Finding the area of an annulus with one measurement Suppose you have a circular pool of lava (the reason for the contents will be clear in a moment) and in the center of the pool is a circular lawn. With a single straight-line measurement, determine the area of the lawn lava. 
The measuring instrument you have is a laser transit, with which you can look through a telescope, point it to a pole your assistant is holding, and read off the distance from the transit to the pole. Because you and your assistant can't stand in the lava, you can't use the obvious strategy of measuring the diameter of the lawn, but you can find the distance between any two points outside of the lava. How can you find the area of the lawn lava?
EDIT:
An apology is in order here. I was apparently passing a brain stone when I posted the original question. I should have asked what the area of the lava was. The comments have already answered the question.
My real purpose was to observe that measuring the chord will give you the area of the lava, regardless of what the diameter of the pool and the lava were, so you could shrink the lawn diameter to zero and the chord would allow you to compute the area of the lava. In this way, the problem is like the question spatial geometry hole in sphere.
My eventual question was to be, does anyone know of similar problems, where the answer doesn't require one of parameters of the problem and so can be solved by setting that parameter to zero?
 A: 
By Pythagoras, $(l/2)^2+r_1^2=r_2^2$.  So the area of the lava (shaded) is $$\begin{align}
\pi r^2_2 - \pi r^2_1 &= \pi (r^2_2-r^2_1)\\
&= \pi (l/2)^2\\
&=\frac{\pi l^2}{4}
\end{align}$$
A: First find the lengths $\ell$ of two chords of the large circle sharing a common endpoint $A$ and just touching the small circle.  Then find the distance $k$ between their opposite endpoints.
Below, I show that (surprisingly) $k$ cancels out, so you really only need $\ell$.
You have an isosceles triangle with two sides of length $\ell$, and the small circle touches those two sides at their midpoints.  Let $C$ be the center of the circles.  Let $B$ be the midpoint of one chord of length $\ell$.  Then $ABC$ is a triangle with a right angle at $B$.  The height of the big isosceles triangle is $\sqrt{\ell^2-\left(\frac k2\right)^2}$.  The triangle $ABC$ is similar to the half of the isosceles triangle with legs $k/2$ and $\sqrt{\ell^2-\left(\frac k2\right)^2}$ and hypotenuse $\ell$.  Triangle $ABC$ has legs $\ell/2$ and $r$, where the latter is the radius of the small circle.  Therefore
$$
\frac{r}{\ell/2} = \frac{k/2}{\sqrt{\ell^2-\left(\frac k2\right)^2}}
$$
Hence
$$
r = \frac{k\ell}{2\sqrt{4\ell^2-k^2}}.
$$
If one also wants the radius of the big circle, that's the radius of the circumscribed circle of a triangle with side lengths $\ell$, $\ell$, and $k$.  If I'm not mistaken that's
$$
R = \frac {\ell^2}{\sqrt{4\ell^2-k^2}}.
$$
Generally, a triangle with sides $a,b,c$ has circumscribed circle radius
$$
\frac{abc}{\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}}.
$$
The area between the circles is
$$
\pi(R^2-r^2) = \pi\left(\frac \ell 2\right)^2.
$$
Therefore you don't need to know $k$.  You need only $\ell$, so you really can do this with just one measurement.
A: A similar problem in three dimensions is the "Hole in the Sphere" problem. If you drill a cylindrical hole in a sphere, the volume remaining in the sphere depends only on the height of the hole - it is independent of the diameter of the hole or diameter of the sphere.
