Final answer:
The perimeter of ∆ABC with vertices at A(2, 8), B(16, 2), and C(6, 2) is approximately 32.44 units, and the area is approximately 866.11 square units, calculated using the distance formula and Heron's formula.
Step-by-step explanation:
To calculate the perimeter and area of ∆ABC with vertices at A(2, 8), B(16, 2), and C(6, 2), we use the distance formula to find the lengths of the sides of the triangle and then apply those values to find the perimeter and area.
The distance formula is √((x2-x1)^2 + (y2-y1)^2), which gives us the lengths of AB, BC, and AC:
- AB: √((16-2)^2 + (2-8)^2) = √(196 + 36) = √232
- BC: √((6-16)^2 + (2-2)^2) = √(100 + 0) = 10
- AC: √((6-2)^2 + (2-8)^2) = √(16 + 36) = √52
The perimeter (P) is AB + BC + AC, and to find the area (A) we could use Heron's formula, which requires the semi-perimeter (s = P/2).
Calculating the perimeter:
P = √232 + 10 + √52 = (√232) + 10 + (√52) ≈ 15.23 + 10 + 7.21 ≈ 32.44 units
For the area, we first calculate the semi-perimeter:
s = P/2 ≈ 32.44/2 ≈ 16.22 units
Then apply Heron's formula:
A = √(s(s-AB)(s-BC)(s-AC)) = √(16.22(16.22-√232)(16.22-10)(16.22-√52))
After calculations:
A ≈ √(16.22(1)(6.22)(9)) ≈ √866.11 square units
Therefore, the perimeter of ∆ABC is approximately 32.44 units and the area is approximately 866.11 square units.