Difference between revisions of "2021 AMC 10A Problems/Problem 13"
m (→Solution) |
m (→Solution) |
||
Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
− | Drawing the tetrahedron out and testing side lengths, we realize that the <math>\triangle ABD, ABC, and ABD</math> are right triangles by the Converse of the Pythagorean Theorem. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid. If we take <math>\triangle ADC</math> as the base, then <math>AB</math> must be the height. <math>\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2</math>, so we have an answer of <math>\boxed{\textbf{(C) } 4}</math>. | + | Drawing the tetrahedron out and testing side lengths, we realize that the <math>\triangle ABD, \triangle ABC,</math> and <math>\triangle ABD</math> are right triangles by the Converse of the Pythagorean Theorem. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid. If we take <math>\triangle ADC</math> as the base, then <math>AB</math> must be the height. <math>\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2</math>, so we have an answer of <math>\boxed{\textbf{(C) } 4}</math>. |
==Similar Problem== | ==Similar Problem== |
Revision as of 10:10, 4 April 2021
Contents
[hide]Problem
What is the volume of tetrahedron with edge lengths , , , , , and ?
Solution
Drawing the tetrahedron out and testing side lengths, we realize that the and are right triangles by the Converse of the Pythagorean Theorem. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid. If we take as the base, then must be the height. , so we have an answer of .
Similar Problem
https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_21
Video Solution (Simple & Quick)
~ Education, the Study of Everything
Video Solution (Using Pythagorean Theorem, 3D Geometry - Tetrahedron)
~ pi_is_3.14
Video Solution by TheBeautyofMath
https://youtu.be/t-EEP2V4nAE?t=813
~IceMatrix
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.