# 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**