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 | + | 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>. | + | 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 == | ||
| Line 11: | Line 11: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:40, 14 March 2017
Problem
Faces 
and 
of tetrahedron 
meet at an angle of 
. The area of face is 
, the area of face is 
, and 
. Find the volume of the tetrahedron.
Solution
Since the area 
, the perpendicular from 
to 
has length 
.
The perpendicular from to 
is 
. Therefore, the volume is 
.
See also
| 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 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
