Question

Q. What length of tarpaulin 3 m wide will be required to make conical tent of height 8m 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 t = 3.14).​

Answer 1

Step-by-step explanation:

Given a conical tent with

Height ( h ) = 8m

Radius ( r ) = 6m

So, its length = l2 = h2 + r2

= l2 = ( 8 )2m + ( 6 )m2

= l2 = 64m + 36m

= l2 = 100m

= l = 10m

Curved surface area of the cone = TTrl

= 3.14 x 6m x 10m

= 188.4 cm2

The extra length of material that will be required for stiching margins and wastages in cutting is approx 20 cm = 0.2m

So, its length = l - 0.2m

and its width = 3m

area of sheet = CSA of tent

so the area of the trapaulln = l x b

= ( l - 0.2m ) x 3m = 188.4m2

= l - 0.2m = 62.8m2

= l = 63m

Therefore, length of the trapaulin sheet required is 63m.

Answer 2

Trapaulian will be curved surface area of tent

Lets find curved surface area first

Curved surface area of tent=$\pi$rl

r=6m , h=8m

Let slant height be=l

We know that

$\rm {l}^{2} = { h}^{2} + {r}^{2} \\ \rm {l}^{2} = {(8)}^{2} + ( {6)}^{2} \\ \rm {l}^{2} = 64 + 36 \\ \rm {l}^{2} = 100 \\ \rm l = \sqrt{100} \\ \rm l = \sqrt{ {10}^{2} } \\ \rm l = 10 \:m$

Curved surface area of tent=$\pi$rl

$\rm = (3.14 \times 6 \times 10) \: {m}^{2} \\ \rm = 188.4 \: {m}^{2}$

Now,

Area of tarpaulin material=Area of tent

Length×Breadth=Area of tent

Length×3=Area of tent

Length=$\frac{1}{3}$(Area of tent)

$\rm = \frac{188.4}{3} \\ \rm = 62.8 \: m$

Length=62.8 m

Now, given that margin is 20 cm

Total length=Length calculated + Margin

=62.8m+20cm

$\rm = 62.8m + 20 \times \dfrac{1}{100} m$

=62.8m+0.2m

=63m

$\begin{gathered}\rule{190pt}{2pt} \\ \end{gathered}$

${ \red{ \rm{Additional\:Information}}}$

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