It is easy to prove that RT △ ABC ≌ RT △ ABA1≌ RT△ABB 1 so there are: RT △ ABB1area /Rt△ABA 1 area = (1/2 * aa/. A1b654438bb1) = the square of (ab/a1b) * (ab/a1b) = the square of AC/BC = 5.
Rt △ a1b1a 2/rt △ a2b1b 2 area =25/ 16.
Wait a minute.
So all white areas/shaded areas =25/ 16.
That is, the sum of the areas of all shadow triangles is 16/(25+ 16)* total area =16/41* 6 = 96/41.
It may seem a bit troublesome, but the idea is relatively simple.