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Difference between revisions of "1994 AHSME Problems/Problem 8"

(Created page with "==Problem== In the polygon shown, each side is perpendicular to its adjacent sides, and all 28 of the sides are congruent. The perimeter of the polygon is <math>56</math>. The ar...")
 
(→‎Solution)
Line 6: Line 6:
 
<math> \textbf{(A)}\ 84 \qquad\textbf{(B)}\ 96 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 112 \qquad\textbf{(E)}\ 196 </math>
 
<math> \textbf{(A)}\ 84 \qquad\textbf{(B)}\ 96 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 112 \qquad\textbf{(E)}\ 196 </math>
 
==Solution==
 
==Solution==
 +
Since the perimeter is <math>56</math> and all of the sides are congruent, the length of each side is <math>2</math>. We break the figure into squares as shown below.
 +
 +
<asy>
 +
unitsize(0.8cm);
 +
draw(shift(3,4)*((0,-1)--(1,-1)));
 +
draw(shift(3,4)*((-1,-2)--(2,-2)));
 +
draw(shift(3,4)*((-2,-3)--(3,-3)));
 +
draw(shift(3,4)*((-2,-4)--(3,-4)));
 +
draw(shift(3,4)*((-1,-5)--(2,-5)));
 +
draw(shift(3,4)*((0,-6)--(1,-6)));
 +
draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle);
 +
draw((1,0)--(1,1));
 +
draw((2,-1)--(2,2));
 +
draw((3,-2)--(3,3));
 +
draw((4,-2)--(4,3));
 +
draw((5,-1)--(5,2));
 +
draw((6,0)--(6,1));
 +
</asy>
 +
 +
We see that there are a total of <math>2(1+3+5)+7=25</math> squares with side length <math>2</math>. Therefore, the total area is <math>4\cdot 25=\boxed{\textbf{(C) }100.}</math>
 +
 +
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]

Revision as of 15:36, 28 June 2014

Problem

In the polygon shown, each side is perpendicular to its adjacent sides, and all 28 of the sides are congruent. The perimeter of the polygon is $56$. The area of the region bounded by the polygon is [asy] draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle); [/asy] $\textbf{(A)}\ 84 \qquad\textbf{(B)}\ 96 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 112 \qquad\textbf{(E)}\ 196$

Solution

Since the perimeter is $56$ and all of the sides are congruent, the length of each side is $2$. We break the figure into squares as shown below.

[asy] unitsize(0.8cm); draw(shift(3,4)*((0,-1)--(1,-1))); draw(shift(3,4)*((-1,-2)--(2,-2))); draw(shift(3,4)*((-2,-3)--(3,-3))); draw(shift(3,4)*((-2,-4)--(3,-4))); draw(shift(3,4)*((-1,-5)--(2,-5))); draw(shift(3,4)*((0,-6)--(1,-6))); draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle); draw((1,0)--(1,1)); draw((2,-1)--(2,2)); draw((3,-2)--(3,3)); draw((4,-2)--(4,3)); draw((5,-1)--(5,2)); draw((6,0)--(6,1)); [/asy]

We see that there are a total of $2(1+3+5)+7=25$ squares with side length $2$. Therefore, the total area is $4\cdot 25=\boxed{\textbf{(C) }100.}$

--Solution by TheMaskedMagician

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