Difference between revisions of "1984 AIME Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | + | [[Image:1984_AIME-9.png]] | |
| − | Position face <math>ABC</math> on the bottom. Since <math>[\triangle ABD] = 12 = \frac{1}{2} AB \cdot h_{ABD}</math>, we find that <math>h_{ABD} = 8</math>. The height of <math>ABD</math> forms a <math>30-60-90</math> with the height of the tetrahedron, so <math>h = \frac{1}{2} 8 = 4</math>. The volume of the tetrahedron is thus <math>\frac{1}{3}Bh = \frac{1}{3} 15 \cdot 4 = 020</math>. | + | Position face <math>ABC</math> on the bottom. Since <math>[\triangle ABD] = 12 = \frac{1}{2} AB \cdot h_{ABD}</math>, we find that <math>h_{ABD} = 8</math>. The height of <math>ABD</math> forms a <math>30-60-90</math> with the height of the tetrahedron, so <math>h = \frac{1}{2} 8 = 4</math>. The volume of the tetrahedron is thus <math>\frac{1}{3}Bh = \frac{1}{3} 15 \cdot 4 = 020</math>. |
== See also == | == See also == | ||
Revision as of 20:26, 10 September 2007
Problem
In tetrahedron 
, edge 
has length 3 cm. The area of face 
is 
and the area of face 
is 
. These two faces meet each other at a 
angle. Find the volume of the tetrahedron in 
.
Solution
Position face 
on the bottom. Since ![$[\triangle ABD] = 12 = \frac{1}{2} AB \cdot h_{ABD}$](http://latex.artofproblemsolving.com/3/9/d/39d2315c1b1256bfd9a313d066d4f858c586d5ea.png)
, we find that 
. The height of 
forms a 
with the height of the tetrahedron, so 
. The volume of the tetrahedron is thus 
.
See also
| 1984 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
