Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 8 Problems/Problem 24"
Line 3: Line 3:
 
The figure below shows a polygon <math>ABCDEFGH</math>, consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that <math>AH = EF = 8</math> and <math>GH = 14</math>. What is the volume of the prism?
 
The figure below shows a polygon <math>ABCDEFGH</math>, consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that <math>AH = EF = 8</math> and <math>GH = 14</math>. 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("<math>A</math>",A,N); dot("<math>B</math>",B,1.2*N); dot("<math>C</math>",C,N); dot("<math>D</math>",D,dir(0)); dot("<math>E</math>",E,S); dot("<math>F</math>",F,1.5*S); dot("<math>G</math>",G,S); dot("<math>H</math>",H,W); dot("<math>I</math>",I,NE); dot("<math>J</math>",J,1.5*S); [/asy]
+
<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>
  
 
<math>\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288\qquad</math>
 
<math>\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288\qquad</math>
Line 12: Line 33:
  
 
Since <math>\overline{GH}=14,</math> then <math>\overline{JG}=14-8=6.</math> So, the area of <math>\triangle BJG</math> is <math>\frac{8\cdot6}{2}=24.</math> If we let <math>\triangle BJG</math> be the base, then the height is <math>\overline{FG}=8.</math> So, the volume is <math>24\cdot8=\boxed{192~\textbf{(C)}}.</math>
 
Since <math>\overline{GH}=14,</math> then <math>\overline{JG}=14-8=6.</math> So, the area of <math>\triangle BJG</math> is <math>\frac{8\cdot6}{2}=24.</math> If we let <math>\triangle BJG</math> be the base, then the height is <math>\overline{FG}=8.</math> So, the volume is <math>24\cdot8=\boxed{192~\textbf{(C)}}.</math>
 +
 +
Solution by aops-g5-gethsemanea2

Revision as of 19:00, 28 January 2022

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 $24\cdot8=\boxed{192~\textbf{(C)}}.$

Solution by aops-g5-gethsemanea2

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_24&oldid=170597"