Difference between revisions of "2008 AMC 12A Problems/Problem 18"
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<cmath> abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95}, </cmath> | <cmath> abc/6 = \sqrt{a^2b^2c^2}/6 = \frac{\sqrt{19 \cdot 6 \cdot 30}}{6} = \sqrt{95}, </cmath> | ||
which is answer choice <math>\boxed{\text{C}}</math>. | which is answer choice <math>\boxed{\text{C}}</math>. | ||
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+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/MOcX5BFbcwU?t=1320 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == |
Latest revision as of 03:51, 16 January 2023
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
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 because we can consider the tetrahedron to be a right triangular pyramid. 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.