дан треугольник ABC:если A(4;0),B(12;-2)C(5;-9)Найти пиримитр ABC. Найти пиримитр ABC

ответ
пЕриметр
x (AB) = x(B) - x(A) = 12 - 4 = 8
y (AB) = y(A) + /x(B)/ = 0 +/-2/ = 2
AB^2 = {x (AB)}^2 + {y(AB)}^2 = 8^2 + 2^2 = 64 + 4 = 68 = (8,246... )^2= 8,2^2
AB = 8,2
x (BC) = x(B) - x(C) = 12 - 5 = 7
y (BC) = /y(C)/ - /y(B)/ = /-9/ - /-2/ = 7
BC^2 = {x(BC)}^2 + {y(BC)} = 7^2 + 7^2 = 98 = (9,899...)^2 = 9,9^2
BC = 9,9
x(AC) = x(C0 - x(A) = 5 - 4 = 1
y(AC) = /y(C)/ - /y(A)/ = /-9/ - 0 = 9
AC^2 = {x(AC)}^2 + {y(AC)}^2 = 1^2 + 9^2 = 82 = (9,055...)^2 = 9,0^2
AC = 9,0
P = AB + BC + AC = 8,2 + 9,9 + 9,0 = 27,1