Question

What is the perimeter of a polygon with vertices at (−2, 1) , ​ (−2, 4) ​, (2, 7) , ​ (6, 4) ​, and (6, 1) ​?

Answer 1

The other answer was using the formula correctly through most of the problem, but made a mistake when finding the last distance (#5). The person took away (-2-6 = - 4), which is incorrect. The correct answer is -8 because your adding two negatives. So, the correct answer for the whole problem is 24 units.

Good job to the person who answer they did the whole math problem - just made a little mistake at the end.

Answer 2

First we need the distances of the sides of the polygon, because perimeter = sum of all sides.
$d = \sqrt{{(y2 - y1)}^{2} + {(x2 - x1)}^{2} }$
$d1 = \sqrt{{(1 - 4)}^{2} + {( - 2 - - 2)}^{2} } \\ = \sqrt{{(- 3)}^{2} + {(0)}^{2} } = \sqrt{9} \: = 3$
$d2 = \sqrt{{(4 - 7)}^{2} + {( - 2 - 2)}^{2} } \\ = \sqrt{{(- 3)}^{2} + {( -4)}^{2} } = \sqrt{9 + 16} \\ = \sqrt{25} \: = 5$
$d3 \: = \sqrt{{(7 - 4)}^{2} + {(2 - 6)}^{2} } \\ = \sqrt{{(3)}^{2} + {( - 4)}^{2} } = \sqrt{9 + 16} \\ = \sqrt{25} \: = 5$
$d4 = \sqrt{{(4 - 1)}^{2} + {(6 - 6)}^{2} } \\ = \sqrt{ {(3)}^{2} + {(0)}^{2} } = \sqrt{9} \: = 3$
$d5 = \sqrt{{(1 - 1)}^{2} + {(-2 - 6)}^{2} } \\ = \sqrt{ {(0)}^{2} + {( - 4)}^{2}} = \sqrt{16} \: = 4$
Now, we add all sides for the perimeter:
p = d1 + d2 + d3 + d4 + d5
p = 3+5+5+3+4 = 20 units