2025 AMC 12A Problem 20

Attempt Problem 20 of the 2025 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12A solutions, or check the answer key.

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20.

The base of the pentahedron shown below is a 13×813 \times 8 rectangle, and its lateral faces are two isosceles triangles with base of length 88 and congruent sides of length 13,13, and two isosceles trapezoids with bases of lengths 77 and 1313 and nonparallel sides of length 13.13.

What is the volume of the pentahedron?

416416

520520

528528

676676

832832

Answer: C
Concepts:3D geometryvolumepolyhedron
Difficulty rating: 2110
Solution:

The top is a ridge of length 7,7, centered above the base at some height h.h. Its endpoints sit above (3,4)(3, 4) and (10,4)(10, 4) of the 13×813 \times 8 base. A slant edge to a base corner has length 32+42+h2=13,\sqrt{3^2 + 4^2 + h^2} = 13, so h=12.h = 12.

At height z,z, the horizontal cross-section is a rectangle measuring (13z2)\left(13 - \dfrac{z}{2}\right) by (82z3).\left(8 - \dfrac{2z}{3}\right). At z=0z = 0 its area is 104104; at z=6z = 6 it is 104=4010 \cdot 4 = 40; at z=12z = 12 the ridge has area 0.0.

By the prismatoid formula, V=126(104+440+0)=2(264)=528. \begin{aligned} V &= \frac{12}{6}\left(104 + 4 \cdot 40 + 0\right) \\ &= 2(264) = 528. \end{aligned}

Thus, the correct answer is C.

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