# 1994 AHSME Problems/Problem 8

## 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 $4\cdot 25=\boxed{\textbf{(C) }100.}$