Expected area of triangle inscribed in a circle On  a  unit  circle , BC  is  a  chord  with  length  4/π . Point  A  is  picked   randomly at  its circumference .
Find  the  expected  area  of  △  ABC .

For easy reference let chord BC sits in the lower portion of the unit circle.  It lies horizontally making angle $2\phi$ at the centre of the circle symmetric to the y axis.
Then the perpendicular distance of the chord from the centre is $D = \sqrt{(1 - (\frac{2}{\pi})^2}$ and $sin\phi = \frac{2}{\pi}$
Let the line joining A and the centre makes an angle $\theta$ with the x axis, then
For $\theta$ between $0$ and $\frac{\pi}{2}$, 
$$Area = \frac{1}{2}\frac{4}{\pi}(sin\theta + D)$$
For $\theta$ between  $0$ and $\frac{\pi}{2} - \phi$
$$Area = \frac{1}{2}\frac{4}{\pi}(D - sin\theta)$$
For $\theta$ between $\frac{\pi}{2} - \phi$ and $\frac{\pi}{2}$
$$Area = \frac{1}{2}\frac{4}{\pi}(sin\theta - D)$$
Hence the average area is
$$\frac{4}{2\pi^2}(\int_0^{\frac{\pi}{2}}(sin\theta + D)d\theta + \int_0^{\frac{\pi}{2} - \phi}(D - sin\theta)d\theta + \int_{\frac{\pi}{2} - \phi}^{\frac{\pi}{2}}(sin\theta - D)d\theta)$$
$$= \frac{2}{\pi^2}(\frac{2}{\pi} + D\pi - 2D\phi + sin\phi) = 0.533$$
I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.
Also if the area is $\frac{3}{2\pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.
