Geometry - Area Chasing 5

Recall that the areas of triangles with equal altitudes are proportional to the bases of the triangles. We have

$\dfrac{EB}{EA}=\dfrac{A_{CEB}}{A_{CEA}}=\dfrac{A_{GEB}}{A_{GEA}}$

$\dfrac{b+3+7.5}{a+7+10.5}=\dfrac{b}{a}$

$\dfrac{b+10.5}{a+17.5}=\dfrac{b}{a}$

$10.5a-17.5b=0$ $\color{#D61F06}(1)$

$\dfrac{DA}{DC}=\dfrac{A_{BDA}}{A_{BDC}}=\dfrac{A_{GDA}}{A_{GDC}}$

$\dfrac{a+b+7}{3+7.5+10.5}=\dfrac{7}{10.5}$

$\dfrac{a+b+7}{21}=\dfrac{7}{10.5}$

$10.5a+10.5b=73.5$ $\color{#D61F06}(2)$

Subtracting $\color{#D61F06}(1)$ from $\color{#D61F06}(2)$ , we get

$28b=73.5$

$b=2.625$

It follows that

$10.5a-17.5b=0$

$10.5a-(17.5)(2.625)=0$

$a=4.375$

Finally, the area of the blue region is $4.375+2.625=$ $\boxed{7}$ .