2025 AMC 12A Problems/Problem 20: Difference between revisions
m bro forgot dx |
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Note that the height is <imath>12</imath> from the previous method \n | Note that the height is <imath>12</imath> from the previous method \n | ||
Note that as you go up, the length and width both decrease linearly and reach <imath>0</imath> at the end \n | Note that as you go up, the length and width both decrease linearly and reach <imath>0</imath> at the end \n | ||
So the answer is <imath>\int_0^{12} (8-\ | So the answer is <imath>\int_0^{12} (8 - \tfrac{2}{3}x)(13 - \tfrac{1}{2}x) , dx = \boxed{528}</imath> | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== | ||
Revision as of 13:31, 7 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 .
[Diagram]
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 \n
So the answer is 
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
