Difference between revisions of "1992 AIME Problems/Problem 7"

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== Solution ==
 
== Solution ==
Since the area <math>BCD=80=\frac{1}{2}\cdot10\cdot16</math>, the perpendicular from D to BC has length 16.
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Since the area <math>BCD=80=\frac{1}{2}\cdot10\cdot16</math>, the perpendicular from <math>D</math> to <math>BC</math> has length <math>16</math>.
  
The perpendicular from D to ABC is <math>16 \cdot \sin 30^\circ=8</math>. Therefore, the volume is <math>\frac{8\cdot120}{3}=\boxed{320}</math>.
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The perpendicular from <math>D</math> to <math>ABC</math> is <math>16 \cdot \sin 30^\circ=8</math>. Therefore, the volume is <math>\frac{8\cdot120}{3}=\boxed{320}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 17:40, 14 March 2017

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

See also

1992 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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