Question

In the adjoining figure ABCD
is a rectangle with the sides 4cm and 8cm. Taking 8cm as the diameter two semicircles ate drawm. Find the area overlapped by two semicircles.

Answer 1

Are overlapped = area of rectangle-2*area of semicircles
=l*b-2*1/2*pie*r^2
=8*4-(22/7*4*4)
=32-(22/7*16)

Just calculate it
Hope it help

Answer 2

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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