I just don't get it...

Check the gradient of the hypotenuse each triangle.

A easier to understand explanation: The hypotenuse of neither large triangle is a straight line. The large figures shown are quadrangles, not right triangles.

Saith, I cut and pasted and there were no tricks. It still came out as the pictures depict. However, looking at the topology of the areas of the blue and purple configuration, in one the area is 3 x 5 = 15 as opposed to 2 x 8 = 16, but the orientation of the triangles, even though their area is the same in both figures, the topology of the geometrical distribution of the areas isn't the same in both cases; the rectangular pieces cannot topologically match in both cases becuase of the surface topology of the two triangles do not leave the same area for the rectangles to fit into: 2x8 opposed to 3x15 is not availabe for insertion of the distributed rectangular poeces in both cases.

Considering the triangles as 1/2(5 x 13) = 65/2 = 32.5 total square area.
Adding the triangles separately 1/2(3 x 8 + 2 x 10) = 12 + 5 = 17
and 32.5 - 17 = 15.5 which leaves the two areas 3 x 5 = 15, or 2 x 8 = 16 and their average is 31/2 = 15.5 added to the aea of the triangles is 17 + 15.2 = 32.5. The difference comes in cutting the large rectangle in half and then producing the triangles. One cannot make two equal triangles with the long side an odd number of units without using part of a square as opposed to producing the same over all area of the two triangles by not using a fraction of a unit square. Instead, the topology is shared by the two rectangular sections that if the unit squares weren't in the same ratio as we see, and instead we had 16 square units as the total triangular area it would all change.
Draw out a 6 x 14 rectangle and cut it in half with a diagonal line.There looks like two maybe three points where the diagonal goes through a corner which would allow two separate triangles of unequal areas, and would allow two separate rectangles of unequal areas by using the unit squares without cutting a fraction of a square when slicing up the rectangles, which you can't do in the case of the triangles.

Me thinks the problem is somewhat illusory because the difference is hidden in the fractional squares determined when the triangles were made by the diagonal line.

I don't know if all I said was all bullshit or not.
Geistkiesel

It seemed intuitively obvious to me that the large composite figures were quadrangles rather than triangles because the two small triangles (red & green) are not similar. Id est: Compare the smallest angles of the two little triangles and note that they are not equal.

If you have some graph paper to help you make a precise drawing you will see the light. Without graph paper it takes a little more work to do the following.

• Draw the two legs of the composite figure without drawing the hypotenuse. The 13 unit leg horizontally & the 5 unit leg vertically on the right.
  
• Make a dot 8 units from the left of the 13-unit leg and 3 units up. Connect that dot with the ends of the two legs.
  
• Make a dot 5 units from the left of the 13 unit leg and 2 units up. Connect that dot with the ends of the two legs.

If you do the above construction accurately, you will see a very slim quadrangle formed by 4 line segments. The hypotenuse of a right triangle with legs 5 & 13 would go through that slim quadrangle, whose area is one square unit.

Perhaps there is somebody here who can post such a construction.

Using Pete’s advice: the gradient or tangent (rise/run) of the smallest (green) triangle is 2/5 (0.400); The gradient or tangent of the medium (red) triangle is 3/8 (0.375).

The drawings imply that there are two large triangles whose gradients or tangents are 5/13 (0.384615). These are actually quadrangles. What appears to be the hypotenuse consists of two line segments, rather than being a straight line.

When the green triangle forms the smaller angle of the composite figure, what appears to be the hypotenuse of the composite figure is a pair of lines convex to (bulging above) the hypotenuse of a triangle with a 5/13 gradient. When the red triangle forms the smaller angle of the composite figure, what appears to be the hypotenuse of the composite figure is a pair of ones concave to (sinking below) the hypotenuse of a triangle with a 5/13 gradient.

Those here who know trigonometry can calculate the angles of the various triangles.

• The green right triangle with legs 2 & 5: 21.801409 degrees.
  
  
• The red triangle with legs 3 & 8: 20.556045 degrees.
  
  
• The composite figures alleged to be triangles with legs 5 & 13: 21.037511 degrees.

I just noticed that the figures shown in the original post to this thread are precise enough to show the discrepancy.

Study the figures in that post. Focus closely on the hypotenuse of the red triangle in each of the composite figures. Count squares to determine where the green triangle would be if drawn on top of each red triangle. In one composite figure the hypotenuse of the red triangle is above the apex of the green triangle. In the other composite figure, it is below.

Yeah, if you rotate the 2nd triangle by 180 degress and place it on top of the 1st triangle you get a rectangle (5X13) with a comibnation of :

1. red rectangle (3X8) at left bottom (total 24 small rectangles)

2. green rectangle (2X5) at right top (total 10 small rectanlges)

3. blue, voilet, and white rectangle at left top (7,8 and 1 small rectangles; total 16 small rectangles)

4. blue & violet rectangle (3X5) at right bottom (7 and 8 small rectangles ; total 15 small rectangles)

So total colored rectangle are 64; one less than total number of rectangles (5X13 = 65) , the remaining 1 being the one with no color.