MATH SOLVE

5 months ago

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
[tex]A(-4,9)\\B(11,9)\\C(5,-1) \\D(-10,-1)\\E(-4.-1)[/tex] 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
[tex]AB=CD \\AD=BC[/tex]
The area of a parallelogram is equal to
[tex]A=B*h[/tex] 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
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
[tex]A(-4,9)\\B(11,9)[/tex] substitute the values
[tex]d=\sqrt{(9-9)^{2}+(11+4)^{2}}[/tex]
[tex]d=\sqrt{(0)^{2}+(15)^{2}}[/tex]
[tex]dAB=15\ units[/tex]
Step 2
Find the distance AE
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
[tex]A(-4,9)\\E(-4.-1)[/tex] substitute the values
[tex]d=\sqrt{(-1-9)^{2}+(-4+4)^{2}}[/tex]
[tex]d=\sqrt{(-10)^{2}+(0)^{2}}[/tex]
[tex]dAE=10\ units[/tex]
Step 3
Find the area of the parallelogram
The area of a parallelogram is equal to
[tex]A=B*h[/tex]
[tex]A=AB*AE[/tex]
substitute the values
[tex]A=15*10=150\ units^{2}[/tex]
therefore
the answer is
the area of the parallelogram is [tex]150\ units^{2}[/tex]