Difference between revisions of "1992 AIME Problems/Problem 7"
m (→Solution) |
|||
| Line 6: | Line 6: | ||
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>. | 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>. | ||
| + | |||
| + | ==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>. | ||
| + | |||
== See also == | == See also == | ||
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. 
