Question

Hallar el perimetro del triangulo cuyos vertices son los (-1.-2)(4,2)-3,5)

Answer 1

AB= Xb - Xa = 4 + 1 = 5 BC=√(Xc - Xb)^2 + (Yc -Yb)^2 BC= √(-3 -4)^2 + (5 - 2)^2 BC= √49 + 9 = √58 = 7.6 CA= Ya - Yc = -2 -5= -7 = 7 P= AB + BC + CA P= 5 + 7.6 + 7 P=19.6