# 1986 AJHSME Problems/Problem 13

## Problem

The perimeter of the polygon shown is $[asy] draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label("6",(0,3),W); label("8",(4,6),N); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((0.5,6)--(0.5,5.5)--(0,5.5)); draw((7.5,6)--(7.5,5.5)--(8,5.5)); draw((7.5,3)--(7.5,3.5)--(8,3.5)); draw((2.2,0)--(2.2,0.5)--(2.7,0.5)); draw((2.7,2.5)--(3.2,2.5)--(3.2,3)); [/asy]$ $\text{(A)}\ 14 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 48$ $\text{(E)}\ \text{cannot be determined from the information given}$

## Solution

### Solution 1

You might have not seen this coming but there is a very simple way to do this. If we try to make a rectangle out of this, we have to take out both of the lines that are taking out part of the rectangle we want to make. but now we see that to finish the rectangle, we have to use those same irregular lines! So all we have to do is find the perimeter of the shape as if it would be a rectangle. After that, we get $\boxed{\text{C 28}}$.

### Solution 2 $[asy] unitsize(12); draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label("6",(0,3),W); label("8",(4,6),N); draw((8,3)--(8,0)--(2.7,0),dashed); [/asy]$

The perimeter of the requested region is the same as the perimeter of the rectangle with the dashed portion. This makes the answer $2(6+8)=28\rightarrow \boxed{\text{C}}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 