Difference between revisions of "1997 AHSME Problems/Problem 2"
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Revision as of 16:23, 8 August 2011
Problem
The adjacent sides of the decagon shown meet at right angles. What is its perimeter?
![[asy] defaultpen(linewidth(.8pt)); dotfactor=4; dot(origin);dot((12,0));dot((12,1));dot((9,1));dot((9,7));dot((7,7));dot((7,10));dot((3,10));dot((3,8));dot((0,8)); draw(origin--(12,0)--(12,1)--(9,1)--(9,7)--(7,7)--(7,10)--(3,10)--(3,8)--(0,8)--cycle); label("$8$",midpoint(origin--(0,8)),W); label("$2$",midpoint((3,8)--(3,10)),W); label("$12$",midpoint(origin--(12,0)),S);[/asy]](http://latex.artofproblemsolving.com/2/a/b/2ab8d1d178253954086e567a3212c8ba9b1ae2a8.png)
Solution
The three unlabelled vertical sides have the same sum as the two labelled vertical sides, which is 
.
The four unlabelled horizontal sides have the same sum as the one large horizontal side, which is 
.
Thus, the perimeter is 
, which is option 
.
