2022 AMC 8 Problems/Problem 24

Problem

The figure below shows a polygon $ABCDEFGH$ , consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that $AH = EF = 8$ and $GH = 14$ . What is the volume of the prism?

[asy] usepackage("mathptmx"); unitsize(1cm); defaultpen(linewidth(0.7)+fontsize(11)); real r = 2, s = 2.5, theta = 14; pair G = (0,0), F = (r,0), C = (r,s), B = (0,s), M = (C+F)/2, I = M + s/2 * dir(-theta); pair N = (B+G)/2, J = N + s/2 * dir(180+theta); pair E = F + r * dir(- 45 - theta/2), D = I+E-F; pair H = J + r * dir(135 + theta/2), A = B+H-J; draw(A--B--C--I--D--E--F--G--J--H--cycle^^rightanglemark(F,I,C)^^rightanglemark(G,J,B)); draw(J--B--G^^C--F--I,linetype ("4 4")); dot(" $A$ ",A,N); dot(" $B$ ",B,1.2*N); dot(" $C$ ",C,N); dot(" $D$ ",D,dir(0)); dot(" $E$ ",E,S); dot(" $F$ ",F,1.5*S); dot(" $G$ ",G,S); dot(" $H$ ",H,W); dot(" $I$ ",I,NE); dot(" $J$ ",J,1.5*S); [/asy]

$\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288\qquad$

Solution

We try to visualize the prism by folding it in our heads. Then, $\overline{AB}$ goes on $\overline{BC},$ $\overline{AH}$ goes on $\overline{CI},$ and $\overline{EF}$ goes on $\overline{FG}.$ So, $\overline{BJ}=\overline{CI}=8$ and $\overline{FG}=\overline{BC}=8.$ Also, $\overline{HJ}$ becomes an edge parallel to $\overline{FG},$ so that means $\overline{HJ}=8.$

Since $\overline{GH}=14,$ then $\overline{JG}=14-8=6.$ So, the area of $\triangle BJG$ is $\frac{8\cdot6}{2}=24.$ If we let $\triangle BJG$ be the base, then the height is $\overline{FG}=8.$ So, the volume is $24\cdot8=\boxed{192~\textbf{(C)}}.$

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2022 AMC 8 Problems/Problem 24

Problem

Solution