2025 AMC 12A Problem 20

Question

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 lengths

7

and

13

and nonparallel sides of length

13.

What is the volume of the pentahedron?

416

520

528

676

832

Place the ridge of length

7

at height

h

; a slant edge gives

\sqrt{3^2 + 4^2 + h^2} = 13,

so

h = 12

The cross-section at height

z

is a

\left(13 - \tfrac{z}{2}\right) \times \left(8 - \tfrac{2z}{3}\right)

rectangle

The top is a ridge of length

7,

centered above the base at some height

h.

Its endpoints sit above

(3, 4)

and

(10, 4)

of the

13 \times 8

base. A slant edge to a base corner has length

\sqrt{3^2 + 4^2 + h^2} = 13,

so

h = 12.

At height

z,

the horizontal cross-section is a rectangle measuring

\left(13 - \dfrac{z}{2}\right)

by

\left(8 - \dfrac{2z}{3}\right).

At

z = 0

its area is

104

; at

z = 6

it is

10 \cdot 4 = 40

; at

z = 12

the ridge has area

0.

By the prismatoid formula,

\begin{aligned} V &= \frac{12}{6}\left(104 + 4 \cdot 40 + 0\right) \\ &= 2(264) = 528. \end{aligned}

Thus, the correct answer is C.

Problem 20 in other years