Question Number 47250 by Raj Singh last updated on 07/Nov/18
Answered by MJS last updated on 07/Nov/18
$$\mathrm{surface}\:\mathrm{of}\:\mathrm{cuboid} \\ $$$$\mathrm{2}×\left(\mathrm{15}×\mathrm{10}+\mathrm{15}×\mathrm{5}+\mathrm{10}×\mathrm{5}\right)=\mathrm{550} \\ $$$$\mathrm{surface}\:\mathrm{of}\:\mathrm{cylinder}\:\left(\mathrm{without}\:\mathrm{bottom\&top}\right) \\ $$$$\mathrm{2}×\mathrm{3}.\mathrm{5}\pi×\mathrm{5}=\mathrm{35}\pi \\ $$$$\mathrm{surface}\:\mathrm{of}\:\mathrm{bottom\&top} \\ $$$$\mathrm{2}×\left(\mathrm{3}.\mathrm{5}\right)^{\mathrm{2}} \pi=\frac{\mathrm{49}}{\mathrm{2}}\pi \\ $$$$\mathrm{surface}\:\mathrm{of}\:\mathrm{remaining}\:\mathrm{body} \\ $$$$\mathrm{550}+\mathrm{35}\pi−\frac{\mathrm{49}}{\mathrm{2}}\pi=\mathrm{550}+\frac{\mathrm{21}}{\mathrm{2}}\pi\:\approx\mathrm{582}.\mathrm{99cm}^{\mathrm{2}} \\ $$