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