Answer:
Option A. [tex]10\ units^{2}[/tex]
Step-by-step explanation:
we know that
The area of the rectangle is equal to
A=LW
where
L is the length of rectangle
W is the width of rectangle
we have
[tex]A(-3,1),B(-2,-1),C(2,1),D(1,3)[/tex]
Plot the vertices
see the attached figure
L=AD=BC
W=AB=DC
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
Find the distance AD
[tex]A(-3,1),D(1,3)[/tex]
substitute in the formula
[tex]AD=\sqrt{(3-1)^{2}+(1+3)^{2}}[/tex]
[tex]AD=\sqrt{(2)^{2}+(4)^{2}}[/tex]
[tex]AD=\sqrt{20}[/tex]
[tex]AD=2\sqrt{5}\ units[/tex]
Find the distance AB
[tex]A(-3,1),B(-2,-1)[/tex]
substitute in the formula
[tex]AB=\sqrt{(-1-1)^{2}+(-2+3)^{2}}[/tex]
[tex]AB=\sqrt{(-2)^{2}+(1)^{2}}[/tex]
[tex]AB=\sqrt{5}[/tex]
[tex]AB=\sqrt{5}\ units[/tex]
Find the area
[tex]A=(2\sqrt{5})*(\sqrt{5})=10\ units^{2}[/tex]
