Question

What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use = 3.14).​

Answer 1

Answer:

h = Height of tent = 8 m

r = Base radius = 6 m

Therefore,

l= √r²+h² =10 m

Curved surface area of the cone = πrl=3.14×6×10m²

Therefore, length of 3 m wide tarpaulin required = 3.14×6×10/3=62.8 m

Extra length required = 0.2 m

Therefore, total length required = 63m

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Answer 2

$\huge { \boxed{\mathbb{ANSWER}}}$

$\large \underline{ \underline{ \frak{ \color{magenta}{Given:}}}}$

$\mapsto \sf{Radius(r) \: of \: conical \: tent=6m}$

$\mapsto \sf{Height(h) \: of \: conical \: tent=8m}$

$\large \underline{ \underline{ \frak{ \color{black}{To \: find:}}}}$

$\leadsto \sf{Total \: length \: of \: tarpaulin. }$

$\large \underline{ \underline{ \frak{ \color{orange}{Formula \: required:}}}}$

$\dag\underline{ \green{\boxed{ \pink{ \rm{Curved \: surface \: area \: of \: cone=\pi rl}}}}}$

$\dag\underline{ \green{\boxed{ \pink{ \rm{Slant \: height (l) \: of \: conical \: tent= \sqrt{ {r}^{2} + {h}^{2} } }}}}}$

$\large \underline{ \underline{ \frak{ \color{silver}{According \: to \: Question:}}}}$

$\bf \underline{Let's \: start \: now!!!}$

${ \dashrightarrow \sf{Slant \: height (l) \: of \: conical \: tent= \sqrt{ {r}^{2} + {h}^{2} } }}$

${ \dashrightarrow \sf{Slant \: height (l) \: of \: conical \: tent= \sqrt{ {(8)}^{2} + {(6)}^{2} } }}$

${ \dashrightarrow \sf{Slant \: height (l) \: of \: conical \: tent= \sqrt{ 64 + 36 } }}$

${ \dashrightarrow \sf{Slant \: height (l) \: of \: conical \: tent= \sqrt{ 100 } }}$

${ \dashrightarrow \sf{Slant \: height (l) \: of \: conical \: tent= 10m }}$

$\bf \underline{Then, }$

${ \dashrightarrow \sf{Curved \: surface \: area \: of \: conical \: tent=\pi rl}}$

${ \dashrightarrow \sf{Curved \: surface \: area \: of \: conical \: tent= (3.14 \times 6 \times 10) {m}^{2} }}$

${ \dashrightarrow \sf{Curved \: surface \: area \: of \: conical \: tent= 188.4{m}^{2} }}$

$\bf \underline{Now, }$

$\dashrightarrow \sf{Area_{(tent)}=Area_{(tarpaulin)}}$

${ \dashrightarrow \sf{Area_{(tent)} = (length) \times breadth}}$

${ \dashrightarrow \sf{Area_{(tent)} = length \times 3}}$

${ \dashrightarrow \sf{length \: of \: tarpaulin = \frac{1}{3}(Area \: of \: tent) }}$

${ \dashrightarrow \sf{length \: of \: tarpaulin = \frac{1}{3} \times 188.4m }}$

${ \dashrightarrow \sf{length \: of \: tarpaulin = 62.8m }}$

$\bf \underline{Again, }$

${\dashrightarrow \sf{Total \: length \: of \: tarpaulin=length \: of \: tarpaulin+Margin}}$

${\dashrightarrow \sf{Total \: length \: of \: tarpaulin=62.8m + 20cm}}$

${\dashrightarrow \sf{Total \: length \: of \: tarpaulin=62.8m + 20 \times \frac{1}{100}m }}$

${\dashrightarrow \sf{Total \: length \: of \: tarpaulin=62.8m + 2 \cancel0 \times \frac{1}{10 \cancel0}m }}$

${\dashrightarrow \sf{Total \: length \: of \: tarpaulin=62.8m +0.2m}}$

${\dashrightarrow \sf{Total \: length \: of \: tarpaulin=63.0m}}$

$\bf \underline{Hence, }$

$\: \: \: \: \: \star\underline{ \green {\boxed{ \red{ \rm{Total \: length \: of \: tarpaulin=63.0m}}}}}$

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