2025 AMC 12A Problems/Problem 20: Difference between revisions
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So the answer is <imath>\int_0^{12} (8 - \tfrac{2}{3}x)(13 - \tfrac{1}{2}x) dx = \boxed{528}</imath> | So the answer is <imath>\int_0^{12} (8 - \tfrac{2}{3}x)(13 - \tfrac{1}{2}x) dx = \boxed{528}</imath> | ||
== Solution 3 == | |||
It just reminds about one of old aime problem.{https://artofproblemsolving.com/wiki/index.php title=1983_AIME_Problems/Problem_11} | |||
~DRA777 | |||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== | ||
Revision as of 15:51, 9 November 2025
Problem
The base of the pentahedron shown below is a
rectangle, and its lateral faces are two isosceles triangles with base of length
and congruent sides of length
, and two isosceles trapezoids with bases of length
and
and nonparallel sides of length
.
What is the volume of the pentahedron?
Solution 1 (Split Into Three Parts)
Notice that the triangular faces have a slant height of
and that the height is therefore
. Then we can split the pentahedron into a triangular prism and two pyramids. The pyramids each have a volume of
and the prism has a volume of
. Thus the answer is
~ Shadowleafy
Solution 2
Note that the height is
from the previous method \n
Note that as you go up, the length and width both decrease linearly and reach
at the end
So the answer is
Solution 3
It just reminds about one of old aime problem.{https://artofproblemsolving.com/wiki/index.php title=1983_AIME_Problems/Problem_11} ~DRA777
Video Solution 1 by OmegaLearn
See Also
| 2025 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 19 |
Followed by Problem 21 |
| 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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