Difference between revisions of "1997 AHSME Problems/Problem 2"
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Thus, the perimeter is <math>2(12+10) = 44</math>, which is option <math>\boxed{D}</math>. | Thus, the perimeter is <math>2(12+10) = 44</math>, which is option <math>\boxed{D}</math>. | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME box|year=1997|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:11, 5 July 2013
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 
.
See also
| 1997 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
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