Let the height of triangle ABC be h.
[asy]
pair A, B, C, D, E;
A = (0,0);
B = (16,0);
C = (8,h);
D = (0,h-5);
E = (16,h-11);
draw(A--B--C--cycle);
draw(D--E);
draw(A--D,dashed);
draw(B--E,dashed);
label("A", A, SW);
label("B", B, SE);
label("C", C, N);
label("D", D, SW);
label("E", E, SE);
label("5", (A+D)/2, W);
label("11", (B+E)/2, E);
label("h", (C+D)/2, W);
label("h−5", (C+E)/2, E);
[/asy]
Since the shaded portions have the same area, we have the following equation:
(1/2) * 11 * h = (1/2) * (h - 5) * 16
Solving for h, we get:
h = 15
Therefore, the height of triangle ABC is 15 units.
TYSM!! @Alan! It's correct. Also, thanks for leaving some of the work for me to solve. Really appreciate it.