Triangle ABC has vertices at (-4,0) , (-1, 6) and (3,-1) perimeter of triangle ABC, rounded to the nearest tenth ?

Answer:The perimeter of triangle ABC is [tex]P=21.8\ units[/tex]Step-by-step explanation:The perimeter of triangle ABC is equal to[tex]P=AB+BC+AC[/tex]the formula to calculate the distance between two points is equal to [tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex] we have[tex]A(-4,0),B(-1, 6),C(3,-1)[/tex]step 1Find the distance AB[tex]A(-4,0),B(-1, 6)[/tex]substitute in the formula[tex]d_A_B=\sqrt{(6-0)^{2}+(-1+4)^{2}}[/tex] [tex]d_A_B=\sqrt{(6)^{2}+(3)^{2}}[/tex] [tex]d_A_B=\sqrt{45}\ units[/tex] step 2Find the distance BC[tex]B(-1, 6),C(3,-1)[/tex]substitute in the formula[tex]d_B_C=\sqrt{(-1-6)^{2}+(3+1)^{2}}[/tex] [tex]d_B_C=\sqrt{(-7)^{2}+(4)^{2}}[/tex] [tex]d_B_C=\sqrt{65}\ units[/tex] step 3Find the distance AC[tex]A(-4,0),C(3,-1)[/tex]substitute in the formula[tex]d_A_C=\sqrt{(-1-0)^{2}+(3+4)^{2}}[/tex] [tex]d_A_C=\sqrt{(-1)^{2}+(7)^{2}}[/tex] [tex]d_A_C=\sqrt{50}\ units[/tex] step 4Find the perimeter[tex]P=AB+BC+AC[/tex]substitute the values[tex]P=\sqrt{45}+\sqrt{65}+\sqrt{50}=21.8\ units[/tex]