Type the correct answer in each box. if necessary, round your answers to the nearest hundredth. the vertices of ∆abc are a(2, 8), b(16, 2), and c(6, 2). the perimeter of ∆abc is (blank) units, and its area is (blank) square units.

The perimeter of ∆ABC is 32.44 units, and its area is 30 square units

Step-by-step explanation:

The perimeter is the sum of three distances.

To find the distances  BC, CD, DB use the distance formula:

D=√(y2-y1)^2+(x2-x1)^2

We have given A(2, 8), B(16, 2), and C(6, 2).

AB= √(16-2)^2+(2-8)^2

AB= √(14)^2+(-6)^2

AB=√196+36

AB=√232 = 15.23

BC=√(6-16)^2+(2-2)^2

BC=√(-10)^2+0

BC=√100 = 10

CA = √(2-6)^2+(8-2)^2

CA=√(-4)^2+(6)^2

CA=√16+36

CA=√52 = 7.21

Perimeter = AB+BC+CA

Perimeter = 15.23+10+7.21

Perimeter = 32.44 units

For the area BC is the parallel to x-axis

area = (1/2)base * height

=(1/2)10*(ya-yb)

=(1/2)10*(8-2)

=(1/2)10*6

=30 unit²

The perimeter of ∆ABC is 32.44 units, and its area is 30 square units

i have no clue at all sorry

step-by-step explanation: