Answer: Option C $$\eqalign{
& l = 10\,m \cr
& h = 8\,m \cr
& So,\,r = \sqrt {{l^2} - {h^2}} = \sqrt {{{\left( {10} \right)}^2} - {8^2}} = 6\,m \cr
& \therefore {\text{Curved}}\,{\text{surface}}\,{\text{area}} \cr
& \pi \,rl = \left( {\pi \times 6 \times 10} \right){m^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 60\pi \,{m^2} \cr} $$
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