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>. | ||
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+ | == See also == | ||
+ | {{AHSME box|year=1997|num-b=1|num-a=3}} |
Revision as of 16:24, 8 August 2011
Problem
The adjacent sides of the decagon shown meet at right angles. What is its perimeter?
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 |