Difference between revisions of "1986 AJHSME Problems/Problem 13"
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Revision as of 20:16, 3 July 2013
Contents
[hide]Problem
The perimeter of the polygon shown is
Solution
Solution 1
For the segments parallel to the side with side length 8, let's call those two segments and , the longer segment being , the shorter one being .
For the segments parallel to the side with side length 6, let's call those two segments and , the longer segment being , the shorter one being .
So the perimeter of the polygon would be...
Note that , and .
Now we plug those in:
28 is .
Solution 2
The perimeter of the requested region is the same as the perimeter of the rectangle with the dashed portion. This makes the answer
See Also
1986 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.