olymmath

AMC 12 2025a Problem 20

  • Paper

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 1313, and two isosceles trapezoids with bases of length 77 and 1313 and nonparallel sides of length 1313.

/* Refined by Mathemagician108 */
import graph3;
size(200);
real l = 13;
real w = 8;
real offset = (l - 7)/2; // 3
real midy = w/2; // 4
real h = 12;
triple O1 = (0,0,0);
triple O2 = (l,0,0);
triple O3 = (l,w,0);
triple O4 = (0,w,0);
triple T1 = (offset, midy, h);
triple T2 = (l - offset, midy, h);
currentprojection=orthographic((-4,-6,3));
draw(O4--O1--O2, linewidth(1));
draw(O2--O3--O4, dashed + linewidth(1));
draw(O3--T2, dashed + linewidth(1));
draw(O1--T1, linewidth(1));
draw(O4--T1, linewidth(1));
draw(O2--T2, linewidth(1));
draw(T1--T2, linewidth(1));
label("13", (O1+O2)/2, 3*-Y); // Bottom length
label("13", (O2+T2)/2, 1.5*X);
label("13", (O4+T1)/2, 2*-X);
label("8", (O1+O4)/2, 2*-X); // Width
label("7", (T1+T2)/2, 1.5*Z); // Top length

What is the volume of the pentahedron?

\textbf{(A)}~416 \textbf{(B)}~520 \textbf{(C)}~528 \textbf{(D)}~676 \textbf{(E)}~832

Dataset text AMC 12 2000 to 2025. Quoted with source.

Show the answer

C

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Similar problems

These problems have the most words in common with this statement.

Locator amc12/2025a/p20 · Tag YPBNJHHS56 · Edit this page on GitHub · Keys: j next, k previous