Math, asked by snehalthakre220407, 16 days ago

A conical tent is made of 4.5m wide tarpaulin. If the height of the tent is 4m and the radius of the base is 3m then find the length of the tarpaulin used, assuming that the extra length of material that will be required for the stitching margins and wastage in cutting is approximately 20cm. (Pi = 3.14) ​

Answers

Answered by mathdude500
6

\large\underline{\sf{Solution-}}

Let r, h and L represents the radius, height and slant height of the cone respectively.

Given that,

  • Radius of cone, r = 3 m

  • Height of cone, h = 4 m

Now, we know, slant height L, height h and radius r is connected by the relationship

\rm \:  {L}^{2} =  {r}^{2} +  {h}^{2} \\

On substituting the values of r and h, we get

\rm \:  {L}^{2} =  {3}^{2} +  {4}^{2} \\

\rm \:  {L}^{2} =  9 + 16 \\

\rm \:  {L}^{2} =  25 \\

\rm \:  {L}^{2} =   {5}^{2}  \\

\rm\implies \:L = 5 \: m \\

Now, area of canvas required to make the tent is equals to Curved surface area of cone of slant height L and radius r.

So,

\rm \: Area_{(Canvas\:required)} = CSA_{(Cone)} \\

\rm \: Area_{(Canvas\:required)} = \pi \: r \: L \\

\rm \: Area_{(Canvas\:required)} = 3.14 \times 3 \times 5 \\

\rm\implies \:Area_{(Canvas\:required)} = 47.1 \:  {m}^{2}  \\

Now,

\rm \: Length \: of \: taruplin \: required = \dfrac{Area_{(Canvas\:required)}}{Width \: of \: tarpaulin}  \\

\rm \: Length \: of \: taruplin \: required =  \dfrac{47.1}{4.5}   \\

\rm\implies \: Length \: of \: taruplin \: required =  10.47 \: m \:  \{approx. \}  \\

Now, further given that the extra length of material that will be required for the stitching margins and wastage in cutting is approximately 20 cm = 0.2 m

Hence,

\rm\implies \: Length \: of \: taruplin \: required =  10.47 + 0.2 = 10.67 \: m  \\

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r  \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} =  \dfrac{4}{3}\pi {r}^{3}  }\\ \\ \bigstar \: \bf{Volume_{(cube)} =  {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}

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