Respuesta :
Answer:
The area of the rectangle is 20 sq .units.
Given:
(-8, -2), (-3,-2), (-3,-6), and (-8, -6)
Solution:
The area of the rectangle ‘A’ is given by the formula:
Area = Length × Width
Now, we have to find the sides of the rectangle.
The sides of the rectangle include s1, s2, s3, and s4.
Let’s now assume the points as:
[tex](x1, y1) = (-8, -2)[/tex]
[tex](x2, y2) = (-3,-2)[/tex]
[tex](x3, y3) = (-3,-6)[/tex]
[tex](x4, y4) = (-8, -6)[/tex]
The side s1 is:
[tex]s 1=\sqrt{(x 2-x 1)^{2}+(y 2-y 1)^{2}}[/tex]
On substituting the values,
[tex]\Rightarrow s 1=\sqrt{(-3+8)^{2}+(-2+2)^{2}}[/tex]
[tex]\Rightarrow s 1=\sqrt{(5)^{2}+(0)^{2}}[/tex]
[tex]\Rightarrow s 1=\sqrt{25}[/tex]
[tex]\therefore s 1=5 \text { units }[/tex]
The side s2 is:
[tex]s 2=\sqrt{(x 3-x 2)^{2}+(y 3-y 2)^{2}}[/tex]
On substituting the values,
[tex]\Rightarrow s 2=\sqrt{(-3+3)^{2}+(-6+2)^{2}}[/tex]
[tex]\Rightarrow s 2=\sqrt{(0)^{2}+(4)^{2}}[/tex]
[tex]\Rightarrow s 2=\sqrt{16}[/tex]
[tex]\therefore s 2=4 \text { units }[/tex]
The side s3 is:
[tex]s 3=\sqrt{(x 4-x 3)^{2}+(y 4-y 3)^{2}}[/tex]
On substituting the values,
[tex]\Rightarrow s 3=\sqrt{(-8+3)^{2}+(-6+6)^{2}}[/tex]
[tex]\Rightarrow s 3=\sqrt{(5)^{2}+(0)^{2}}[/tex]
[tex]\Rightarrow s 3=\sqrt{25}[/tex]
[tex]\therefore s 3=5 \text { units }[/tex]
The side s4 is:
[tex]s 4=\sqrt{(x 1-x 4)^{2}+(y 1-y 4)^{2}}[/tex]
On substituting the values,
[tex]\Rightarrow s 4=\sqrt{(-8+8)^{2}+(-2+6)^{2}}[/tex]
[tex]\Rightarrow s 4=\sqrt{(0)^{2}+(4)^{2}}[/tex]
[tex]\Rightarrow s 4=\sqrt{16}[/tex]
[tex]\therefore s 4=4 \text { units }[/tex]
Now, the length of the given rectangle is 5 units and width of the given rectangle is 4 units.
The area of the rectangle is:
[tex]\Rightarrow A r e a=4 \times 5[/tex]
[tex]\therefore A r e a=20 \ s q . \text { units }[/tex]
