2025 AMC 12A Problems/Problem 20

Problem

The base of the pentahedron shown below is a $13 \times 8$ rectangle, and its lateral faces are two isosceles triangles with base of length $8$ and congruent sides of length $13$ , and two isosceles trapezoids with bases of length $7$ and $13$ and nonparallel sides of length $13$ .

[Diagram]

What is the volume of the pentahedron?

Solution 1 (Split Into Three Parts)

Notice that the triangular faces have a slant height of $\sqrt{13^2-4^2}=\sqrt{153}$ and that the height is therefore $\sqrt{153-(\frac{13-7}{2})^2} = 12$ . Then we can split the pentahedron into a triangular prism and two pyramids. The pyramids each have a volume of $\frac{1}{3}(3)(8)(12) = 96$ and the prism has a volume of $\frac{1}{2}(8)(12)(7) = 336$ . Thus the answer is $336+96 \cdot 2 = \boxed{\text{(C) } 528}$

~ Shadowleafy

2025 AMC 12A (Problems • Answer Key • Resources)
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2025 AMC 12A Problems/Problem 20

Problem

Solution 1 (Split Into Three Parts)

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