The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is
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:
Perimeter of the ΔABC is 22.815 units .
Step-by-step 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
\(AB=\sqrt{(6-(-2))^{2}+(2-2)^{2}}\)
\(AB=\sqrt{(6+2)^{2}}\)
\(AB=\sqrt{8^{2}}\)
\(AB=\sqrt{64}\)
\(\sqrt{64}=8\)
AB = 8 units
\(BC= \sqrt{(0-6)^{2}+(8-2)^{2}}\)
\(BC= \sqrt{(6)^{2}+(6)^{2}}\)
\(BC= \sqrt{36+36}\)
\(BC= \sqrt{72}\)
\(\sqrt{72} = 8.49\ (Approx)\)
BC = 8.49 units
\(CA = \sqrt{(-2-0)^{2}+(2-8)^{2}}\)
\(CA = \sqrt{(-2)^{2}+(-6)^{2}}\)
\(CA = \sqrt{4+36}\)
\(CA = \sqrt{40}\)
\(\sqrt{40} =6.325 \ (Approx)\)
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 .