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?

$416$

$520$

$528$

$676$

$832$

Solution:

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, $V = \frac{12}{6}\left(104 + 4 \cdot 40 + 0\right) = 2(264) = 528.$

Thus, the correct answer is C.

2025 AMC 12A Problem 20

Solution:

Problem 20 in Other Years