Let
[tex]A(-12,2)\\B(6,2)\\C(-2,-3) \\D(-20.-3)\\E(-12.-3)[/tex]
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
[tex]AB=CD \\AD=BC[/tex]
The area of a parallelogram is equal to
[tex]A=B*h[/tex]
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
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
[tex]A(-12,2)\\B(6,2)[/tex]
substitute the values
[tex]d=\sqrt{(2-2)^{2}+(6+12)^{2}}[/tex]
[tex]d=\sqrt{(0)^{2}+(18)^{2}}[/tex]
[tex]dAB=18\ units[/tex]
Step 2
Find the distance AE
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
[tex]A(-12,2)\\E(-12.-3)[/tex]
substitute the values
[tex]d=\sqrt{(-3-2)^{2}+(-12+12)^{2}}[/tex]
[tex]d=\sqrt{(-5)^{2}+(0)^{2}}[/tex]
[tex]dAE=5\ units[/tex]
Step 3
Find the area of the parallelogram
The area of a parallelogram is equal to
[tex]A=B*h[/tex]
[tex]A=AB*AE[/tex]
substitute the values
[tex]A=18*5=90\ units^{2}[/tex]
therefore
the answer is
the area of the parallelogram is [tex]90\ units^{2}[/tex]
The area of the parallelogram is [tex]\boxed{\bf 90\text{\bf square units}}[/tex].
Further explanation:
Given:
The vertices of a parallelogram are [tex]\text{A}(-12,2)[/tex], [tex]\text{B}(6,2)[/tex], [tex]\text{C}(-2,-3)[/tex], and [tex]\text{D}(-20,-3)[/tex] as shown in attached Figure 1.
Calculation:
As we know that a parallelogram is a quadrilateral having opposite sides parallel and equal in length such that [tex]\text{AB}=\text{CD}[/tex] and [tex]\text{AD}=\text{BC}[/tex].
So the area of a parallelogram is calculated as follows:
[tex]\boxed{\text{A}=\text{b}\cdot \text{h}}[/tex]
Where [tex]\text{b}[/tex] is the base and [tex]\text{h}[/tex] is the height.
Draw perpendicular from [tex]\text{A}[/tex] on [tex]\text{CD}[/tex] which meets at point [tex]\text{E}[/tex] as shown in attached Figure 2.
So we have to find the distance [tex]\text{AB}[/tex] and [tex]\text{AD}[/tex] by the formula as shown below:
[tex]\boxed{d=\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}}[/tex]
Where, [tex]d[/tex] is the distance between the points and [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex].
The point of [tex]\text{A}[/tex] and [tex]\text{B}[/tex] are [tex](-12, 2)[/tex] and [tex](6, 2)[/tex] respectively and distance [tex]\text{AB}[/tex] is calculated as, shown below:
[tex]\begin{aligned}\text{AB}&=\sqrt{(2-2)^{2}+(6-(-12))^{2}}\\&=\sqrt{0^{2}+18^{2}}\\&=\sqrt{18^{2}}\\&=18\end{aligned}[/tex]
The coordinates of [tex]A[/tex] and [tex]E[/tex] are [tex](-12,2)[/tex] and [tex](-12,-3)[/tex] respectively and distance [tex]\text{AE}[/tex] is calculated as shown below:
[tex]\begin{aligned}\text{AE}&=\sqrt{(-3-2)^{2}+(-12-(-12))^{2}}\\&=\sqrt{(-5)^{2}+0^{2}}\\&=5\end{aligned}[/tex]
Since, [tex]\text{AB}=\text{CD}[/tex] which is [tex]18[/tex] units and [tex]\text{CD}[/tex] is the base and [tex]\text{AE}[/tex] is the height which is [tex]5[/tex] units as shown in attached figure 2.
Then the area of parallelogram is calculated as shown below:
[tex]\begin{aligned}\text{Area}&=\text{Base}\cdot \text{Height}\\&=\text{CD}\cdot \text{AE}\\&=18\cdot 5\\&=90\end{aligned}[/tex]
Therefore, area of the parallelogram is [tex]\boxed{\bf 90\text{\bf square units}}[/tex].
Learn more:
1. Learn more about the quantity in meters expressed in centimeters https://brainly.com/question/2867118
2. Learn more about the 3 digits in the units period https://brainly.com/question/558692
3. Learn more about the definition of an angle uses the undefined term https://brainly.com/question/3413207
Answer detail:
Grade: Middle school
Subject: Mathematics
Chapter: Coordinate geometry
Keywords: parallelogram, A(−12,2), B(6,2), C(−2,−3), and D(−20,−3), 90 square units, area, height, base, Coordinate geometry, vertices, perpendicular.
