Find the Area of Triangle Formed Between Two Semicircle
Geometry Math Problem: Two semicircles are touching each other and the diameter of both semicircles are in a line, find the area of the triangle when the triangle is formed by the tangent of semicircles and touching point between them

From Figure, Find the Area of Triangle
MAB and BCN are semicircles
PA = 2 cm and QC = 4 cm
AC is the tangent of both semicircles
Then find the area of triangle ABC
Solution to This Viral Circle Geometry Problem
We can solve this problem in two different ways
Method 1: Find the area of quadrilateral then remove unshaded area of triangles
Method 2: Directly find the area using sine rule
Method 1

From figure
Area Of Blue Triangle = Area of quadrilateral APQC – Area of triangle BQC – Area of triangle PAB
We can redraw this figure like this

From above figure
∠PQX = ∠AXC {Because ∠PAC = ∠ACQ}
From triangle ACX
cos θ = CX/AX = 2/6
⇒ cos θ = ⅓
AC² = AX² – CX² = 6² – 2² = 36 – 4 = 32
⇒ AC = 4√2 cm
sin θ = AC/AX = 4√2/6
⇒ sin θ = AC/AX = ⅔√2
Area of quadrilateral APQC = Area of parallelogram PQXA + Area of triangle ACX
Area of parallelogram PQXA = PQ × QX × sin θ = 6 × 2 × ⅔√2
⇒ Area of parallelogram PQXA = 8√2 cm²
Area of △ACX = ½ × AC × CX = ½ × 4√2 × 2 = 4√2 cm²
Area of quadrilateral APQC = 8√2 + 4√2
⇒ Area of quadrilateral APQC = 12√2 cm²

Area of △ BQC = ½ × QC × QB × sin θ = ½ × 4 × 4 × ⅔√2 = 16√2/3 cm²
Area of △ BQC = ½ × PB × PA × sin (180 – θ) = ½ × PB × PA × sin θ = ½ × 2 × 2 × ⅔√2 = 4√2/3 cm²
Area of Blue triangle = 16√2/3 cm²
Method 2: Using sine rule

Area of Triangle ABC = ½ × AB × BC
AB² = PA² + PB² – 2 × PA × PB × cos (180 – θ) = 2² + 2² – 2 × 2 × 2 × ⅓
⇒ AB² = 8 – 8/3 = 16/3
⇒ AB = 4/√3 cm
BC² = QB² + QC² – 2 × QB × QC × cos θ = 4² + 4² – 2 × 4 × 4 × ⅓
⇒ BC² = 32 – 8/3 = 64/3
⇒ BC = 8/√3 cm
Area of △ABC = ½ × 4/√3 × 8/√3
⇒ Area of Triangle ABC = 16/3 cm²