Question

What is the perimeter of a polygon with vertices at (−2, 1) , ​ (−2, 7) ​, (1, 11) , ​ (4, 7) ​ ​, and ​​ ​ (4, 1)?

Please, do not round any side lengths.

Answer 1

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
`d= \sqrt{( x_(2)- x_(1) )^2+( y_(2)- y_(1))^2 }`. Filling in using the endpoints for that segment we have
`d= √((1-(-2))^2+(11-7)^2)`. That simplifies to
`d= √(3^2+4^2)` which simplifies even further to
`d= √(25)` and d = 5. The other slant now...
`d= √((1-4)^2+(11-7)^2)` which simplifies to
`d= √((-3)^2+4^2)` which simplifies further to
`d= √(25)` and d = 5. So the side lengths for our polygon are 6+6+6+5+5 which equals 28 units. There you go!