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The length of a tarpaulin 5m wide will be required to make conical tent of height 9m and radius 10m?Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 120cm(Use π = 3.14).


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Answer 1

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

Given that,

Radius of cone, r = 10 m

Height of cone, h = 9 m

So, Let first evaluate the slant height of cone.

We know, slant height (l) of cone is evaluated as

$\red{ \boxed{ \sf{ \: {l}^{2} = {h}^{2} + {r}^{2}}}}$

Where,

l is slant height of cone

h is height of cone

r is radius of cone.

So, on substituting the values of h and r, we get

$\rm :\longmapsto\: {l}^{2} = {10}^{2} + {9}^{2}$

$\rm :\longmapsto\: {l}^{2} = 100 + 81$

$\rm :\longmapsto\: {l}^{2} = 181$

$\bf\implies \:l = \sqrt{181} \: = \: 13.45 \: m$

Now, We evaluate the Curved surface area of cone.

We know, Curved Surface Area of cone is given by

$\red{ \boxed{ \sf{ \:CSA_{(cone)} \: = \: \pi \: r \: l \: }}}$

So, on substituting the values, we get

$\rm :\longmapsto\:CSA_{(cone)} = 3.14 \times 10 \times 13.45$

$\bf :\longmapsto\:CSA_{(cone)} =422.33 \: {m}^{2}$

Now,

Tarpaulin of 5 m wide and of length L m is required to make a conical tent of height of 9 m and radius 10 m.

So,

Area of tarpaulin = Curved Surface Area of cone

$\rm :\longmapsto\:L \times 5 = 422.33$

$\bf\implies \:L = \dfrac{422.33}{5} = 84.466 \: m$

Also,

The extra length of material that will be required for stitching margins and wastage in cutting is approximately 120cm = 1.2 m

So,

$\rm :\longmapsto\:Length \: of \: tarpaulin = 84.466 + 1.2 = 85.666 \approx \: 85.67 \: m$

Additional Information :

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²