# 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?

$[asy] usepackage("mathptmx"); size(275); defaultpen(linewidth(0.8)); real r = 2, s = 2.5, theta = 14; pair G = (0,0), F = (r,0), C = (r,s), B = (0,s), M = (C+F)/2, I = M + s/2 * dir(-theta); pair N = (B+G)/2, J = N + s/2 * dir(180+theta); pair E = F + r * dir(- 45 - theta/2), D = I+E-F; pair H = J + r * dir(135 + theta/2), A = B+H-J; draw(A--B--C--I--D--E--F--G--J--H--cycle^^rightanglemark(F,I,C)^^rightanglemark(G,J,B)); draw(J--B--G^^C--F--I,linetype ("4 4")); dot("A",A,N); dot("B",B,1.2*N); dot("C",C,N); dot("D",D,dir(0)); dot("E",E,S); dot("F",F,1.5*dir(-100)); dot("G",G,S); dot("H",H,W); dot("I",I,NE); dot("J",J,1.5*S); [/asy]$

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

## Solution

While imagining the folding, $\overline{AB}$ goes on $\overline{BC},$ $\overline{AH}$ goes on $\overline{CI},$ and $\overline{EF}$ goes on $\overline{FG}.$ So, $BJ=CI=8$ and $FG=BC=8.$ Also, $\overline{HJ}$ becomes an edge parallel to $\overline{FG},$ so that means $HJ=8.$

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

~aops-g5-gethsemanea2

## Remark

After folding polygon $ABCDEFGH$ on the dotted lines, we obtain the following triangular prism: $[asy] /* Made by MRENTHUSIASM */ usepackage("mathptmx"); size(200); defaultpen(linewidth(0.8)); import graph3; import solids; currentprojection=orthographic((0.3,-0.3,0.3)); triple J, G, B, A, H, F; J = (0,0,0); G = (6,0,0); B = (0,8,0); A = (0,8,8); H = (0,0,8); F = (6,0,8); draw(surface(B--J--G--cycle),yellow); draw(surface(H--A--F--cycle),yellow); draw(B--J,dashed); draw(G--J--H--A--B^^A--F--H^^F--G^^B--G); draw((0.5,0,0)--(0.5,0.5,0)--(0,0.5,0)^^(0.5,0,8)--(0.5,0.5,8)--(0,0.5,8)); dot("A=C",A,1.5*E); dot("B",B,1.5*E); dot("D=J",J,1.5*W); dot("F",F,1.5*E); dot("H=I",H,1.5*W); dot("E=G",G,1.5*E); label("8",midpoint(A--H),1.5*NW,red); label("6",midpoint(H--F),1.5*S,red); label("8",midpoint(H--J),1.5*W,red); [/asy]$ ~MRENTHUSIASM

~ pi_is_3.14

## Video Solution

~Mathematical Dexterity

~Interstigation

~STEMbreezy

~savannahsolver