Area between two triangles/squares/pentagons (Polygon ring) Introduction
During a class we got this mathematical problem: Calculate the area of a ring, formed by 2 concentric circles, knowing that the outer circle chord is 10 cm, and it is tangent to the inner circle. Here is a drawing:
Link to someone else who shared this problem: https://www.quora.com/How-do-I-calculate-the-area-of-a-ring-formed-by-2-concentric-circles-knowing-that-the-outer-circle-chord-is-long-10-cm-and-it-is-tangent-to-the-inner-circle
The main problem, area of the polygon ring
After solving it my teacher wanted me to extend this issue to triangles, squares or pentagons. I then asked where the chord would be and he didn't want to tell me. After working on this for quite a while I'm none the wiser. Do anyone have a clue on how one could find the area between these figures using only a "chord" of 10 cm without anything else given? 
He mentioned something about having the "chord" go through one of the lines of the inner square in the square case, but I didn't quite understand how.
Solved for a special case
After Jean Marie's hint I tried drawing a line through the diagonal of the chord and another through on of the vertices at 45 degrees. Only the latter gave me something meaningful to work on. I've managed to find the area of a special case where the  "chord" is 8 units since the intersection points of the chord and the outer square then meet. When the chord is 8 units the intersection points meet and enables the pythagorean theorem to be applied.

This is also true for a triangle as long as the intersection points meet, but it is a particular case and far from general.
Solved for the general square case by YNK
YNK managed to find the area between two squares by using the inscribed circles of these squares. From this construction the lengths of the small and large square is 2r and 2R respectively. The chord is given as 10 cm. The area between the squares would then be (2R)^2 - (2r)^2 which can be written as 4(R^2-r^2). We substitute R^2-r^2 by 5^2 from the original problem, and the area between them turns out to 4*25=100 square cm.

Current problem, the general case for triangles or pentagons
The problem now remains to study whether or not this method of only using a chord of 10 cm to find the areas betweem figures works for triangles and pentagons. I tried to do the same for a triangle, but no success yet.
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Thanks for all the help so far everyone! :)
 A: 
In this answer, we hope to analyse the n-gon to show that the area of a polygon ring created by placing a smaller regular polygon with $n \left(\ge 3\right)$ sides completely inside a larger regular polygon having the same number of sides. We also assume without loss of generality that the centers of the two polygons coincide at $O$.
The diagrams shows a part of the polygon under investigation. Please pay your attention to $\mathrm{Fig.\space 1}$. Let the sidelengths of the two polygons be $b$ and $kb$, where $k\gt1$. The radii of the inscribed circles of the two polygons are $r_{\text{i}}$ and $R_{\text{i}}$. The chord of the larger incircle is a tangent to the smaller incircle at $K$ and it has the length $c_{\text{i}}$. The line segment $ON$ is drawn perpendicular to the sides $A_1A_2$ and $\bar{A_1}\bar{A_2}$ to make $M$ and $N$ are their midpoints respectively. We also have,
$$\theta=\dfrac{\pi}{n}.\tag{0}$$
Considering the two right-angled triangles $A_1MO$ and $\bar{A_1}NO$, we can write,
$$r_{\text{i}}=\dfrac{b}{2}\cot\left(\theta\right) \qquad\text{and}\qquad R_{\text{i}}=\dfrac{kb}{2}\cot\left(\theta\right).\tag{1}$$
Using (1) we can express the area of the polygon ring $A$ as,
$$A=\dfrac{nkb R_{\text{i}}}{2}-\dfrac{nb r_{\text{i}}}{2}=\dfrac{nb^2}{4}\left(k^2-1\right) \cot\left(\theta\right).\tag{2}$$
An expression for $c_{\text{i}}$ can derived using the right-angled triangle $POK$ as shown below.
$$ c_{\text{i}}=2\sqrt{R_{\text{i}}^2- r_{\text{i}}^2} = b\sqrt{k^2-1}\cot\left(\theta\right)\tag{3}$$
By substituting (0) and (3) in (2), we can obtain a formula for $A$ in terms of $c_{\text{i}}$ and $n$.
$$A=\dfrac{n}{4}\tan\left(\dfrac{\pi}{n}\right) c_{\text{i}}^2\tag{4}$$
According to (4), the area of the polygon ring depends only on the number of sides and the length of the chord $PQ$.
We kindly request you to go through this text to see whether the description given there answers your question. We hope to supplement this answer by providing some more details a few hours from now. In the mean time you can try to derive an expression for the area of a n-gon ring using the circumcircles.
