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In a circular table cover of radius 32 cm, adesign is formed leaving an equilateraltriangle ABC in the middle as shown inFig 12.24. Find the area of the design.10Fig. 1224 |
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Answer» Let ABC be the eq./\and let o be the centre of the circle of r=32cmArea of circle =πr^2=(22/7×32×32)cm2=22528/7 cm2 Draw OM_|_BCNow, /_ BOM= 1/2×120°=60°So,From /\BOM,we haveOM/OB=cos 60°(1/2)i.e., OM= 16 cmAlso, BM/OB= cos60°(1/2)i.e., BM= 16√3 cmBC = 2 BM =32√3 cmHence, area of /\BOC = 1/2 BC .OM=1/2×32√3×16area of /\ ABC = 3× area of /\ BOC= 3×1/2×32√3×16= 768√3 cm^2Area of design= area of O - area of /\ ABC= (22528/7 - 768√3)cm^2 |
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