Another Pointy Quadrilateral

As an alternative approach...

Draw a vertical from $G$ to the extension of $EH$ and let the point of intersection be $A$ , Next, draw a horizontal from $G$ to the extension of $EF$ and call the point of intersection $B$ . Now let $|HA| = x$ and $|FB| = y$ . Then by Pythagoras we have that

$(6 + x)^{2} + y^{2} = 13^{2} \Longrightarrow x^{2} + 12x + y^{2} = 133$ , and

$x^{2} + (y + 3)^{2} = 10^{2} \Longrightarrow x^{2} + 6y + y^{2} = 91$ .

Upon subtracting the second from the first of these equations we have that $12x - 6y = 42 \Longrightarrow y = 2x - 7$ .

Substituting this result into $x^{2} + (y + 3)^{2} = 10^{2}$ and simplifying yields that

$x^{2} + (2x - 4)^{2} = 100 \Longrightarrow 5x^{2} - 16x - 84 = (5x + 14)(x - 6) = 0$ ,

from which we can conclude that $x = 6$ , (as we must have $x \gt 0$ ), and thus $y = 2x - 7 = 5$ .

The green area will then be the area of rectangle $EBGA$ minus the areas of triangles $\Delta HAG$ and $\Delta FBG$ , which comes out to

$(x + 6)(y + 3) - \dfrac{x(y + 3)}{2} - \dfrac{y(x + 6)}{2} = 12 \times 8 - 3 \times 8 - 5 \times 6 = \boxed{42}$ .

The hypotenuse $FH=\sqrt{3^2+6^2}=3\sqrt{5}$

Half perimeter of $FGH$ is then $s=\frac12(3\sqrt{5}+23)$

Heron's formula for the area $[FGH]=\sqrt{s(s-3\sqrt{5})(s-10(s-13)}=33$

$[EFGH]=[EFH]+[FGH]=\frac12\times3\times6+33=\boxed{42}$