F.4 Trigonometry

For the outer square, named from the upper left corner clockwisely, A, B, C and D respectively. Also named the intersections inside clockwisely, where E is the "north" point, and then F, G and H.

There are 4 regions with blue colour.
And the angle EDF = 30 degree (It requires proof but not list here as it is difficult to explain here.)
The area of each of them
= sector DEF - triangle DEF
= pi * (2^2) * 30 / 360 - 1/2 * 2 * 2 * sin 30
= pi / 3 - 1

Due to the rotational symmetry, quadrilateral EFGH should be a square. (As sum of all the angles are 360 and each of them are equal, valued 90)

Let M be the mid-point of EF.
EM = DE sin (30 / 2)
= 2 sin 15

Therefore,
EF = 2EM = 4 sin 15

Shaded region = 4(pi / 3 - 1) + (4 sin 15)^2
= 4 * pi / 3 - 4 + 16 (sin 15)^2

where do u get ur questions?
it is so familiar...