1997 AHSME Problems/Problem 2

Revision as of 16:24, 8 August 2011 by Talkinaway (talk | contribs) (Solution)

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]

$\mathrm{(A)\ } 22 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \  } 34 \qquad \mathrm{(D) \  } 44 \qquad \mathrm{(E) \  }50$

Solution

The three unlabelled vertical sides have the same sum as the two labelled vertical sides, which is $10$.

The four unlabelled horizontal sides have the same sum as the one large horizontal side, which is $12$.

Thus, the perimeter is $2(12+10) = 44$, which is option $\boxed{D}$.

See also

1997 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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