Percentage of the total area?

Let the co-ordinates of F be (0,0). The co-ordinates of O are (3,3).

The equation of the circle with center at (3,3) and radius 3 can be written as

(x-​3)^​2 + (y-​3)^​2 = 9 --------------------------------(1)

Equation of the diagonal FG can be written as y = 0.6*x---------------------------(2)

Substituting (2) in (1) and solving for x will yield the x coordinates of points A and B. Y coordinates of A and B can be found using equation (2).

Substituting (2) in (1) yields

1.36 x^​2 -9.6 x + 9 = 0

Solving the quadratic equation yields two values of X ( x1 , x2) which are equal to x1= 5.9458 and x2 = 1.113

The co-ordinates A and B can be determined as y = 0.6*x and so the co-ordinates of A are (1.113, 0.6678) and B are (5.9458,3.56748)

The length of AB can be calculated as =sqrt((x2-x1)^2+(y2-y1)^2) =sqrt((5.9458-1.113)^2+(3.56748-0.6678)^2) =sqrt(23.355955840000007+8.4081441024) =sqrt(31.76409994240001) =5.635964863481674=5.636

The area of the triangle AOB can be calculated using Heron's formula The area is given by: Area = sqrt( p* ( p−a ) (p−b) ( p−c) )
where p is half the perimeter, or
p = (a + b + c )/2

Sides: a = 3   b = 3   c = 5.636

Area of Triangle AOB = 2.9 -------------------------(3) Perimeter: p = 11.636 Semiperimeter: s = 5.818

Angle ∠ ABO = α = 20.06° = 20°3'36″ = 0.35 rad Angle ∠ BAO = β = 20.06° = 20°3'36″ = 0.35 rad Angle ∠AOB = γ = 139.88° = 139°52'47″ = 2.441 rad

Area of the sector AOB which subtends 139°52'47″ in the center is pi* 9 * 139.88/360 = 10.986 sqm.

Area of the circle below the chord AB = Area of Sector AOB - Area of Triangle AOB = 10.986 - 2.9 = 8.086 sqm.

Hence area of the blue colored region = Area of Triangle FGH - Area of the circle below the chord AB = 30 - 8.086 = 21.914

So percent of the blue colored area = (21.914/60) * 100 = 36.52 percent