What is the area of the green region?

Let f be the formula for the line that passes trough (0,4) and (10,0), then f(x) = - 0.4x + 4. So the green area = 0.5(f(4)+f(7))*3 = 5.4.

We note that there are four similar triangles. Let the areas from the smallest triangle (blue) to the largest one be $A_1$ , $A_2$ , $A_3$ and $A_4$ respectively. Then $A_4 = \dfrac 12 \times (1+2+3+4)\times 4 = 20$ . For similar shape figures, the area $A$ is directly proportional to the square of a side length $a$ . That is $A \propto a^2$ . Then $\dfrac {A_3}{A_4} = \dfrac {(1+2+3)^2}{(1+2+3+4)^2} = 0.36$ $\implies A_3 = 0.36 A_4 = 7.2$ . Similarly, $A_2 = \dfrac {3^2}{10^2} A_4 = 0.9 \times 20 = 1.8$ . The area of the green region $A_{\color{#20A900}\text{green}} = A_3-A_2 = 7.2-1.8 = \boxed{5.4}$ .

First we need to take the ratio of proportionality: $\frac{4}{1+2+3+4}=0.4$ .

Then we can calculate the sides of the trapezoid of the green figure as indicated in the given picture.

Thus, the area of the green region is: $\frac{(2.4+1.2)\times 3}{2} = \boxed{5.4}$