Difference between revisions of "1984 AIME Problems/Problem 9"
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− | Position face <math>ABC</math> on the bottom. Since <math>[\triangle ABD] = 12 = \frac{1}{2} \cdot AB \cdot h_{ABD}</math>, we find that <math>h_{ABD} = 8</math>. | + | Position face <math>ABC</math> on the bottom. Since <math>[\triangle ABD] = 12 = \frac{1}{2} \cdot AB \cdot h_{ABD}</math>, we find that <math>h_{ABD} = 8</math>. Because the problem does not specify, we can safely assume both <math>ABC</math> and <math>ABD</math> are isosceles triangles. Thus, 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 = \boxed{020}</math>. |
== See also == | == See also == |
Revision as of 15:48, 16 September 2015
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 , we find that . Because the problem does not specify, we can safely assume both and are isosceles triangles. Thus, 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 |