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Question

# From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. [CBSE 2014]

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Solution

## $\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{cone}=\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{cylinder}=h=2.8\mathrm{cm}\mathrm{and}\phantom{\rule{0ex}{0ex}}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{base},r=\frac{4.2}{2}=2.1\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{slant}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{cone},l=\sqrt{{r}^{2}+{h}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{2.{1}^{2}+2.{8}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4.41+7.84}\phantom{\rule{0ex}{0ex}}=\sqrt{12.25}\phantom{\rule{0ex}{0ex}}=3.5\mathrm{cm}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{the}\mathrm{total}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{solid}=\mathrm{CSA}\mathrm{of}\mathrm{cylinder}+\mathrm{CSA}\mathrm{of}\mathrm{cone}+\mathrm{Area}\mathrm{of}\mathrm{a}\mathrm{base}\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi }rh+\mathrm{\pi }rl+\mathrm{\pi }{r}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }r\left(2h+l+r\right)\phantom{\rule{0ex}{0ex}}=\frac{22}{7}×2.1×\left(2×2.8+3.5+2.1\right)\phantom{\rule{0ex}{0ex}}=22×0.3×\left(5.6+5.6\right)\phantom{\rule{0ex}{0ex}}=6.6×11.2\phantom{\rule{0ex}{0ex}}=73.92{\mathrm{cm}}^{2}$ So, the total surface area of the remaining solid is 73.92 cm2.

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