Question

The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is

Answer 1

use the distance formula to find the length between points, and then add them all up, like so . . .

L(AB) = √[(6 - -2)² + (2 - 2)²] = 8
L(BC) = √[(0 - 6)² + (8 - 2)²] = 8.4853
L(AC) = √[(0 - -2)² + (8 - 2)²] = 6.3246

8 + 8.4853 + 6.3246 = 14.8099

. . . answer is 14.8099 (rounded to the ten thousandths place)

Answer 2

Perimeter of the ΔABC is 22.815 units .

Explanation:

Formula

`Distance\ formula = \sqrt{(x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2)}`

As given

The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8).

Thus

`AB=\sqrt{(6-(-2))^(2)+(2-2)^(2)}`

`AB=\sqrt{(6+2)^(2)}`

`AB=\sqrt{8^(2)}`

`AB=√(64)`

`√(64)=8`

AB = 8 units

`BC= \sqrt{(0-6)^(2)+(8-2)^(2)}`

`BC= \sqrt{(6)^(2)+(6)^(2)}`

`BC= √(36+36)`

`BC= √(72)`

`√(72) = 8.49\ (Approx)`

BC = 8.49 units

`CA = \sqrt{(-2-0)^(2)+(2-8)^(2)}`

`CA = \sqrt{(-2)^(2)+(-6)^(2)}`

`CA = √(4+36)`

`CA = √(40)`

`√(40) =6.325 \ (Approx)`

CA = 6.325 units

Total perimeter of ΔABC = AB +BC + CA

= 8 + 8.49 + 6.325

= 22.815 units

Therefore the perimeter of the ΔABC is 22.815 units .