∫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.