smc

$△EFG\sim △EAB$ because $FG\parallel AB.$ The ratio of $△EFG$ to $△EAB$ is $2:5$ since $AB=5$ and $FG=2$ from subtraction. If we let $h$ be the height of $△EAB,$ $$\frac{2}{5}=\frac{h-3}{h}$$ $$2h=5h-15$$ $$3h=15$$ $$h=5$$ The height is $5$ so the area of $△EAB$ is $\frac{1}{2}(5)(5)=\boxed{\frac{25}{2}}$.