Geometry Math Problem

The hypotenuse of the right triangle is equal to 4 cm and an acute angle is 30 degrees. Right triangle is inscribed inside a quarter circle as shown in the figure, then find the area of the quarter circle

From figure AQ = AP, OP = 4 cm and ∠PQC = 30°, Find the area of a quarter circle

Solution: Area of the quarter circle

From triangle PQC

sin 30 = PC / 4

⇒ ½ = QC / 4

⇒ QC = 2 cm

∠QPC = 30°

⇒ ∠PQC = 180 – (90 + 30) = 60°

Apply Pythagoras theorem in the triangle APQ

AQ² + AP² = PQ²

We know AQ = AP, then

2AQ² = PQ²

⇒ √2 × AQ = PQ

⇒ AQ = PQ / √2 = 4 / √2 = 2√2 cm

Connect AC, then we get a triangle ACQ

From triangle APQ

AQ = AP so ∠APQ = ∠AQP

⇒ ∠AQP = 45°

From triangle ACQ

∠AQC = 45 + 60 = 105°

Apply cosine rule in triangle ACQ

AC² = AQ² + CQ² − 2 × AQ × CQ × cos 105

⇒ AC² = (2√2)² + 2² − 2 × 2√2 × 2 × (−(√3 − 1) / (2√2))

⇒ AC² = 8 + 4 + 4√3 − 4

⇒ AC² = 8 + 4√3

Where AC is the radius of the quarter circle so

Area of quarter circle = ¼ × π AC² = ¼ × π (8 + 4√3) cm²

⇒ Area of quarter circle = 2π + π √3 cm²

⇒ Area of quarter circle ≈ 11.725 cm²