What is the perimeter of a polygon with vertices at (−2, 1) , (−2, 7) , (1, 11) , (4, 7) , and (4, 1)? , do not round any side lengths.

If you plot these points on a piece of graph paper, you will see that the point (-2,7) is directly above the point (-2,1) because they share the same x value.  They both fall on a perfectly vertical line.  If they both fall on a perfectly vertical line, all you have to do to find the length of that segment is to count the number of units 1 is above the other.  That's 6.  The same goes for the points (4,7) and (4,1).  They are also on the same perfectly vertical line so they are also 6 units apart and the length of that segment is 6.  Btw, this polygon looks like a house that a kid would draw with a pointy roof.  The side lengths I just gave you were the sides of the house.  In order to find the length of the base of the house, we count the x units between the 2 points (-2,1) and (4,1).  The distance between -2 and 4 is 6.  So far we have 3 measurements for this house, and all 3 are 6.  Now we come to the pointy roof on the house.  It's a line that is a diagonal, so we will use the distance formula here to get the correct measure.  The distance formula is .  Filling in using the endpoints for that segment we have .  That simplifies to which simplifies even further to and d = 5.  The other slant now... which simplifies to which simplifies further to and d = 5.  So the side lengths for our polygon are 6+6+6+5+5 which equals 28 units.  There you go!