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artcher [175]
3 years ago
8

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

Mathematics
2 answers:
Alinara [238K]3 years ago
8 0

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.

alekssr [168]3 years ago
5 0
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





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