What is the perimeter of a triangle with vertices located at (1 3) (26) and (0 4) rounded to the nearest hundredth?

Perimeter of the triangle is ##7.40## unit
Side##AB=sqrt((1-2)^2+(3-6)^2)=sqrt10=3.16##
Side##BC=sqrt((2-0)^2+(6-4)^2)=sqrt8=2.83##
Side##CA=sqrt((0-1)^2+(4-3)^2)=sqrt2=1.41##
Perimeter of the triangle is ##3.16+2.83+1.41 =7.40##unit[Ans]