Jake draws Δ
ABC on the coordinate plane.
Triangle A B C on a coordinate grid with vertices as follows. A at (negative 4, 2), B at (negative 1, 7), and C at (2, 2).
What is the approximate perimeter of Δ
ABC to the nearest hundredth?
a 5.83 units
b 17.66 units
c 7.62 units
d 21.23 units
To find the perimeter of triangle ABC, we need to calculate the length of each side using the distance formula:
Side AB: √[(-1 - (-4))^2 + (7 - 2)^2] = √[3^2 + 5^2] = √(9 + 25) = √34 ≈ 5.83 units
Side BC: √[(2 - (-1))^2 + (2 - 7)^2] = √[3^2 + 5^2] = √(9 + 25) = √34 ≈ 5.83 units
Side AC: √[(2 - (-4))^2 + (2 - 2)^2] = √[6^2 + 0] = √36 = 6 units
Perimeter = AB + BC + AC ≈ 5.83 + 5.83 + 6 ≈ 17.66 units
Therefore, the approximate perimeter of triangle ABC is 17.66 units, which is option b.