Find the Area of the Blue Geometric Shape

Geometry Math Problem: Find the area of the blue geometric shape formed by the intersection of the tangent of the quarter circle with a semicircle

From the figure, the radius of the quarter circle is 12 cm and the diameter of the semi-circle is 12 cm. BQ is the tangent of the quarter circle, Then what is the area of the blue geometric shape?
Solution: Area of the geometric shape

From the figure,
Area of the blue geometric shape = Area of the triangle ABQ – Area of red circular sector – Area of the yellow circular sector – Area of the green triangle

From triangle ABQ
cos A = AQ/AB
⇒ cos A = 12/24
⇒ cos A = 1/2
Then, ∠A = 60°
So we get ∠B = 180 – (∠A + ∠Q) = 180 – (60 + 90)
⇒ ∠B = 30°
From triangle BRP
∠B = ∠P = 30°
⇒ ∠BRP = 180 – (∠B + ∠P) = 180 – (30 + 30)
⇒ ∠BRP = 120°
then, ∠ORP = 180 – 120
⇒ ∠ORP = 60°

From above figure
Area of the red sector
Area of the sector = (60/360) × π × 12²
⇒ Area of the sector = 1/6 × π × 144
⇒ Area of the sector = 24π cm²
Area of the yellow sector
Area of the sector = (60/360) × π × 6²
⇒ Area of the sector = 1/6 × π × 36
⇒ Area of the sector = 6π cm²
Area of the green triangle BRP
Area of the triangle = ½ × RB × RP × sin 120
⇒ Area of the triangle = ½ × 6 × 6 × √3/2
⇒ Area of the triangle = 9√3 cm²
Area of the green triangle ABQ
Area of the triangle = ½ × AB × AQ × sin 60
⇒ Area of the triangle = ½ × 12 × 24 × √3/2
⇒ Area of the triangle = 72√3 cm²
So
Blue area = 72√3 – 6π – 24π – 9√3
Thus, the Area of the geometric shape = 63√3 – 30π cm²