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Difference between revisions of "2002 AMC 8 Problems/Problem 22"

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==Solution 2==
 
==Solution 2==
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We can count the number of showing faces from each side. One thing that we notice is that the front face has the same number of squares as the back face, the side faces have the same surface area, etc. Therefore, we are looking for <math>2(</math>front surface area + side surface area <math>+</math> top surface area<math>)</math>. We find that this is <math>2(5 + 4 + 4) = 2 * 13 = \boxed{\text{(C)}\ 26}</math>.
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We can count the number of showing faces from each side. One thing that we notice is that the front face has the same number of squares as the back face, the side faces have the same surface area, etc. Therefore, we are looking for <math>2(</math>front surface area <math>+</math> side surface area <math>+</math> top surface area<math>)</math>. We find that this is <math>2(5 + 4 + 4) = 2 * 13 = \boxed{\text{(C)}\ 26}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2002|num-b=21|num-a=23}}
 
{{AMC8 box|year=2002|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:45, 6 January 2023

Problem

Six cubes, each an inch on an edge, are fastened together, as shown. Find the total surface area in square inches. Include the top, bottom, and sides.

[asy] /* AMC8 2002 #22 Problem */ draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1,1)); draw((1,0)--(1.5,0.5)--(1.5,1.5)); draw((0.5,1.5)--(1,2)--(1.5,2)); draw((1.5,1.5)--(1.5,3.5)--(2,4)--(3,4)--(2.5,3.5)--(2.5,0.5)--(1.5,.5)); draw((1.5,3.5)--(2.5,3.5)); draw((1.5,1.5)--(3.5,1.5)--(3.5,2.5)--(1.5,2.5)); draw((3,4)--(3,3)--(2.5,2.5)); draw((3,3)--(4,3)--(4,2)--(3.5,1.5)); draw((4,3)--(3.5,2.5)); draw((2.5,.5)--(3,1)--(3,1.5));[/asy]

$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36$

Solution

Count the number of sides that are not exposed, where a cube is connected to another cube and subtract it from the total number of faces. There are $5$ places with two adjacent cubes, covering $10$ sides, and $(6)(6)=36$ faces. The exposed surface area is $36-10 = \boxed{\text{(C)}\ 26}$.

Solution 2

We can count the number of showing faces from each side. One thing that we notice is that the front face has the same number of squares as the back face, the side faces have the same surface area, etc. Therefore, we are looking for $2($front surface area $+$ side surface area $+$ top surface area$)$. We find that this is $2(5 + 4 + 4) = 2 * 13 = \boxed{\text{(C)}\ 26}$.

See Also

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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