A tailor has a piece of canvas whose area is 552 msquare.He uses it to make a conical tent with base radius 7m. Assuming that all stitching margins and wastage is 2msquare. Find the volume of the tent that can be made with it

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area of cloth wasted = 2 m

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area left for making tent = 550 m

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radius of base, r = 7 m

Let height = h

Let slant height =

*l*

$CSA=550\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\pi rl}=550\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{22}{7}\times 7\times \mathrm{l}=550\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{l}=25\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{now},\mathrm{h}=\sqrt{{\mathrm{l}}^{2}-{\mathrm{h}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{h}=\sqrt{625-49}=\sqrt{576}=24\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{volume}\mathrm{of}\mathrm{tent}=\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}=\frac{1}{3}\times \frac{22}{7}\times 49\times 24=1232{\mathrm{m}}^{3}$

$CSA=550\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\pi rl}=550\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{22}{7}\times 7\times \mathrm{l}=550\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{l}=25\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{now},\mathrm{h}=\sqrt{{\mathrm{l}}^{2}-{\mathrm{h}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{h}=\sqrt{625-49}=\sqrt{576}=24\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{volume}\mathrm{of}\mathrm{tent}=\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}=\frac{1}{3}\times \frac{22}{7}\times 49\times 24=1232{\mathrm{m}}^{3}$

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