Let (x, y) be the coordinates of D and (x',y') be thee coordinates of E. since, the diagonals of a parallelogram bisect each other at the same point, therefore
(x+8)/2 = (6+9)/2
x = 7
(y+2)/2 = (1+4)/2
y=3
Thus, the coordinates of D are (7,3)
E is the midpoint of DC, therefore
x' = (7+9)/2
x'=8
y' = (3+4)/2
y'=7/2
Thus, the coordinates of E are (8,7/2)
Let A (x1,y1) = A (6,1),E(x2,y2) = E(8,7/2) and D (x3,y3) = D(7,3) Now
Area (ΔABC) = 1/2[x1(y2 -y3)+x2(y3 -y1) + x3(y1- y2)]
=1/2[6(7/2 - 3) + 8(3 - 1) + 7(1 - 7/2)]
=1/2[3/2]
=3/4sq. unit
Hence, the area of the treangle ΔADE is 3/4sq. unit
