Question

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

Answer 1

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)`