Question

7.
What length of tarpaulin 4 m wide will be required to make conical tent of height 24 IR
and base diameter 14 m? Assume that the extra length of material that will be required
for stitching margins and wastage in cutting is approximately 50 cm. Also, find total
cost of tarpaulin if 1 m² cost Rs. 11.

​

Answer 1

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Given :-

• Width of Tarpaulin = 4 m
• Diameter of conical tent = 14m => radius, r = 7 m
• Height of conical tent, h = 24 m.
• Extra length of material that will be required for stitching margins and wastage in cutting is approximately 50 cm.

To find :-

☆ Length of Tarpaulin

☆ Total cost of Tarpaulin @ Rs 11 per m² .

$\large\bold\red{Formula \: used :-}$

$\small\bold\blue{\small\bold\blue{slant \: height \: of \: cone \: \: \: \: \: \: {l}^{2} = {r}^{2} + {h}^{2} } }$

$\small\bold\blue{curved \: surface \: area \: of \: cone \: = \pi \: rl}$

$\large\bold\red{Solution:- }$

$slant \: height \: of \: cone \\ {l}^{2} = {r}^{2} + {h}^{2} \\ {l}^{2} = {24}^{2} + {7}^{2} \\ {l}^{2} = 576 + 49 \\ {l}^{2} = 625 \\ l = 25 \: m$

$Curved \: Surface \: area \: of \: Cone \: = \pi \: rl \\ = \frac{22}{7} \times 7 \times 25 \\ = 22 \times 25 \\ = 550 \: {m}^{2}$

$Length \: of \: Tarpaulin = \\ \frac{Curved \: Surface \: area \: of \: Cone }{breadth \: of \: Tarpaulin } + extra \: wastage \: length \\ = \frac{550}{4} + \frac{50}{100} \\ = \frac{275}{2} + \frac{1}{2} \\ = \frac{276}{2} = 138 \: m$

$area \: of \: tarpaulin \: = \: l \times b = 138 \times 4 = 552 \: {m}^{2} \\ cost \: of \: tarpaulin \: at \: the \: rate \: of \: Rs\: 11 \: per \: {m}^{2} = \\ 138 \times 4 \times 11 \\ = Rs \: 6072$

$\huge \fcolorbox{black}{cyan}{♛hope \: it \: helps \: you♛}$