Q:

# What is the area of a parallelogram whose vertices are A(−4, 9) , B(11, 9) , C(5, −1) , and D(−10, −1) ?

Accepted Solution

A:
Let $$A(-4,9)\\B(11,9)\\C(5,-1) \\D(-10,-1)\\E(-4.-1)$$  using a graphing tool  see the attached figure to better understand the problem we know that Parallelogram is a quadrilateral with opposite sides parallel and equal in length so $$AB=CD \\AD=BC$$ The area of a parallelogram is equal to $$A=B*h$$  where  B is the base h is the height  the base B is equal to the distance AB the height h is equal to the distance AE  Step 1 Find the distance AB the formula to calculate the distance between two points is equal to $$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$$ $$A(-4,9)\\B(11,9)$$  substitute the values $$d=\sqrt{(9-9)^{2}+(11+4)^{2}}$$ $$d=\sqrt{(0)^{2}+(15)^{2}}$$ $$dAB=15\ units$$ Step 2 Find the distance AE the formula to calculate the distance between two points is equal to $$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$$ $$A(-4,9)\\E(-4.-1)$$  substitute the values $$d=\sqrt{(-1-9)^{2}+(-4+4)^{2}}$$ $$d=\sqrt{(-10)^{2}+(0)^{2}}$$ $$dAE=10\ units$$ Step 3 Find the area of the parallelogram The area of a parallelogram is equal to $$A=B*h$$ $$A=AB*AE$$ substitute the values $$A=15*10=150\ units^{2}$$ therefore the answer is the area of the parallelogram is $$150\ units^{2}$$