3. Find the perimeter of quadrilateral ABCD with vortices 4(-2,-2), B(-1,3),c(5, 3), and D(4, -2).

Answer:[tex]12 + 2 \sqrt{26}[/tex]         Step-by-step explanation:To find the perimeter of quadrilateral we need to calculate the length of its four sides which is distance between its four vertices.where [tex]A(x_{1}, y_{1}) = (-2, -2)[/tex],[tex]B(x_{2}, y_{2}) = (-1, 3)[/tex][tex]C(x_{3}, y_{3}) = (5, 3)[/tex][tex]D(x_{4}, y_{4}) = (4, -2)[/tex]Distance of AB = [tex]\sqrt{\left (x^{_{2}}-x^{_{1}} \right )^{2}+\left (y_{2}-y_{1} \right )^{2}}[/tex]= [tex]\sqrt{\left (-1+ 2 \right )^{2}+ (3 + 2)^{^{2}}}[/tex]= [tex]\sqrt{1^{^{2}}+5^{^{2}}}[/tex]=[tex]\sqrt{26}[/tex]Distance of BC = [tex]\sqrt{\left (x_{3}-x_{2} \right )^{2}+ \left (y_{3}-y_{2}\right )^{2}}[/tex]= [tex]\sqrt{(5+1)^{2}+ (3-3)^{2}} = \sqrt{36} = 6[/tex]Distance of CD = [tex]\sqrt{\left (x_{4}-x_{3} \right )^{2}+ \left (y_{4}-y_{3}\right )^{2}[/tex] = [tex]\sqrt{(4-5)^{^{2}}+(-2-3)^{^{2}}} = \sqrt{1 + 25} =\sqrt{26}[/tex]Distance of AD = [tex]\sqrt{\left (x_{4}-x_{1} \right )^{2}+ \left (y_{4}-y_{1} \right )^{2}[/tex]= [tex]\sqrt{(4 + 2)^{_{2}}+ (-2 + 2)^{^{2}}}=\sqrt{36 + 0} = 6[/tex]Perimeter of quadrilateral ABCD = Dist. ( AB + BC + CD + AD) = [tex]\sqrt{26} +6 +\sqrt{26} +6[/tex]=[tex]12 +2 \sqrt{26}[/tex]