Difference between revisions of "2022 AMC 8 Problems/Problem 24"
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We try to visualize the prism by folding it in our heads. Then, <math>\overline{AB}</math> goes on <math>\overline{BC},</math> <math>\overline{AH}</math> goes on <math>\overline{CI},</math> and <math>\overline{EF}</math> goes on <math>\overline{FG}.</math> So, <math>\overline{BJ}=\overline{CI}=8</math> and <math>\overline{FG}=\overline{BC}=8.</math> Also, <math>\overline{HJ}</math> becomes an edge parallel to <math>\overline{FG},</math> so that means <math>\overline{HJ}=8.</math> | We try to visualize the prism by folding it in our heads. Then, <math>\overline{AB}</math> goes on <math>\overline{BC},</math> <math>\overline{AH}</math> goes on <math>\overline{CI},</math> and <math>\overline{EF}</math> goes on <math>\overline{FG}.</math> So, <math>\overline{BJ}=\overline{CI}=8</math> and <math>\overline{FG}=\overline{BC}=8.</math> Also, <math>\overline{HJ}</math> becomes an edge parallel to <math>\overline{FG},</math> so that means <math>\overline{HJ}=8.</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{ | + | 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{\textbf{(C) }192}.</math> |
Solution by aops-g5-gethsemanea2 | Solution by aops-g5-gethsemanea2 | ||
Revision as of 19:01, 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("$A$",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]](http://latex.artofproblemsolving.com/a/c/f/acf1e7169bedc87ab283bbf301ba10ff1c74d62e.png)
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 
Solution by aops-g5-gethsemanea2
