Dragan

Let O be a center of a semicircle and T the tangent point on XY

∠XYO = arccos( YO / XY )

Radius  TO = sin( ∠XYO ) * YO 

Area = 1/2 (TO² *pi )  

When I realized that the 2 triangles between purple squares have sides 3-4-5, it was a piece of cake to solve it.

The yellow square area is  54.76 cm² 

1)  Extend line segment QS to point T on a circle.

QT = PQ² / RQ              RT = QT - RQ         

M is a midpoint of line segment RT. Now we have a right triangle OMS.

MO = sqrt( OS² - MS² )

Radius  OR = sqrt( MR² + MO² )

2)   Let's place a point P on a circle opposite to point M. Now we have a right triangle CMP

∠ABC = arctan( AC / BC )

∠CMP = ∠ABC - ∠BCM

CM = cos(∠CMP) * MP

Use the law of cosines to find angles B and C.

 ∠ B = 53.13°           ∠ C = 67.38°

x / tan(B) + x + x / tan(C) = 14   

x ≈ 6.461      

Let the center of a circle be O and the midpoint of BC be M

Angle B = 45º

AM = sin( 45º) * 6

XY = 2/3 * AM            (Segment XY is a side of a square inscribed in a circle.)

Area of an inscribed square    A_s = XY²

r = sin(45º) * XY

Area of a circle     A_c = r²pi

Area of a shaded segment    A = (A_c - A_s) / 4 = pi - 2   

6*cos75º = 1.55291427

1.55291427 * 5.7955495 = 9

angle DEA = 62º

angle ADB = 45º

Hi, Melody! Please correct your diagram; CQ = 7     and    QD = 3 

Thanks!

AR:AC = 6 / 13 

I used 10 cm square and the result is the same as Melody's

Point B travels 3/4pi inches more than point A.