Calculus question about related rates

Here's the question:
The top of a 10-foot ladder is sliding down a wall at a rate of 12 feet per second. How fast is the base of the ladder sliding away from the wall at the instant when the top of the ladder is 6 feet from the ground?

I've been trying to figure this out but I'm not having much luck. Any help is appreciated!

Let me try it...

x is the height and y be the distance of base from the wall

x^2 = 10^2 - y^2

2 x (dx/dt) = - 2 y (dy/dt)

at x = 6, we have y = 8 (i.e., you get sqrt(10^2 - 6^2) = 8 )

and given dx/dt = 12 ft/sec

2 * 6 * 12 = -2 * 8 * (dy/dt)

dy/dt = -12*12/16 = -9 ft/sec

(negative means it is moving away from the wall, i.e., towards to -y direction.)

b = base distance
h = top distance
b^2 = 100-h^2
b = (100-h^2)^(1/2)
db/dt = db/dh* dh/dt
db/dt = (1/2)(100-h^2)^(-1/2)*(-2h)*(-12)
When h = 6
db/dt = (1/2)(64)^(-1/2)*24*6
db/dt = 9 ft/s