ABCD is a rectangle with sides 8cm and 4cm. Two semicircles are drawn taking the 8cm sides as diameters. Find the area overlapped by the two semicircles.
Answers
Solution:
For Sector PQXR , PQ = QR = 4 cm
now for the sector PYRQ = PR= PQ = 4cm
therefore in triangle PQR , PQ = QR =PR = 4cm
∴ Δ PQR is an equilateral triangle and have side of 4cm
∴∠PQR =∠QPR =∠PRQ = 60°
now the area of shaded portion = 2× area of PYRXQP
=( area of sector PQXRMP + area of segment PYMPR)
2(60°/360°×π(4)² + area of sector PQNRYP - area of Equilateral triangle PQR)
=2( 60°/360°×π×16 +60°/360°×π×16 -√3/4×16)
=2(2×1/6×π×16 -4√3)
=2(16/3×3.141-4×1.732) ( ∵π = 3.141 and √3 = 1.732 )
=2×(16.75-6.92)
=2×9.82
=19.64cm²
Hence the shaded portion = 19.64cm²
Area of overlapping by two semicircle = 8(4π/3 - √3) = 19.6 cm²
Step-by-step explanation:
Diameter = 8 cm
Radius = 8/2 = 4 cm
PQ = 4cm ( = AB & CD)
PR = QR = 4 cm ( Radius of Two circle)
Hence ΔPQR is an equilateral triangle
Hence each angle = 60°
Area of PRXQ = (60/360)π4² = 8π/3 cm²
=> Area of Δ PQR + Area of RXQ = 8π/3 cm²
Simialrly
Area of QRYP = (60/360)π4² = 8π/3 cm²
=> Area of Δ PQR + Area of RYP = 8π/3 cm²
area of Δ PQR = (√3 / 4 )4² = 4√3 cm²
Area of half shaded portion = Area of Δ PQR + Area of RXQ + Area of Δ PQR + Area of RYP - Area of Δ PQR
= 8π/3 + 8π/3 - 4√3
= 16π/3 - 4√3
Area of overlapping by two semicircle = 2 (16π/3 - 4√3 )
= 8(4π/3 - √3)
= 19.6 cm²
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