ABCD is a rectangle in which AB=20cm and BC=10cm. A semicircle is drawn with centre O and radius 10root2. cm. It passes through A and ,touching the ends. Find the area of shaded region .





In OAB, OA = OB = radii of circle = 102 cmNow, AB = 20 cmNow, AB2 = 202 = 400and OA2 + OB2 = 1022 + 1022 = 200 +200 = 400Since, OA2 + OB2 =AB2AOB is a right  in which AOB = 90°   Converse of Pythagoras TheoremTake any point X on the arc AB of the semicircle.Consider the sector OAXB,radius,r = 102 cm; central angle, θ = AOB = 90°area of sector OAXB = θ360°×πr2 = 90°360°×π×200 = 50π cm2arAOB = 12×OA×OB = 2002 = 100 cm2area of segment AXB = area of sector OAXB - arAOBarea of segment AXB = 50π-100 cm2 area of shaded region = area of ABCD- area of segment AXB = 20×10 -50π-100  = 300-50π cm2

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