Difference between revisions of "2021 AMC 12B Problems/Problem 14"
Jamess2022 (talk | contribs) (→Solution) |
|||
Line 29: | Line 29: | ||
Solving for other values, we get <math>b = 2\sqrt{10}</math>, <math>a = 6\sqrt{2}</math>. | Solving for other values, we get <math>b = 2\sqrt{10}</math>, <math>a = 6\sqrt{2}</math>. | ||
The volume is then <cmath>\frac{1}{3} abx = \boxed{\textbf{(A)}24\sqrt{5}}</cmath> ~jamess2022(burntTacos) | The volume is then <cmath>\frac{1}{3} abx = \boxed{\textbf{(A)}24\sqrt{5}}</cmath> ~jamess2022(burntTacos) | ||
+ | |||
+ | ==Video Solution by Hawk Math== | ||
+ | https://www.youtube.com/watch?v=p4iCAZRUESs | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2021|ab=B|num-b=13|num-a=15}} | {{AMC12 box|year=2021|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:55, 12 February 2021
Contents
[hide]Problem
Let be a rectangle and let be a segment perpendicular to the plane of . Suppose that has integer length, and the lengths of and are consecutive odd positive integers (in this order). What is the volume of pyramid
Solution
Solution 1
This question is just about pythagorean theorem With these calculation, we find out answer to be ~Lopkiloinm
Solution 2
Let be , be , be , , , be , , respectively.
We have three equations:
Subbing in the first and third equation into the second equation, we get: Therefore, , Solving for other values, we get , . The volume is then ~jamess2022(burntTacos)
Video Solution by Hawk Math
https://www.youtube.com/watch?v=p4iCAZRUESs
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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.