Triangle Cascade

With $BC || DE || FG$ , we can conclude that $\triangle ABC \sim \triangle ADE \sim \triangle AFG$ .

Suppose the ratio of $\dfrac{DE}{BC} = x$ and that of $\dfrac{FG}{BC} = y$ for some positive numbers $x,y$ .

We can then set up equations of the known area ratios:

$x(x-1)(9) = 10$

$y(y-x)(9) = 24$

Solving for $x$ , we will obtain:

$9x^2 -9x -10 = 0$

$(3x-5)(3x+2) = 0$

Thus, $x = \dfrac{5}{3}$ . Substituting this into the second equation, we will obtain:

$y(3y - 5) = 8$

$3y^2 - 5y - 8 = 0$

$(3y-8)(y+1) = 0$

Thus, $y = \dfrac{8}{3}$ .

Now we can calculate the area of $\triangle BCD$ = $9x^2 - 10 - 9 = 25-19 = 6$ .

Then the area of $\triangle DEF$ = $9y^2 - 9x^2 - 24 = 64-25-24 = 15$ .

Finally, the total yellow area = $15+6 = \boxed{21}$ .

We note that $\triangle ABC$ , $\triangle ADE$ , and $\triangle AFG$ are similar. Let $BC = x$ , $DE=y$ , and $FG = z$ . Then the ratio of the three similar triangles is $A_{\color{#D61F06}ABC} : A_{ADE} : A_{AFG} = x^2 : y^2:z^2$ . We can also consider the ratios of areas of the red, green, and blue triangles as follows:

$\begin{aligned} \frac {y(y-x)}{x^2} & = \frac {A_{\color{#20A900}CDE}}{A_{\color{#D61F06}ABC}} = \frac {10}9 \\ \left(\frac yx \right)^2 - \frac yx - \frac {10}9 & = 0 \\ \left(\frac yx - \frac 53\right)\left(\frac yx + \frac 23\right) & = 0 \\ \implies \frac yx & = \frac 53 & \small \color{#3D99F6} \text{Since }\frac yx > 0 \end{aligned}$

Similarly,

$\begin{aligned} \frac {z(z-y)}{x^2} & = \frac {A_{\color{#3D99F6}EFG}}{A_{\color{#D61F06}ABC}} = \frac {24}9 \\ \left(\frac zx \right)^2 - \frac {yz}{x^2} & = \frac 83 \\ \left(\frac zx \right)^2 - \frac 53 \frac zx - \frac 83 & = 0 \\ \left(\frac zx - \frac 83\right)\left(\frac yx + 1\right) & = 0 \\ \implies \frac zx & = \frac 83 & \small \color{#3D99F6} \text{Since }\frac zx > 0 \end{aligned}$

Now, we have $\dfrac {A_{ADE}}{A_{\color{#D61F06}ABC}} = \left(\dfrac 53\right)^2 \implies \dfrac {A_{\color{#CEBB00}BCD}+{\color{#D61F06}9}+\color{#20A900}10}{\color{#D61F06}9} = \dfrac {25}9 \implies A_{\color{#CEBB00}BCD} = 6$

Similarly, $\dfrac {A_{AFG}}{A_{\color{#D61F06}ABC}} = \left(\dfrac 83\right)^2 \implies \dfrac {A_{\color{#CEBB00}DEF}+A_{ADE}+\color{#3D99F6}24}{\color{#D61F06}9} = \dfrac {A_{\color{#CEBB00}DEF}+25+\color{#3D99F6}24}{\color{#D61F06}9} = \dfrac {64}9 \implies A_{\color{#CEBB00}DEF} = 15$

Therefore $A_{\color{#CEBB00}BCD} + A_{\color{#CEBB00}DEF} = 6 + 15 = \boxed{21}$ .