Area of an annulus knowing certain chord lengths We have two concentric circles - what's the blue area?

 A: 
Let $A = (-8, 0), B = (-2, 0), C = (2, 0), D = (8, 0)$ so that $AB = CD = 6, BC = 4$. $P$ must lie on the perpendicular bisector of $BC$, so $P = (0, p)$ where $p \in \mathbb R$.
Then the blue area is just $\pi AP^2 - \pi BP^2 = \pi \left((8^2 + p^2) - (2^2 + p^2) \right) = 60 \pi$ and is invariant regardless of $P$.
A: I will give also my answer (but I will not "mark" it as solution)

Using Pythagorean theorem we have
$$R^2 = x^2 + 8^2$$
$$r^2 = x^2 + 2^2$$
after subtraction we have
$$ R^2 - r^2 = 8^2 - 2^2  = 60 $$
So result is $60\pi$ and it depends ONLY on values $a=6, b=4$. General formula is
$$ Area = \left[(a+\frac{b}{2})^2 - (\frac{b}{2})^2\right]\pi = a(a+b)\pi $$
so using it we have $ 6(6+4)\pi = 60\pi$
A: Alternative answer

*

*If the task is correctly formulated then the answer (area) is one and equal to $A$

*But we can choose infinite number of cases (pair of two circles) which have chord  6-4-6 (look on picture)

*According to 1. each case should have the same area $A$

*So lets choose case when chord is circle diameter (draw chord as line and mark its segments and at its middle use calipers to draw big and small circle) - then

$$R=(6+4+6)/2=8$$
$$r=4/2=2$$
$$A=\pi R^2 - \pi r^2=60\pi$$

