What is the area of a parallelogram whose vertices are A(−12, 2) , B(6, 2) , C(−2, −3) , and D(−20, −3) ? Enter your answer in the box. units²

Let[tex]A(-12,2)\\B(6,2)\\C(-2,-3) \\D(-20.-3)\\E(-12.-3)[/tex]using a graphing tool see the attached figure to better understand the problemwe know that Parallelogram is a quadrilateral with opposite sides parallel and equal in lengthso[tex]AB=CD \\AD=BC[/tex]The area of a parallelogram is equal to[tex]A=B*h[/tex]where B is the baseh is the height the base B is equal to the distance ABthe height h is equal to the distance AE Step 1Find the distance ABthe formula to calculate the distance between two points is equal to[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex][tex]A(-12,2)\\B(6,2)[/tex]substitute the values[tex]d=\sqrt{(2-2)^{2}+(6+12)^{2}}[/tex][tex]d=\sqrt{(0)^{2}+(18)^{2}}[/tex][tex]dAB=18\ units[/tex]Step 2Find the distance AEthe formula to calculate the distance between two points is equal to[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex][tex]A(-12,2)\\E(-12.-3)[/tex]substitute the values[tex]d=\sqrt{(-3-2)^{2}+(-12+12)^{2}}[/tex][tex]d=\sqrt{(-5)^{2}+(0)^{2}}[/tex][tex]dAE=5\ units[/tex]Step 3Find the area of the parallelogramThe area of a parallelogram is equal to[tex]A=B*h[/tex][tex]A=AB*AE[/tex]substitute the values[tex]A=18*5=90\ units^{2}[/tex]thereforethe answer isthe area of the parallelogram is [tex]90\ units^{2}[/tex]