2025 AMC 12A Problems/Problem 20: Difference between revisions
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== Solution 1 (Split Into Three Parts) == | == Solution 1 (Split Into Three Parts) == | ||
Notice that the triangular faces have a slant height of <imath>\sqrt{13^2-4^2}=\sqrt{153}</imath> and that the height is therefore <imath>\sqrt{153-(\frac{13-7}{2})^2} = 12</imath>. Then we can split the pentahedron into a triangular prism and two pyramids. The pyramids each have a volume of <imath>\frac{1}{3}(3)(8)(12) = 96</imath> and the prism has a volume of <imath>\frac{1}{2}(8)(12)(7) = 336</imath>. Thus the answer is <imath>336+96 \cdot 2 = \boxed{(C) | Notice that the triangular faces have a slant height of <imath>\sqrt{13^2-4^2}=\sqrt{153}</imath> and that the height is therefore <imath>\sqrt{153-(\frac{13-7}{2})^2} = 12</imath>. Then we can split the pentahedron into a triangular prism and two pyramids. The pyramids each have a volume of <imath>\frac{1}{3}(3)(8)(12) = 96</imath> and the prism has a volume of <imath>\frac{1}{2}(8)(12)(7) = 336</imath>. Thus the answer is <imath>336+96 \cdot 2 = \boxed{\text{(C) } 528}</imath> | ||
~ Shadowleafy | ~ Shadowleafy | ||
Revision as of 16:55, 6 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
