<h3>Given</h3>
- a rectangle x units wide and y units high divided into unit squares
<h3>Find</h3>
- The total perimeter of the unit squares, counting each line segment once
<h3>Solution</h3>
For each of the y rows of squares, there are x segments at the top, plus another x segments at the bottom. The total number of horizontal segments is then
... horizontal segment count = (y +1)x
Likewise, for each of the x columns of squares, there are y segments to the left, plus another y segments to the right of the entire area. Then the total number of vertical segments is
... vertical segment count = (x+1)y
The total segment count is ...
... total segments = horizontal segments + vertical segments
.. = (y+1)x +(x+1)y
... total segments = 2xy +x +y
_____
<u>Check</u>
We know a square (1×1) has 4 segments surrounding it.
... count = 2·1·1 +1 +1 = 4 . . . . (correct)
We know the 3×3 window in the problem statement has 24 segments.
... count = 2·3·3 +3 +3 = 18 +3 + 3 = 24 . . . . (correct)
We know a 1×3 row of panes will have 10 frame elements.
... count = 2·1·3 +1 +3 = 6 +1 +3 = 10
It looks like our formula works well.
