In ΔABC shown below, point A is at (0, 0), point B is at (x2, 0), point C is at (x1, y1), point D is at x sub 1 over 2, y sub 1 over 2, and point E is at the quantity of x sub 1 plus x sub 2 over 2, y sub 1 over 2: Triangle ABC is shown. Point D lies on segment AC and point E lies on segment BC. A segment is drawn between points D and E. Point A is at the origin. Prove that segment DE is parallel to segment AB.

option: d is correct

step-by-step explanation:

18

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring).   so find the distance between (1,2) and (4,2).   that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3.   it is a regular polygon which means all sides are congruent.   you have 6 sides (hexa) so multiply 6 by 3 to receive 18.

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3