:3D geometry trigonometry volume

Difficulty rating: 2040

Small Hint:

Find the altitudes from $A$ and $D$ to the common edge $BC$

Big Hint:

The height from $D$ to plane $ABC$ is its face altitude multiplied by $\sin30^\circ$

Solution:

The altitudes to $BC$ in faces $ABC$ and $BCD$ are $\frac{2(120)}{10}=24$ and $\frac{2(80)}{10}=16,$ respectively. Because the dihedral angle is $30^\circ,$ the perpendicular height from $D$ to plane $ABC$ is $16\sin30^\circ=8.$ Using face $ABC$ as the base, the volume is $\frac13(120)(8)=320.$

1992 AIME Problem 7

Answer: 320

Difficulty rating: 2040

Small Hint:

Big Hint:

Solution:

Problem 7 in Other Years