Question

what is the length of the longest side of a triangle that has the vertices (2, 6), (-4, 6), and (-4, 4)? a. 2/10 units b. 40 units c. /10 units d. 4/10 units

Answer 1

Length (2, 6) to (-4, 6) is sqrt((x2 - x1))^2 + (y2 - y1)^2) = sqrt((-4 -2)^2 + (6 - 6)^2) = sqrt((-6)^2 + 0) = 6

Length (2, 6) to (-4, 4) is sqrt((-4 - 2)^2 + (4 - 6)^2) = sqrt((-6)^2 + (-2)^2) = sqrt(36 + 4) = sqrt(40) = 2sqrt(10) units

Length (-4, 6) to (-4, 4) is sqrt((-4 - (-4))^2 + (4 - 6)^2) = sqrt(0^2 + (-2)^2) = 2

Therefore, the length of the longest side is 2sqrt(10) units