Octagon inside a circle An octagon which has side lengths 3, 3, 11, 11, 15, 15, 15 and 15  is inscribed
in a circle. What is the area of the octagon?
I tried using the cosine law on the triangles made when connected with the center but the numbers became really hard to use when I solved for the sine of the angles to easily get the triangle area.
This is a problem from BIMC 2017 individual question 6, where the answer is 567. ;)
 A: Arrange the triangles so there is a $15$ in each quadrant; the $3$ sides are vertical and the $11$ sides are horizontal.  Then the distance between $(\sqrt{r^2-9/4},3/2)$ and $(11/2,\sqrt{r^2-121/4})$ is 15.
A: 
Arranging the sides in the order $3,15,11,15,3,15,11,15$, we obtain two overlapping, perpendicular inscribed rectangles of sides $3×h$ and $11×k$ respectively, where $h,k$ are to be determined. The Pythagorean theorem gives
$$3^2+h^2=11^2+k^2$$
$$h^2-k^2=112$$
Furthermore, the rectangles are aligned and have a common centroid. This gives
$$\left(\frac{h-11}2\right)^2+\left(\frac{k-3}2\right)^2=15^2$$
$$(h-11)^2+(k-3)^2=30^2$$
This immediately suggests finding Pythagorean triples $(a,b,30)$, of which there is only one: $18^2+24^2=30^2$. If we let $h-11=18$ and $k-3=24$, we see that the other equation $h^2-k^2=112$ is "miraculously" satisfied.
Thus $h=29$ and $k=27$. The octagon's area can then be calculated easily by adding the rectangles' areas, subtracting their intersection and adding the triangles' areas:
$$3h+11k-33+2\left(\frac{h-11}2\right)\left(\frac{k-3}2\right)=87+297-33+2\cdot9\cdot12=567$$
