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

Solution 2

Note that the height is $12$ from the previous method \n Note that as you go up, the length and width both decrease linearly and reach $0$ at the end

So the answer is $\int_0^{12} (8 - \tfrac{2}{3}x)(13 - \tfrac{1}{2}x) dx = \boxed{528}$

Video Solution 1 by OmegaLearn

https://youtu.be/WwMDpnuZrJ4

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

Contents

Problem

Solution 1 (Split Into Three Parts)

Solution 2

Video Solution 1 by OmegaLearn

See Also