My approach here was :
<ACB=30 . So, AB=8
Area of whole parallellogram = 16*8 = 128
AC is the diagonal. ( drawn from opposite ends) So it divides the parallelograms into two equal halves.
Area of each half = 128/2 = 64
i.e. Area of triangle ACD = 64
i.e. (1/2) * AD * CH = 64
i.e. 1/2 * 16 * CH =64
Therefore CH =8 and Quantity A is greater. But answer is B.
16 isn’t the height?
Flip triangle ACD horizontally counter-clockwise . 16 becomes the base and CH becomes the height.
@Leaderboard is referring to the height of the parallelogram.
This part is not correct