Difference between revisions of "2002 AMC 8 Problems/Problem 22"
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==Solution== | ==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 <math>5</math> places with two adjacent cubes, covering <math>10</math> sides, and <math>(6)(6)=36</math> faces. The exposed surface area is <math>36-10 = \boxed{\text{(C)}\ 26}</math>. | 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 <math>5</math> places with two adjacent cubes, covering <math>10</math> sides, and <math>(6)(6)=36</math> faces. The exposed surface area is <math>36-10 = \boxed{\text{(C)}\ 26}</math>. | ||
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+ | ==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 <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>. | ||
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+ | Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]] | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=PyvyO5JMgHI ~David | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/cyuum8M3hEg | ||
==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}} |
Latest revision as of 13:34, 29 October 2024
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.
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 places with two adjacent cubes, covering sides, and faces. The exposed surface area is .
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 front surface area side surface area top surface area. We find that this is .
Solution by ILoveMath31415926535
Video Solution
https://www.youtube.com/watch?v=PyvyO5JMgHI ~David
Video Solution by WhyMath
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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