Difference between revisions of "2021 AMC 12B Problems/Problem 14"
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==Solution 1== | ==Solution 1== | ||
− | This question is just about | + | This question is just about Pythagorean theorem |
− | <cmath>a^2+(a+2)^2-b^2 = (a+4)^2 | + | <cmath>\begin{align*} |
− | + | a^2+(a+2)^2-b^2 &= (a+4)^2 \ | |
− | + | 2a^2+4a+4-b^2 &= a^2+8a+16 \ | |
− | + | a^2-4a+4-b^2 &= 16 \ | |
− | < | + | (a-2+b)(a-2-b) &= 16, |
− | With these calculation, we find out answer to be <math>\boxed{\textbf{(A) }24\sqrt5}</math> ~Lopkiloinm | + | \end{align*}</cmath> |
+ | from which <math>(a,b)=(3,7).</math> With these calculation, we find out answer to be <math>\boxed{\textbf{(A) }24\sqrt5}</math> | ||
+ | |||
+ | ~Lopkiloinm | ||
+ | |||
==Solution 2== | ==Solution 2== | ||
Let <math>\overline{AD}</math> be <math>b</math>, <math>\overline{CD}</math> be <math>a</math>, <math>\overline{MD}</math> be <math>x</math>, | Let <math>\overline{AD}</math> be <math>b</math>, <math>\overline{CD}</math> be <math>a</math>, <math>\overline{MD}</math> be <math>x</math>, |
Revision as of 06:24, 12 June 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 1
This question is just about Pythagorean theorem from which 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
Video Solution by OmegaLearn (Pythagorean Theorem and Volume of Pyramid)
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 |
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