Mathe

Umfang des Dreiecks ABC 

A(2/1); B(5/4); C(3/5)

Hallo Gast!

$U=a+b+c\\ \ a\ =\ \sqrt{(y_B-y_C)^2+(x_B-x_C)^2}=\sqrt{(4-5)^2+(5-3)^2}=2,236\\ +b=\sqrt{(y_A-y_C)^2+(x_A-x_C)^2}=\sqrt{(1-5)^2+(2-3)^2}=4,123\\ +\underline{c=\sqrt{(y_A-y_B)^2+(x_A-x_B)^2}=\sqrt{(1-4)^2+(2-5)^2}=4,243}\\ \color{blue}\text{ U = a + b + c = 10,602}$

  !