Question

The coordinates of the vertices of triangle ABC are A(4,-3), B(2,5), and C(-4,2)? Find the perimeter of triangle ABC to the nearest tenth

Answer 1

Answer: 24.4

Explanation:

`AB=√((4-2)^2 +(-3-5)^2)=√(4+64)=√(68)\\\\BC=√((2-(-4))^2 +(5-2)^2)=√(6^2 +3^2) =√(45)\\\\AC=√((4-(-4))^2 +(-3-2)^2)=√(64+25)=√(89)\\\\\therefore \text{Perimeter }=√(68)+√(45)+√(89) \approx 24.4`