Difference between revisions of "1993 AJHSME Problems/Problem 17"
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==See Also== | ==See Also== | ||
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Revision as of 23:11, 4 July 2013
Problem
Square corners, 
units on a side, are removed from a 
unit by 
unit rectangular sheet of cardboard. The sides are then folded to form an open box. The surface area, in square units, of the interior of the box is
![[asy] fill((0,0)--(20,0)--(20,5)--(0,5)--cycle,lightgray); fill((20,0)--(20+5*sqrt(2),5*sqrt(2))--(20+5*sqrt(2),5+5*sqrt(2))--(20,5)--cycle,lightgray); draw((0,0)--(20,0)--(20,5)--(0,5)--cycle); draw((0,5)--(5*sqrt(2),5+5*sqrt(2))--(20+5*sqrt(2),5+5*sqrt(2))--(20,5)); draw((20+5*sqrt(2),5+5*sqrt(2))--(20+5*sqrt(2),5*sqrt(2))--(20,0)); draw((5*sqrt(2),5+5*sqrt(2))--(5*sqrt(2),5*sqrt(2))--(5,5),dashed); draw((5*sqrt(2),5*sqrt(2))--(15+5*sqrt(2),5*sqrt(2)),dashed); [/asy]](http://latex.artofproblemsolving.com/0/8/1/081723c76105ecf289d9aa0b7b20c7579ab57b17.png)
Solution
If the sides of the open box are folded down so that a flat sheet with four corners cur out remains, then the revealed surface would have the same area as the interior of the box. This is equal to the area of the four corners subtracted from the area of the original sheet, which is 
.
See Also
| 1993 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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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