Dragan

Red area             A_r = 12*16 = 192 u²

Green area         A_g = (25 * 192)/6 = 800 u²

The area of entire rectangle  A_e = A_r + A_g = 992 u²  

Let us denote a crosspoint of AF and DG with a letter "M" and an angle AGD fith ''q''.

EF + BF = 15       BF = AG                   AG = 10  

q = ?         sin(q) = AM/AG                   q = 26.565°  

AM = ?             AM = sin(q) * AG          AM = 4.472  

GM = ?      GM = sqrt(AG)² - (AM)²       GM = 8.944  

Area (AGM) = (AM * GM)*1/2               Area = 20 u²    

Angle BCD = 80°, but don't ask me how I've done it.  

All you need is a bottom row:

Circle diameter   >  D = 1

Radius                >  R = 0.5

Triangle side       >  a = 4D + 2[R / tan(30°)]     a = 5.732    a/2 = 2.866

Triangle height    >  H = sqrt[ a² - (a/2)²]      H = 4.964

Triangle area       >  A = H * a/2              A = 14.227 u²   

You can solve this one in your head in less then a minute.

                    A = 4*6 + 4*15    

It's interesting that the sum of the internal angles of the shape OXYZ is equal to the angle XYZ.

                           17.5° +65° + 40° = 122.5°     

Chord               >   PB = 8 

Triangle side    >   AB = 7 

Angle ABD  = 30°

Let the circle's diameter be   BD          ( Point D is 180° from point B)

Diameter         >  BD = ?                BD = AB / cos(30°)           BD = 8.083  

Chord              >  PD = ?             PD = sqrt [(BD)² - (PB)²]      PD = 1.155  

Angle      PBD > b                 cos(b) = PB / BD                       ∠ b = 8.213°   

Angle      ABP  > a                a = ∠ABD - ∠b                           ∠ a = 21.787° 

Angle      PBC  > c                c = ∠ABC - ∠a                           ∠ c = 38.213°   

PA = ?                   (PA)² = (AB)²+(PB)² - 2(AB)(PB) cos(a)         PA = 3  

PC = ?                   (PC)² = (BC)²+(PB)² - 2(BC)(PB) cos(c)          PC = 5      

Vertical line must pass through both points of inersection of circles's circumferences, and the line L must pass through the centers of both circles. 

AB and AC must be equal!!  

Let the square side AB = 2

Let's denote the midpoints of a side AB with a letter M, and a side CD with a letter Q

MP = tan(15°) * 1            MP = 0.268   

PQ = BC - MP                 PQ = 1.732   

 ∠DCP => q                    tan(q) = PQ / [(AB)/2]      q = 60°