Another question from my test:
The area of the circle is $157$
$ AE=EB $
$ ABCD$ is a square
The diagonal of the square ($AC$) is equal to the diameter of the circle.
Find the area of BCFE?
I managed to find the area of $ACB$, but I am having a trouble finding $AFE$. Any suggestions?
An idea:
Draw the diagonal $\;BD\;$ and take a close look at triangle $\;\Delta ABD\;$. Deduce that $\;F\;$ is the intersection point of the medians of this triangle and thus
$$AF=\frac23r\;\left(\implies FC=\frac43r\;\right)\;,\;\;r=\text{ the circle's radius}$$
Likewise, using Pythagoras and with $\;x=\sqrt2\,r=$ the square's side:
$$DE=\frac{\sqrt5}2x\implies FE=\frac13DE=\frac13\frac{\sqrt5}2\sqrt2\,r=\frac{\sqrt{10}}6\,r\;,\;\;AE=\frac12x=\frac r{\sqrt2}$$
So now you know all the sides of $\;\Delta AEF\;$ and you can calculate its area, for example with Heron's formula for the semiperimeter.
Hint:
Clearly,
$$\Delta\ ADE=\frac{1}{4}\ \square\ ABCD$$
$$\Delta\ ABC=\frac{1}{2}\ \square\ ABCD$$
Since base $AE=\frac{1}{2}AD$, we get
$$\Delta\ AFE=\frac{1}{2}\ \Delta\ AFD=\frac{1}{3}\ \Delta\ ADE=\frac{1}{12}\ \square\ ABCD$$
Therefore
$$\square\ BCEF=\Delta\ ABC-\Delta\ AFE=\frac{5}{12}\ \square\ ABCD\\ $$