Question

To the nearest tenth, find the perimeter of ∆ABC with vertices A(-1,4), B(-2,1) and C(2,1). Show your work.

PLEASE HELP

Answer 1

A=(-1,4)=(xa,ya)→xa=-1, ya=4
B=(-2,1)=(xb,yb)→xb=-2, yb=1
C=(2,1)=(xc,yc)→xc=2, yc=1

Perimeter of ∆ABC: P=AB+BC+AC

AB=d A-B=sqrt [ (xb-xa)^2+(yb-ya)^2 ]
AB=sqrt [ (-2-(-1))^2+(1-4)^2]
AB=sqrt [ (-2+1)^2+(-3)^2]
AB=sqrt [ (-1)^2+9]
AB=sqrt [ 1+9]
AB=sqrt [10]
AB=3.162277660

BC=d B-C=sqrt [ (xc-xb)^2+(yc-yb)^2 ]
BC=sqrt [ (2-(-2))^2+(1-1)^2]
BC=sqrt [ (2+2)^2+(0)^2]
BC=sqrt [ (4)^2+0]
BC=sqrt [ 16+0]
BC=sqrt [16]
BC=4

AC=d A-C=sqrt [ (xc-xa)^2+(yc-ya)^2 ]
AC=sqrt [ (2-(-1))^2+(1-4)^2]
AC=sqrt [ (2+1)^2+(-3)^2]
AC=sqrt [ (3)^2+9]
AC=sqrt [ 9+9]
AC=sqrt [9*2]
AC=sqrt [9] * sqrt [2]
AC=3 sqrt [2]
AC=3 (1.414213562)
AC=4.242640686

P=AB+BC+AC
P=3.162277660+4+4.242640686
P=11.40491834
To the nearest tenth:
P=11.4

Answer: The perimeter of ∆ABC is 11.4 units