Question

The circumference of the base of 20m high conical tent is 44m.Calculate the length of canvas used in making the tent if width of the canvas is 2m. * The circumference of the base of 20m high conical tent is 44m. Calculate the length of canvas used in making the tent if the width of the canvas is 2m. *​

Answer 1

Appropriate Question :-

The circumference of the base of 20m high conical tent is 44m. Calculate the length of canvas used in making the tent if width of the canvas is 2m.

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

Let assume that radius of base of conical tent be r m

It is given that Circumference of base = 44 m

$\rm\implies \:2\pi \: r \: = \: 44$

$\rm \: 2 \times \dfrac{22}{7} \times r = 44$

$\rm\implies \:\boxed{ \rm{ \:r \: = \: 7 \: m \: \: }}$

Now, we have

• Radius of cone, r = 7 m

• Height of cone, h = 20 m

Let assume that

• Slant height of cone = l m

Now, we know,

$\rm \: l = \sqrt{ {h}^{2} + {r}^{2} }$

$\rm \: l = \sqrt{ {20}^{2} + {7}^{2} }$

$\rm \: l = \sqrt{400 + 49}$

$\rm \: l = \sqrt{449}$

$\rm\implies \:\boxed{ \rm{ \:l \: = \: 21.19 \: m \: }}$

Now, canvas used in making the tent is equals to curved surface area of cone of radius 7 m and slant height 21.19 m

$\rm \: Curved \: Surface \: Area _{(cone)} = \pi \: r \: l$

$\rm \: Curved \: Surface \: Area _{(cone)} = \frac{22}{7} \times 7 \times 21.19$

$\rm\implies\boxed{\rm{Curved \: Surface \: Area _{(cone)} = 466.18 \: {m}^{2} \:}}$

Now, further given that width of canvas used in making length is 2 m.

So,

$\rm \: Length \: of \: canvas \: = \dfrac{Curved \: Surface \: Area _{(cone)}}{width}$

$\rm \: Length \: of \: canvas \: = \dfrac{466.18}{2}$

$\rm\implies \:\boxed{ \rm{ \: Length \: of \: canvas \: = 233.09 \: m \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

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