The vertices of triangle ABC are A(5, -2), B(9, 4), and C(-3, 8). Find the perimeter of the mid-segment triangle. Round your final answer to the nearest tenth.

Answer 1

find the vertices of the inside triangle and calculate distances of the sides

mid-points of the 3 sides of the triangle are (1,3), (3,6), and (7,1)

they form a triangle with sides of lengths sqr40, sqr13 and sqr41

they sum to about 16.3 = perimeter of the inside triangle

find the mid points by averaging the x coordinates and average the y coordinates

(3,6) = ((9-3)/2, (8+4)/2) = (6/2, 12/2)

(1,3) = ((5-3)/2, (8-2)/2) = (2/2, 6/2)

(7,1) = (9+5)/2, (4-2)/2 = (14/2, 2/2)

distance squared = sum of differences squared of the coordinates

for (3,6) to (1,3) d^2 = 4+9 = 13, d=sqr13

for (3,6) to (7,1), d^2 = 16+25, d =sqr41

for (7,1) to (1,3) d^2 = 36+4 d=sqr40