Finding the area, general case with angle $\theta$. Inspired by this question, I am curious to know the more general case.

Given the radius of the large circle as $R$ and the angle $\theta \le \pi$, what is the area of the colored section? 
My initial thoughts are:


*

*the radius of the smaller circle is easy to derive:
$$r\left(\frac{1}{\sin\left( \frac{\theta}{2}\right)}+1 \right)=R \Rightarrow r= \frac{\sin \frac{\theta}{2}}{1+\sin \frac{\theta}{2}}R$$

*For $\theta = \frac{2 \pi}{n}$, One should be able to use a similar approach to this answer; although we will have to find a formula for the area of the n-polygon.

*Or we can try to derive everything algebraically!

 A: Consider the following diagram where $r$ is the radius of the smaller circle and $R$ that of the larger.
$\hspace{3.2cm}$
Looking at the lower left angle of the green triangle, we get
$$
\frac{r}{R-r}=\sin(\theta/2)
$$
which gives
$$
r=R\frac{\sin(\theta/2)}{1+\sin(\theta/2)}
$$
Now, the total area of the three colored regions is
$$
R^2\frac\theta4
$$
The area of the red region is
$$
r^2\left(\frac\pi4+\frac\theta4\right)
$$
The area of the green region is
$$
\frac12r\sqrt{R^2-2Rr}
$$
Thus, the area of the blue region is
$$
R^2\frac\theta4-r^2\left(\frac\pi4+\frac\theta4\right)-\frac12r\sqrt{R^2-2Rr}
$$
A: The area of the shaded region is
$$\begin{gather}
\lvert S(EAC)\rvert - \lvert T(ADH)\rvert - \lvert S(EDH)\rvert\\
\frac{\theta}{4}R^2 - \frac{r(R-r)}{2}\cos \frac{\theta}{2} - \frac{\pi + \theta}{4} r^2
\end{gather}$$
where $\lvert X\rvert$ denotes the area of $X$, $S(XYZ)$ denotes the circular sector determined by the points $x,\,Y,\,Z$ where $Y$ is the centre of the circle, and $T(XYZ)$ denotes the triangle with vertices $x,\,Y,\,Z$.
The area of a circular sector is of course the radius squared times half the angle. The angle of $S(EAC)$ is $\frac{\theta}{2}$ and the radius $R$. Since the triangle $T(ADH)$ has a right angle at $H$ and angle $\frac{\theta}{2}$ at $A$, the angle at $D$ is $\frac{\pi - \theta}{2}$, and hence the angle of the circular sector $S(EDH)$ is $\frac{\pi + \theta}{2}$, the radius is $r$.
In the triangle $ADH$, the hypotenuse has length $R-r$, hence the leg $AH$ has length $(R-r)\cos\frac{\theta}{2}$, and the leg $DH$ has radius $r$.
A: Hint:


*

*The area near the point $A$ is (that is, you can use $|AD|$ and $\theta$ to obtain the area of $\triangle ADH$) 
$$(R-r)^2\frac{2}{2}\sin\frac{\theta}{2}\cos\frac{\theta}{2}-\frac{\pi-\theta}{2\pi}\pi r^2.$$

*The figure is symmetric with axis $AE$.


I hope this helps ;-)
A: The desired area is $1/2$ of the area of the sector minus the area of the circle minus the area of the triangle formed by the vertex of the sector and the points of tangency, plus the area of the segment of the smaller sector included in that triangle. The area of the sector is simply
$$A_1 = \frac12 R^2 \theta$$
The area of the circle is obviously
$$A_2 = \pi r^2$$
where, as was noted, $r = R \sin{(\theta/2)}/(1+\sin{(\theta/2)})$.  The area of the triangle is determined by the length of the tangent line segment between the vertex and the point of tangency.  By using right triangles, I find this length to be simply $\sqrt{R^2-2 R r}$.  Then the area of the triangle is
$$A_3 = \frac12 (R^2-2 R r) \sin{\theta}$$
The area of the circular segment is a little hairy, but simple in principle.  The base of the triangle subtends an angle $\phi$ from the center of the smaller circle.  Using the law of cosines, we may find that
$$\sin{\frac{\phi}{2}} = \sqrt{\left ( \frac{R}{r}\right )^2 - 2 \left ( \frac{R}{r}\right )} \sin{\frac{\theta}{2}}$$
Then the area of the circular segment is
$$A_4 = \frac12 r^2 (\phi-\sin{\phi})$$
and the sought-after area is then
$$A = \frac12 (A_1-A_2-A_3+A_4)$$
