A tent is in the shape …
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Sia ? 6 years, 6 months ago
For conical portion, r = 2m and {tex}l{/tex} = 2.8 m. Let S1 the curved surface area of conical portion. Then,
{tex}S _ { 1 } = \pi r l = \pi \times 2 \times 2.8 \mathrm { m } ^ { 2 } = 5.6 \pi \mathrm { m } ^ { 2 }{/tex}
For cylindrical portion, we have r = 2m ,h = 2.1m
Let S2 be the curved surface area of cylindrical portion. Then
{tex}S _ { 2 } = 2 \pi r h = 2 \pi \times 2 \times 2.1 \mathrm { m } = 8.4 \pi \mathrm { m } ^ { 2 }{/tex}
Let S be the area of the canvas used. Then,
S = S1 + S2 = {tex}( 5.6 \pi + 8.4 \pi ) \mathrm { m } ^ { 2 } = 14 \times \frac { 22 } { 7 } \mathrm { m } ^ { 2 } = 44 \mathrm { m } ^ { 2 }{/tex}
Total cost of the canvas at the rate of Rs 500 per m2 = Rs {tex}( 500 \times 44 ){/tex}= Rs 22000
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