Let ABCD be a parallelogram. Draw QN parallel to AB with Q lying on AD and N lying on BC. Draw another line MS parallel to AD with M lying on AB and S lying on QN. Draw yet another line PT parallel to AD with P lying on CD and T lying on QN.
The area of AMSQ, BNSM, CPTN and DQTP are respectively 12, 36, 48, and 24.
What is the area of quadrilateral BSDT?
This is a very neat problem because it fools one (well, it fooled me) into thinking the object in question has a particular shape. As long as you imagine any parallelogram with total area 120, you have an acceptable shape for the problem. For example, any rectangle of area 120 will do. Consider a 5 by 24 rectangle with the 5 on top and bottom cut through with one line QN 2:3 and the sides cut MS 18:6 and PT 16:8, or a 10 x 12 cut top to bottom 4:6 and sides 9:3, 8:4, etc. etc. - all work. The key is that top and bottom are cut 2:3 and the sides 3:1, 2:1 with some scale factor for each.
Next, you can lean the rectangle over to become any parallelogram of height h following the rule that base x h = 120, the specified area. Again the top and bottom are cut in a 2:3 ratio, the left height is cut again 3:1 and the right is cut 2:1. The actual sides are cut with these ratios as well. The four areas will continue to obtain the values given.
Since we are changing shapes by scaling x and y in way that leaves the area element delta x delta y constant, this will leave the quadrilateral area constant as well. So we can compute the quadrilateral area as it appears in the second simple rectangle given above. (An easy choice).
Edited on March 5, 2019, 9:40 pm