2008 AMC 12A Problems/Problem 18
Problem
Triangle 
, with sides of length 
, 
, and 
, has one vertex on the positive 
-axis, one on the positive 
-axis, and one on the positive 
-axis. Let 
be the origin. What is the volume of tetrahedron 
?
Solution
![[asy] defaultpen(fontsize(8)); draw((0,10)--(0,0)--(8,0));draw((-3,-4)--(0,0));draw((0,10)--(-3,-4)--(8,0)--cycle); label("A",(8,0),(1,0));label("B",(0,10),(0,1));label("C",(-3,-4),(-1,-1));label("O",(0,0),(1,1)); label("$a$",(4,0),(0,1));label("$b$",(0,5),(1,0));label("$c$",(-1.5,-2),(1,0)); label("$5$",(4,5),(1,1));label("$6$",(-1.5,3),(-1,0));label("$7$",(2.5,-2),(1,-1)); [/asy]](http://latex.artofproblemsolving.com/b/4/8/b48d9f2414f67d68d2e53886e0726e31626cd919.png)
Without loss of generality, let 
be on the axis, 
be on the axis, and 
be on the axis, and let 
have respective lengths of 5, 6, and 7. Let 
denote the lengths of segments 
respectively. Then by the Pythagorean Theorem,
so 
; similarly, 
and 
. Since 
, 
, and 
are mutually perpendicular, the tetrahedron's volume is ![\[abc/6\]](http://latex.artofproblemsolving.com/b/7/4/b74807ac0ee77d6e400564a0e5622c8fa9371d61.png)
because we can consider the tetrahedron to be a right triangular pyramid.
![\[abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95},\]](http://latex.artofproblemsolving.com/f/a/4/fa42628f26cbd85dad184ed745d0a3699e3afaa9.png)
which is answer choice 
.
Video Solution by OmegaLearn
https://youtu.be/MOcX5BFbcwU?t=1320
~ pi_is_3.14
See also
| 2008 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 17 |
Followed by Problem 19 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
