Circle it Up

Recently I was helping someone with some geometry problems involving circles, and I said to myself I'll bet there is a calculus question I can make up from this.

So here is my question.
Suppose you have a 45, 45, 90 triangle, ABC, inscribed in a circles so that it's hypotenuse is the diameter of the circle. Now suppose that sides AC and BC increase in length at 3m/s and the circle expands to keep the triangle enclosed. Then at what rate is the area of the circle increasing when the radius of the circles is 5m?

So we know the radius that we want the radius of the circle to be:
r = 5m
We know the rate at which the sides of the triangle increase:
ds/dt = 3m/s
It would be nice to know the rate at which the diameter of the circle increases. Since we have a 45 45 90 triangle we know that if one of the sides is length x then the hypotenuse will be length x*sqrt(2). So let's write our equation like this:
The diameter is equal to the length of a side times the square root of two.
D = s*sqrt(2)
To make it a little bit easier later on I'm going to rewrite this equation again except now I'm going to substitute two times the radius in for the diameter so it looks like this:
2*r = s*sqrt(2)
now lets take the derivative with respect to time
2*dr/dt = sqrt(2)*ds/dt = 3*sqrt(2)
dr/dt =(3*sqrt(2))/2 = 1.5*sqrt(2)

We know or at least we can look up the equation for the area of a circle
A = pi*r^2
if we take the derivative of this equation with respect to time we get
dA/dt = 2*pi*r*dr/dt (just use the power rule)
Now we can substitue what we know
dA/dt = 2*pi*5*1.5*sqrt(2) = 21.21*pi m^2/s
So when the radius of the circle is 5, the area of the circle is increasing at 21.21*pi square meters per second.

Once again I've created an Excel spread sheet to allow you to experiment with all kinds of things including central and inscribed angles as a geometry bonus.
CircleandTriangle.xls