Difference between revisions of "1984 AIME Problems/Problem 9"
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== Solution 2 (Rigorous)== | == Solution 2 (Rigorous)== | ||
It is clear that <math>DX=8</math> and <math>CX=10</math> where <math>X</math> is the foot of the perpendicular from <math>D</math> and <math>C</math> to side <math>AB</math>. Thus <math>[DXC]=\frac{ab\sin{c}}{2}=20=5*h \rightarrow h = 4</math> where h is the height of the tetrahedron from <math>D</math>. Hence, the volume of the tetrahedron is <math>bh/3=15*4/3=\boxed{020}</math> | It is clear that <math>DX=8</math> and <math>CX=10</math> where <math>X</math> is the foot of the perpendicular from <math>D</math> and <math>C</math> to side <math>AB</math>. Thus <math>[DXC]=\frac{ab\sin{c}}{2}=20=5*h \rightarrow h = 4</math> where h is the height of the tetrahedron from <math>D</math>. Hence, the volume of the tetrahedron is <math>bh/3=15*4/3=\boxed{020}</math> | ||
+ | ~ ShreyJ | ||
== Solution 3 Kinda Sketchy== | == Solution 3 Kinda Sketchy== |
Revision as of 15:05, 18 June 2018
Contents
[hide]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 1
Position face on the bottom. Since , we find that . Because the problem does not specify, we may assume both and to be isosceles triangles. Thus, the height of forms a with the height of the tetrahedron. So, . The volume of the tetrahedron is thus .
Solution 2 (Rigorous)
It is clear that and where is the foot of the perpendicular from and to side . Thus where h is the height of the tetrahedron from . Hence, the volume of the tetrahedron is ~ ShreyJ
Solution 3 Kinda Sketchy
Make faces and right triangles. This makes everything a lot easier. Then do everything in solution 1.
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 |