∫a 1b 1‖a2 B2‖a3 B3,A2B 1‖A3B2‖A4B3。
You can get:
△A2B 1B2 is similar to △A3B2B3, that is,
△A2B 1B2∽△A3B2B3
The areas of △A2B 1B2 and △A3B2B3 are 1, 4 respectively.
This leads to:
a2b 1/a3 B2 = b 1 B2/B2 B3 = a2 B2/a3 B3 = 1/2
So there are:
△ area of △A3A4B3/△ area of /△A3B2B3/△ A3B3 * h/A3B2 * h = A4B3/A3B2 = 2/1
(H can represent the height of two triangles, because A3B2 and A4B3 are parallel, so the height is equal. )
So the area of △A3A4B3 =2*4=8.
Similarly, you can get:
Delta area of △A2A3B2 =2* 1=2
Delta area of a1a2b1= 1/2* 1=0.5.
So:
The sum of the areas of the three shadow triangles in the figure =8+2+0.5= 10.5.