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The coordinates of the vertices of a polygon are (−4, 2) , (2, 2) , (2, −1) , (−2, −3) , and (−5, −2) . What is the perimeter of the polygon? Enter your answer as a decimal, rounded to the nearest tenth of a unit, in the box.

User Jbm
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check the picture below.

now, the distance form -4,2 to 2,2 you can pretty much get it off the grid by counting the units.

and the distance from 2,2 to 2, -1, you also can get it off the grid by just counting.

now, let's check the other lengths,

from -4, 2 to -5, -2

from -5, -2 to -2, -3

from -2, -3 to 2, -1


\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -4}}\quad ,&{{ 2}})\quad % (c,d) &({{ -5}}\quad ,&{{ -2}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ d=√([-5-(-4)]^2+[-2-2]^2)\implies d=√((-5+4)^2+(-4)^2) \\\\\\ d=√((-1)^2+(-4)^2)\implies d=√(1+16)\implies \boxed{d=√(17)}\\\\ -------------------------------


\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (c,d) &({{ -5}}\quad ,&{{ -2}}) % (a,b) &({{ -2}}\quad ,&{{ -3}})\quad \end{array}\qquad \\\\\\ d=√([-2-(-5)]^2+[-3-(-2)]^2)\\\\\\ d=√((-2+5)^2+(-3+2)^2) \\\\\\ d=√(3^2+(-1)^2)\implies d=√(9+1)\implies \boxed{d=√(10)}\\\\ -------------------------------\\\\


\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -2}}\quad ,&{{ -3}})\quad % (c,d) &({{ 2}}\quad ,&{{ -1}}) \end{array}\qquad \\\\\\ d=√([2-(-2)]^2+[-1-(-3)]^2)\implies d=√((2+2)^2+(-1+3)^2) \\\\\\ d=√(4^2+2^2)\implies d=√(16+4)\implies d=√(20)\implies \boxed{d=2√(5)}


\bf \textit{thus the perimeter is then}\implies 6~+~3~+~√(17)~+~√(10)~+~2√(5)
The coordinates of the vertices of a polygon are (−4, 2) , (2, 2) , (2, −1) , (−2, −3) , and-example-1
User Arunas
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