Question

ABC has vertices at A(0,4),B(2,2) and C(–2,–1) what is the perimeter of ABC?

Answer 1

Answer:

Step-by-step explanation:

Using the distance formula,

$AB=\sqrt{(0-2)^{2}+(4-2)^{2}}=2\sqrt{2}\\AC=\sqrt{(0-(-2))^{2}+(4-(-1))^{2}}=\sqrt{29}\\BC=\sqrt{(2-(-2))^{2}+(2-(-1))^{2}}=5$

So the perimeter is $5+2\sqrt{2}+\sqrt{29}$