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 <math>D</math> to <math>BC</math> has length <math>16</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>. | ||
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| + | ==Solution 2== | ||
| + | The area of <math>ABC</math> is 120 and <math>BC</math>=10, the slant height is 24. Height from <math>A</math> to <math>BCD</math> is <math>24 \cdot \sin 30^\circ=12</math>. Since area of <math>BCD</math> is 80, the volume of tetrahedron <math>ABCD</math>= <math>\frac{80\cdot12}{3}=\boxed{320}</math>. | ||
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== See also == | == See also == | ||
| − | + | {{AIME box|year=1992|num-b=6|num-a=8}} | |
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| − | |||
| − | + | [[Category:Intermediate Geometry Problems]] | |
| + | {{MAA Notice}} | ||
Latest revision as of 03:29, 3 February 2025
Contents
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 
.
Solution 2
The area of is 120 and =10, the slant height is 24. Height from 
to 
is 
. Since area of is 80, the volume of tetrahedron 
= 
.
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. 
