Calc the perimeter of a triangle with vertices at (0, 1), (7, 1), and (3, 5). use a reduced radical.

Question

do not understand this

Answer 1

Let A(0,1), B(7,1), C(3,5)

Assume ABC is a triangle.

Find the distances of sides of triangle.

Distance between twopoints = √[(x2-x1)^2+(y2-y1)^2]

AB = √[7-0)^2+(1-1)^2]

= 7

BC = √[(3-7)^2+(5-1)^2]

= √(16+16)

= √32

CA = √[(0-3)^2+(1-5)^2

= √(9+16)

= 5

Perimeter  = AB+BC+CA = 7+√32+5

= 12+√32

= 12+√(16*2)

= 12+4√2

Perimeter= 12+4√2