A tent consist of a frustum …
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Sia ? 6 years, 4 months ago
Radius of base of frustum = 13 m (R)
Radius of base of conical cap = 7 m (r)
Slant height of conical cap = 12 m (l)
height of frustum = 8 m (h)
Slant height of frustum
{tex}L = \sqrt { n ^ { 2 } + ( R - r ) ^ { 2 } }{/tex}
{tex}\Rightarrow \quad L = \sqrt { ( 8 ) ^ { 2 } + ( 13 - 7 ) ^ { 2 } }{/tex}
{tex}\Rightarrow \quad L = \sqrt { 64 + 36 } = \sqrt { 100 } = 10 \mathrm { m }{/tex}
therefore, Canvas required for the tent = Curved Surface area of cone + Curved Surface area of the frustum
{tex}= \pi rl + \pi L ( R + r ){/tex}
{tex}= \frac { 22 } { 7 } \times 7 \times 12 + \frac { 22 } { 7 } \times 10 ( 13 + 7 ){/tex}
{tex}= \frac { 22 } { 7 } \times 84 + \frac { 22 } { 7 } \times 10 \times 20{/tex}
{tex}= \frac { 22 } { 7 } ( 84 + 200 ){/tex}
{tex}= \frac { 22 } { 7 } \times 284{/tex}
= 892.57 m2
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