The perimeter of a triangle is the sum of the lengths of all sides.
Using the distance formula \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) calculate the lengths of the sides and then add them up.
\(A=(1,3) \\
B=(2,6) \\
C=(0,4) \\ \\
\overline{AB}=\sqrt{(2-1)^2+(6-3)^2}=\sqrt{1^2+3^2}=\sqrt{1+9}=\sqrt{10} \\
\overline{AC}=\sqrt{(0-1)^2+(4-3)^2}=\sqrt{(-1)^2+1^2}=\sqrt{1+1}=\sqrt{2} \\
\overline{BC}=\sqrt{(0-2)^2+(4-6)^2}=\sqrt{(-2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{4 \times 2}= \\
=2\sqrt{2} \\ \\
P=\overline{AB}+\overline{AC}+\overline{BC}=\sqrt{10}+\sqrt{2}+2\sqrt{2}=\sqrt{10}+3\sqrt{2} \approx 7.40\)
The answer is A.
