Difference between revisions of "1986 AJHSME Problems/Problem 13"
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| − | + | The perimeter of the requested region is the same as the perimeter of the rectangle with the dashed portion. This makes the answer <math>2(6+8)=28\rightarrow \boxed{\text{C}}</math> | |
==See Also== | ==See Also== | ||
Revision as of 21:15, 8 October 2012
Problem
The perimeter of the polygon shown is
![[asy] draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label("$6$",(0,3),W); label("$8$",(4,6),N); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((0.5,6)--(0.5,5.5)--(0,5.5)); draw((7.5,6)--(7.5,5.5)--(8,5.5)); draw((7.5,3)--(7.5,3.5)--(8,3.5)); draw((2.2,0)--(2.2,0.5)--(2.7,0.5)); draw((2.7,2.5)--(3.2,2.5)--(3.2,3)); [/asy]](http://latex.artofproblemsolving.com/f/1/9/f19e2361261c1f97b4b8c8b6958ec7f30d342c8e.png)
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
![[asy] unitsize(12); draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label("$6$",(0,3),W); label("$8$",(4,6),N); draw((8,3)--(8,0)--(2.7,0),dashed); [/asy]](http://latex.artofproblemsolving.com/9/c/2/9c2c60ec3f042b22a74ebe21e5a64a6233dfa214.png)
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 | ||
