Height of the cylindrical part = H = 8 m

Height of the conical part = h = (13 – 8) m = 5 m

Diameter = 24 m

Its radius

\[\begin{array}{*{35}{l}}

=\text{ }24/2\text{ }=\text{ }12\text{ }m  \\

Then,\text{ }slant\text{ }height\text{ }\left( l \right)\text{ }=\text{ }{{\left( {{r}^{2~}}+\text{ }{{h}^{2}} \right)}^{1/2}}  \\

=\text{ }{{\left( {{12}^{2}}~+\text{ }{{5}^{2}} \right)}^{1/2}}~=\text{ }{{\left( 144\text{ }+\text{ }25 \right)}^{1/2}}~=\text{ }{{\left( 169 \right)}^{1/2}}~=\text{ }13\text{ }m  \\

\left( i \right)\text{ }Total\text{ }surface\text{ }area\text{ }of\text{ }the\text{ }tent\text{ }=\text{ }2\pi rh\text{ }+\text{ }\pi rl\text{ }=\text{ }\pi r\left( 2h\text{ }+\text{ }l \right)  \\

=\text{ }22/7\text{ x }12\text{ x }\left( 2\text{ x }8\text{ }+\text{ }13 \right)  \\

=\text{ }264/7\text{ }\left( 16\text{ }+\text{ }3 \right)  \\

=\text{ }264/7\text{ x }29  \\

=\text{ }7656/7\text{ }{{m}^{2}}  \\

=\text{ }1093.71\text{ }{{m}^{2}}  \\

\end{array}\]

(ii) Area of canvas used in stitching = total area of canvas

Total area of canvas = 7656/7 + total area of canvas/10

Total area of canvas – Total area of canvas/10 = 7656/10

\[\begin{array}{*{35}{l}}

Total\text{ }area\text{ }of\text{ }canvas\text{ }\left( 1\text{ }\text{ }1/10 \right)\text{ }=\text{ }7656/7  \\

Total\text{ }area\text{ }of\text{ }canvas\text{ }x\text{ }9/10\text{ }=\text{ }7656/7  \\

Total\text{ }area\text{ }of\text{ }canvas\text{ }=\text{ }7656/7\text{ x }10/9\text{ }=\text{ }76560/63  \\

Therefore,\text{ }the\text{ }total\text{ }area\text{ }of\text{ }canvas\text{ }=\text{ }1215.23\text{ }{{m}^{2}}  \\

\end{array}\]