Given:
The vertices of the rectangle ABCD are A(0,1), B(2,4), C(6,0), D(4,-3).
To find:
The area of the rectangle.
Solution:
Distance formula:
[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using the distance formula, we get
[tex]AB=\sqrt{(2-0)^2+(4-1)^2}[/tex]
[tex]AB=\sqrt{(2)^2+(3)^2}[/tex]
[tex]AB=\sqrt{4+9}[/tex]
[tex]AB=\sqrt{13}[/tex]
Similarly,
[tex]BC=\sqrt{(6-2)^2+(0-4)^2}[/tex]
[tex]BC=\sqrt{(4)^2+(-4)^2}[/tex]
[tex]BC=\sqrt{16+16}[/tex]
[tex]BC=\sqrt{32}[/tex]
[tex]BC=4\sqrt{2}[/tex]
Now, the length of the rectangle is [tex]AB=\sqrt{13}[/tex] and the width of the rectangle is [tex]BC=4\sqrt{2}[/tex]. So, the area of the rectangle is:
[tex]A=length \times width[/tex]
[tex]A=\sqrt{13}\times 4\sqrt{2}[/tex]
[tex]A=4\sqrt{26}[/tex]
[tex]A\approx 20[/tex]
Therefore, the area of the rectangle is 20 square units.
