A trapezoid has coordinates of (-5, -3), (-2, 5), (2, 5), and (5, -3). What is the approximate perimeter of the trapezoid?Round your answer to the nearest whole number (number that is not a decimal).

Answer:The approximate perimeter of the trapezoid is 31 unitsStep-by-step explanation:step 1Plot the trapezoidLetA(-5, -3), B(-2, 5), C(2, 5), and D(5, -3)see the attached figurestep 2Find the perimeter of trapezoidwe know thatThe perimeter of trapezoid is equal to[tex]P=AB+BC+CD+AD[/tex]the formula to calculate the distance between two points is equal to [tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex] Find the distance ABwe have[tex]A(-5, -3),B(-2, 5)[/tex]substitute in the formula[tex]d=\sqrt{(5+3)^{2}+(-2+5)^{2}}[/tex] [tex]d=\sqrt{(8)^{2}+(3)^{2}}[/tex] [tex]d_A_B=\sqrt{73}\ units[/tex] Find the distance BCwe have[tex]B(-2, 5),C(2, 5)[/tex]substitute in the formula[tex]d=\sqrt{(5-5)^{2}+(2+2)^{2}}[/tex] [tex]d=\sqrt{(0)^{2}+(4)^{2}}[/tex] [tex]d_B_C=4\ units[/tex] Find the distance CDwe have[tex]C(2, 5),D(5, -3)[/tex]substitute in the formula[tex]d=\sqrt{(-3-5)^{2}+(5-2)^{2}}[/tex] [tex]d=\sqrt{(-8)^{2}+(3)^{2}}[/tex] [tex]d_C_D=\sqrt{73}\ units[/tex] Find the distance ADwe have[tex]A(-5, -3),D(5, -3)[/tex]substitute in the formula[tex]d=\sqrt{(-3+3)^{2}+(5+5)^{2}}[/tex] [tex]d=\sqrt{(0)^{2}+(10)^{2}}[/tex] [tex]d_A_D=10\ units[/tex] step 3Find the perimeter[tex]P=AB+BC+CD+AD[/tex]substitute the values[tex]P=\sqrt{73}+4+\sqrt{73}+10[/tex][tex]P=31\ units[/tex]