Given the vertices of the parallelogram:
(3, 3), (7, 3), (2, 1), (6, 1)
Let's find the area of the parallelogram.
We have the parallelogram below:
To find the area of the parallelogram, apply the formula:
Area = base x height.
Let's find the length using the distance formula:
[tex]b=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}^[/tex]
Where:
(x1, y1) ==> (3, 3)
(x2, y2) ==> (7, 3)
Thus, we have:
[tex]\begin{gathered} b=\sqrt{(3-3)^2+(7-3)^2} \\ \\ b=\sqrt{0^2+4^2} \\ \\ b=4 \end{gathered}[/tex]
The length of the parallelogram is 4 units.
Also, let's find the width using the distance formula:
(x1, y1) ==> (3, 3)
(x2, y2) ==> (2, 1)
[tex]\begin{gathered} W=\sqrt{(1-3)^2+(2-3)^2} \\ \\ W=\sqrt{(-2)^2+(-1)^2} \\ \\ W=\sqrt{4+1} \\ \\ W=\sqrt{5} \end{gathered}[/tex]
The width of the parallelogram is √5 units.
Distance from x to y = 1 unit.
To find the height, apply Pythagorean theorem:
[tex]\begin{gathered} h=\sqrt{(\sqrt{5})^2-1^2} \\ \\ h=\sqrt{5-1} \\ \\ h=\sqrt{4} \\ h=2 \\ \end{gathered}[/tex]
The height of the parallelogram is 2 units.
To find the area, we have:
Area = base x height
Area = 4 x 2
Area = 8 square units.
Therefore, the area of the parallelogram is 8 square units.
ANSWER:
C. 8 sq. units.
