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 ...