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
![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](https://aplusreach.com/wp-content/uploads/2021/11/can-you-solve-this-viral-circle-geometry-problem-1.jpg)
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
![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](https://aplusreach.com/wp-content/uploads/2021/11/can-you-solve-this-viral-circle-geometry-problem-1.jpg)
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
![Method 1: Find the area of quadrilateral then remove unshaded area of triangles](https://aplusreach.com/wp-content/uploads/2021/11/can-you-solve-this-viral-circle-geometry-problem-5-2.jpg)
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
![Method 1: Find the area of quadrilateral then remove unshaded area of triangles](https://aplusreach.com/wp-content/uploads/2021/11/can-you-solve-this-viral-circle-geometry-problem-4.jpg)
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
![Method 2: using sine rule](https://aplusreach.com/wp-content/uploads/2021/11/can-you-solve-this-viral-circle-geometry-problem-3.jpg)
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