Question

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



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|__| units²

Answer 1

AB = 15
CD = 15
Distance from AB to CD = 10
the formula of area K= b*h
K= 15*10 = 150

Answer 2

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 $150\ units^{2}$