water in the basin

From the similar triangles on the left side, we get $\dfrac{x}{4} = \dfrac{5}{10} \implies x = 2 \implies R = 12$ .

The wet area is the area of the lower base and the curved surface area of the frustum ; the latter's formula is $\pi (R + r) \sqrt{(R - r)^2 + h^2}$ .

Thus the wet area is

$\pi (R + r) \sqrt{(R - r)^2 + h^2} + \pi r^2 = \pi (12 + 10) \sqrt{(12 - 10)^2 + 4^2} + \pi (10^2) = \boxed{44 \sqrt{5} \pi + 100 \pi}$