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Question

# From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2. (Use π=227).

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Solution

## Given:Height (h) of the conical part = Height (h) of the cylindrical part =2.4 cmDiameter of the cylindrical part =1.4 cmRadius = Diameter2Radius(r) of the cylindrical part =0.7 cmSlant height (l) of conical part = √r2+h2 =√0.72+2.42=√0.49+5.76=√6.25=2.5Total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of cylindrical base =2πrh+πrl+πr2 =2×227×0.7×2.4+227×0.7×2.5+227×0.7×0.7 =4.4×2.4+2.2×2.+2.2×0.7 =10.56+5.50+1.54=17.60 cm2The total surface area of the remaining solid to the nearest cm2 is 18 cm2

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