jmc

Extend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so that you have a rectangle. (We know that $△ADE$ is a $6-8-10$, since $△DEB$ is an $8-15-17$.) The base $CD$ of the rectangle will be $9+6+6=21$. Now, let $E$ be the intersection of $BD$ and $AC$. This means that $△ABE$ and $△DCE$ are with ratio $\frac{21}{9}=\frac{7}{3}$. Set up a proportion, knowing that the two heights add up to 8. We will let $y$ be the height from $E$ to $DC$, and $x$ be the height of $△ABE$.$$\frac{7}{3}=\frac{y}{x}$$$$\frac{7}{3}=\frac{8-x}{x}$$$$7x=24-3x$$$$10x=24$$$$x=\frac{12}{5}$$ This means that the area is $A=\frac{1}{2}(9)(\frac{12}{5})=\frac{54}{5}$. This gets us $54+5=\boxed{59}.$