Answer:
Its perimeter is 23 units
Step-by-step explanation:
The perimeter of any figure is the sum of the lengths of its outline sides
The rule of the distance between two points is:
- [tex]d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]
In the given figure ABCD:
Its perimeter = AB + BC + CD + DE + EA
→ Find the length of each side using the rule above
A = (-4, -2), B = (-1, 2), C = (2, 2), D = (5, -1), E = (2, -4)
→ Substitute them in the rule above to find the lengths of its sides
[tex]AB=\sqrt{(-1--4)^{2}+(2--2)^{2}}=\sqrt{(-1+4)^{2}+(2+2)^{2}}\\\\=\sqrt{(3)^{2}+(4)^{2}}=\sqrt{9+16}=\sqrt{25}[/tex]
∴ AB = 5
[tex]BC=\sqrt{(2--1)^{2}+(2-2)^{2}}=\sqrt{(2+1)^{2}+(0)^{2}}\\\\=\sqrt{(3)^{2}+0}=\sqrt{9+0}=\sqrt{9}[/tex]
∴ BC = 3
[tex]CD=\sqrt{(5-2)^{2}+(-1-2)^{2}}=\sqrt{(3)^{2}+(-3)^{2}}\\\\=\sqrt{9+9}=\sqrt{18}[/tex]
∴ CD = [tex]\sqrt{18}[/tex]
[tex]DE=\sqrt{(2-5)^{2}+(-4--1)^{2}}=\sqrt{(-3)^{2}+(-4+1)^{2}}\\\\=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{9+9}=\sqrt{18}[/tex]
∴ DE = [tex]\sqrt{18}[/tex]
[tex]EA=\sqrt{(-4-2)^{2}+(-2--4)^{2}}=\sqrt{(-6)^{2}+(-2+4)^{2}}\\\\=\sqrt{(-6)^{2}+(2)^{2}}=\sqrt{36+4}=\sqrt{40}[/tex]
∴ EA = [tex]\sqrt{40}[/tex]
→ Add them to find the perimeter of the figure ABCDE
∴ Its perimeter = 5 + 3 + [tex]\sqrt{18}[/tex] + [tex]\sqrt{18}[/tex] +[tex]\sqrt{40}[/tex] ≅ 22.8098
→ Round it to the whole number
∴ Its perimeter = 23 units
