Finding the area of a circle given the inscribed figure inside it 
Note: This image is not to scale. Please look at the lengths of the lines.
The question is rather self explanatory from this image and the title, but to reiterate: I would like to find the exact area of this circle in terms of $\pi$. I thought of maybe estimating the circumference, but that led me nowhere. This is from a problem set which I was given 15 minutes to do, so there should be some trick to finding the solution. Thank you in advance!
 A: I get the same answer as Ross Millikan by a slightly different route.
Draw the six radii from the center of the circle to the vertices of the inscribed hexagon.  You'll have three triangles with sides of length $r$, $r$, and $2$, and three triangles with sides of length $r$, $r$, and $1$.  Let the angles of these two sets at the center be $\theta$ and $\phi$.  Then by the law of cosines we have
$$\begin{align}
4&=2r^2(1-\cos\theta)\\
1&=2r^2(1-\cos\phi)\\
\end{align}$$
with $3\theta+3\phi=2\pi$, hence
$$1-\cos\theta=4(1-\cos\phi)=4(1-\cos{2\pi\over3}\cos\theta-\sin{2\pi\over3}\sin\theta)$$
which simplifies to
$$2\sin\theta=\sqrt3(1+\cos\theta)$$
Squaring both sides and using $\sin^2\theta=1-\cos^2\theta$ leads to the quadratic
$$7\cos^2\theta+6\cos\theta-1=(7\cos\theta-1)(\cos\theta+1)=0$$
from which we can concluse that $\cos\theta=1/7$, which plugs in to give
$$4=2r^2\left(1-{1\over7}\right)={12\over7}r^2$$
which means the requested area of the circle is
$$A=\pi r^2={7\over3}\pi$$
Added later:  The OP has asked if there is a simpler solution that doesn't use any trig.  What I'm about to offer is not exactly simple, but it does avoid trig.
As above, drawing the six radii produces two sets of three identical triangles.  If we rearrange these triangles so that the sides of length $1$ and $2$ alternate instead of being grouped together, the hexagon produced is still inscribed in a circle of the same radius.  So let's look at that figure, and label the points $A$, $B$, $C$, $D$, $E$, and $F$, with $|AB|=|CD|=|EF|=1$ and $|BC|=|DE|=|FA|=2$.
The first thing to notice, if you draw the figure, is that
$$|AC|=|CE|=|EA|=\sqrt3r$$
The second thing to notice is that, by symmetry, the angles at all the vertices of the hexagon are equal, and thus all equal $120^\circ$.
Now consider the triangle $\triangle ABC$.  Extend the line $BC$ and drop a perpendicular to it from $A$.  Let $P$ be the point at which the perpendicular meets the line.  Note that $\angle ABC=120^\circ$ implies $\angle ABP=60^\circ$, hence $|AB|=1$ implies $|BP|={1\over2}$ and $|AP|={\sqrt3\over2}$.  But we now have $|CP|=|BC|+|BP|=2+{1\over2}={5\over2}$ at which point the Pythagorean Theorem tells us
$$3r^2=|AC|^2=|AP|^2+|CP|^2={3\over4}+{25\over4}=7$$
The key idea here was the notion that you can rearrange the sides to alternate without changing the radius of the circle.  Once you accept that, the rest really is fairly simple.  But I rather doubt I would have come up with this as a seventh grader on a 15-minute problem set.
A: Let $\alpha$ (respectively, $\beta$) be the measure of any inscribed angle subtending a chord of length $1$ (respectively, $2$). Let $\overline{PQ}$ be the "middle" chord of length $1$, and $\overline{RS}$ the "middle" chord of length $2$, as shown. Let $X$ be the intersection of $\overline{PS}$ and $\overline{QR}$, and let $\gamma$ be the measure of angle $\angle RXS$ (and $\angle PXQ$). Let $O$ (not shown) be the center of the circle.

By the Inscribed Angle Theorem,
$$\alpha = \frac{1}{2}\angle POQ \qquad \text{and} \qquad \beta = \frac{1}{2}\angle ROS$$
By a related theorem, the measure of an angle formed by two chords is the average of the measure of opposing subtended arcs:
$$\gamma = \frac{1}{2}\left( \; \angle POQ + \angle ROS \; \right) = \alpha + \beta$$
Consequently, $\triangle XRS$ (also $\triangle POQ$) is equiangular, hence equilateral.
Now, in $\triangle PQS$, we have
$$|\overline{PQ}| = 1 \qquad |\overline{PS}| = |\overline{PX}|+|\overline{XS}| = 1 + 2 = 3 \qquad \angle QPS = \frac{\pi}{3}$$
so that, by the Law of Cosines,
$$|\overline{QS}|^2 =  1^2 + 3^2 - 2\cdot 1 \cdot 3 \cos\frac{\pi}{3} = 10 - 3 = 7$$
and by the Law of Sines,
$$ d = \frac{|\overline{QS}|}{\sin\angle QPS} = \frac{\sqrt{7}}{\sqrt{3}/2} = \frac{2\sqrt{7}}{\sqrt{3}}$$
where $d$ is the diameter of the circle. Thus, the radius of the circle is $\sqrt{\frac{7}{3}}$, and the area $\frac{7}{3}\pi$. $\square$
(I believe that there are quicker ways to get from the equilateral triangles to the circle's area.)
A: Let the circle have radius $r$.  If you draw the radii to each point on the circle and through the midpoints of each side you get six $1,\sqrt{r^2-1},r$ right triangles and six $\frac 12, \sqrt{r^2-\frac 14},r$ right triangles.  Using the angles at the center gives $$6\arcsin \frac 1r + 6 \arcsin \frac 1{2r}=2\pi\\ \arcsin \frac 1r + \arcsin \frac 1{2r}=\frac \pi 3\\$$  Take the sine of this, use the angle-sum formula and the fact that $\cos \arcsin x=\sqrt {1-x^2}$ should get you to the Alpha solution $r=\sqrt{\frac 73}$
