Answer:
The perimeter of Δ ABC is 20 + 2[tex]\sqrt{10}[/tex] units ⇒ Last answer
Step-by-step explanation:
The perimeter of any triangle is the sum of the lengths of its three sides
The formula of distance between two points is [tex]d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]
In Δ ABC
∵ A = (3 , 4) , B = (-5 , -2) , C = (5 , -2)
∵ AB = 10 units
∵ AC = 2[tex]\sqrt{10}[/tex]
- To find its perimeter find the length of BC
∵ [tex]x_{1}[/tex] = -5 and [tex]y_{1}[/tex] = -2
∵ [tex]x_{2}[/tex] = 5 and [tex]y_{2}[/tex] = -2
- By using the formula above
∴ [tex]BC=\sqrt{(5--5)^{2}+(-2--2)^{2}}=\sqrt{(5+5)^{2}+(-2+2)^{2}}[/tex]
∴ [tex]BC=\sqrt{(10)^{2}+(0)^{2}}=\sqrt{100}[/tex]
∴ BC = 10 units
To find the perimeter add the lengths of the three sides
∵ P = AB + BC + AC
∴ P = 10 + 10 + 2[tex]\sqrt{10}[/tex]
- Add like terms
∴ P = 20 + 2[tex]\sqrt{10}[/tex]
The perimeter of Δ ABC is 20 + 2[tex]\sqrt{10}[/tex] units
