Difference between revisions of "2000 AMC 8 Problems/Problem 22"
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The percent increase in surface area is <math>\frac{SA_{new} - SA_{old}}{SA_{old}}\cdot 100\% = \frac{28-24}{24}\cdot 100\% \approx 16.67\%</math>, giving the closest answer as <math>\boxed{C}</math>. | The percent increase in surface area is <math>\frac{SA_{new} - SA_{old}}{SA_{old}}\cdot 100\% = \frac{28-24}{24}\cdot 100\% \approx 16.67\%</math>, giving the closest answer as <math>\boxed{C}</math>. | ||
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| + | ==Video Solution== | ||
| + | |||
| + | https://www.youtube.com/watch?v=X5T02J3eNqM ~David | ||
==See Also== | ==See Also== | ||
Latest revision as of 19:06, 15 April 2023
Contents
Problem
A cube has edge length 
. Suppose that we glue a cube of edge length 
on top of the big cube so that one of its faces rests entirely on the top face of the larger cube. The percent increase in the surface area (sides, top, and bottom) from the original cube to the new solid formed is closest to
![[asy] draw((0,0)--(2,0)--(3,1)--(3,3)--(2,2)--(0,2)--cycle); draw((2,0)--(2,2)); draw((0,2)--(1,3)); draw((1,7/3)--(1,10/3)--(2,10/3)--(2,7/3)--cycle); draw((2,7/3)--(5/2,17/6)--(5/2,23/6)--(3/2,23/6)--(1,10/3)); draw((2,10/3)--(5/2,23/6)); draw((3,3)--(5/2,3));[/asy]](http://latex.artofproblemsolving.com/4/c/1/4c1daa4799268f363fcefd996e79452fd6346cfc.png)
Solution
The original cube has 
faces, each with an area of 
square units. Thus the original figure had a total surface area of 
square units.
The new figure has the original surface, with new faces that each have an area of square unit, for a total surface area of of additional square units added to it. But square unit of the top of the bigger cube, and square unit on the bottom of smaller cube, is not on the surface, and does not count towards the surface area.
The total surface area is therefore 
square units.
The percent increase in surface area is 
, giving the closest answer as 
.
Video Solution
https://www.youtube.com/watch?v=X5T02J3eNqM ~David
See Also
| 2000 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
