Question

Find the perimeter of the polygon with vertices A(-6,-4), B(-3,6), C(4,0), and D(2,-1)?

Answer 1

Check the picture below.

so the perimeter of the polygon is the sum of all its sides, namely, AB + BC + CD + DA.

now, let's check how long each side is,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &A&(~{{ -6}} &,&{{ -4}}~) % (c,d) &B&(~{{ -3}} &,&{{ 6}}~) \end{array} \\\\\\ d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\ -------------------------------\\\\ AB=\sqrt{[-3-(-6)]^2+[6-(-4)]^2} \\\\\\ AB=\sqrt{(-3+6)^2+(6+4)^2} \\\\\\ AB=\sqrt{3^2+10^2}\implies \boxed{AB=\sqrt{109}}\\\\ -------------------------------$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &B&(~{{ -3}} &,&{{6}}~) % (c,d) &C&(~{{ 4}} &,&{{ 0}}~) \end{array} \\\\ -------------------------------\\\\ BC=\sqrt{[4-(-3)]^2+[0-6]^2}\implies BC=\sqrt{(4+3)^2+(0-6)^2} \\\\\\ BC=\sqrt{7^2+(-6)^2}\implies \boxed{BC=\sqrt{85}}\\\\ -------------------------------$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &C&(~{{ 4}} &,&{{0}}~) % (c,d) &D&(~{{ 2}} &,&{{ -1}}~) \end{array} \\\\ -------------------------------\\\\ CD=\sqrt{(2-4)^2+(-1-0)^2}\implies CD=\sqrt{(-2)^2+(-1)^2} \\\\\\ \boxed{CD=\sqrt{5}}\\\\ -------------------------------$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &D(~{{ 2}} &,&{{-1}}~) % (c,d) &A&(~{{ -6}} &,&{{ -4}}~) \end{array}\\\\ -------------------------------\\\\ DA=\sqrt{[-6-2]^2+[-4-(-1)]^2}\\\\\\ DA=\sqrt{(-6-2)^2+(-4+1)^2} \\\\\\ DA=\sqrt{(-8)^2+(-3)^2}\implies \boxed{DA=\sqrt{73}}$

sum those sides up, and that's the perimeter of the polygon.