AIME 1992 · 第 7 题

AIME 1992 — Problem 7

专题: Contest Math
难度: L4
来源: AIME

题目详情

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.

解析

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 $ABCD$ = $\frac{80\cdot12}{3}=\boxed{320}$ .