Difference between revisions of "2022 AMC 8 Problems/Problem 24"
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[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("<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] | ||
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<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> | ||
Revision as of 18:59, 28 January 2022
Problem
The figure below shows a polygon , consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that and . 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,N); dot("",B,1.2*N); dot("",C,N); dot("",D,dir(0)); dot("",E,S); dot("",F,1.5*S); dot("",G,S); dot("",H,W); dot("",I,NE); dot("",J,1.5*S); [/asy]
Solution
We try to visualize the prism by folding it in our heads. Then, goes on goes on and goes on So, and Also, becomes an edge parallel to so that means
Since then So, the area of is If we let be the base, then the height is So, the volume is