Question

a comical tent of diameter 7m and height 3m is to be maid by tarpolin how much tarpolin of width 1m is required for tent​

Answer 1

Answer :-

$\:\\ \large{\underline{\underline{\green{Firstly,\;let's\;understand\;the\;concept\;used\;:-}}}}$

Here the concept of CSA of Cone and Area of Rectangular shaped Tarpaulin has been used. We know that Area of Tarpaulin used will be equal to the CSA of Cone. From that we can find out the length of the Tarpaulin.

Let's do it !!

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★ Formula Used :-

$\:\\ \large{\boxed{\sf{CSA\;of\;Cone\;\;=\;\bf{\pi rL}}}}$

$\: \\\large{\boxed{\sf{L^{2}\;\; = \;\bf{H^{2}\;+\;R^{2}}}}}$

$\:\\ \large{\boxed{\sf{Area\;of\;Rectangle\;=\;\bf{Length\;\times\;Breadth}}}}$

$\:\\ \large{\boxed{\sf{Area\;of\;Tarpaulin\;Used\;=\;\bf{CSA\;of\;Tent}}}}$

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★ Question :-

A conical tent of diameter 7m and height 3m is to be maid by tarpaulin. How much length of tarpaulin of width 1m is required for tent ?

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★ Solution :-

Given,

» Diameter of Tent = 7 m

» Radius of Tent = r = ½ × d = 3.5 m

» Height of the Tent = h = 3 m

» Width of Tarpaulin = 1 m

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~ For the Slant Height of the Tent :-

• Let the slant height be L.

$\:\\ \sf{:\Rightarrow\;\;\; L^{2}\;\; = \;\bf{H^{2}\;+\;R^{2}}}$

$\: \\\sf{:\Rightarrow\;\;\; L^{2}\;\; = \;\bf{(3)^{2}\;+\;(3.5)^{2}}}$

$\:\\ \sf{:\Rightarrow\;\;\; L^{2}\;\; = \;\bf{9\;+\;12.25} \: \; = \; \bf{21.25}}$

$\:\\ \sf{:\Rightarrow\;\;\; L\;\; = \;\bf{\sqrt{21.25} \: \; = \; \underline{\underline{4.61\;\;m}}}}$

$\: \\\large{\boxed{\boxed{\sf{Slant\;\:Height\;\;of\;\;the\;\;Tent\;\;=\;\;\bf{4.61\;\;m}}}}}$

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~ For the CSA of Conical Tent :-

$\: \\\qquad \large{\sf{:\Longrightarrow\;\;\; CSA\;of\;Cone_{(Tent)}\;\;=\;\bf{\pi rL}}}$

$\:\\ \qquad \large{\sf{:\Longrightarrow\;\;\; CSA\;of\;Cone_{(Tent)}\;\;=\;\bf{\dfrac{22}{\cancel{7}}\:\times\: \cancel{3.5} \;\times\: 4.61\;\; = \;\; \underline{\underline{50.71\;\;m^{2}}}}}}$

$\: \\\large{\boxed{\boxed{\sf{CSA\;\;of\;\;the\;\;Tent\;\;=\;\;\bf{50.71\;\;m^{2}}}}}}$

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~ For the Length of the Tarpaulin :-

$\: \\\qquad \large{\sf{:\Longrightarrow\;\;\; Area\;of\;Tarpaulin\;Used\;=\;\bf{CSA\;of\;Tent}}}$

$\:\\ \qquad \large{\sf{:\Longrightarrow\;\;\; Length\;\times\;Breadth \;=\;\bf{\pi rL}}}$

$\:\\ \qquad \large{\sf{:\Longrightarrow\;\;\; Length\;\times\;1 \;=\;\bf{50.71}}}$

$\:\\ \qquad \large{\sf{:\Longrightarrow\;\;\; Length\; \;=\;\bf{\underline{\underline{50.71\;\;m}}}}}$

$\:\\ \large{\underline{\underline{\rm{Thus,\;length\;of\;Tarpaulin\;Required\;is\;\;\boxed{\bf{50.71\;\;m}}}}}}$

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$\:\\ \large{\underbrace{\underbrace{\sf{More\;\;Formulas\;\;to\;\;Know\;\;:-}}}}$

• Volume of Cone = ⅓ × πr²h

• Volume of Cylinder = πr²h

• TSA of Cone = πrL + πr²

• Volume of Cuboid = Length × Breadth × Height

• Volume of Cube = (Side)³

Answer 2

Correct question :

A conical tent of diameter 7m and height 3m is to be maid by tarpaulin how much length of tarpaulin of width 1m is required for tent​?

Given,

A conical tent of diameter 7m and height 3m is to be maid by tarpaulin.

To find :

How much tarpaulin of width 1m is required for tent​.

Solution :

Diameter of conical tent = 7m

⇒ Radius of conical tent = 7/2 = 3.5 m

⇒ Height of tent = 3 m

Therefore,

⇒ Slant height, l = √[(3.5)² + (3)²]

⇒ l = √[12.25 + 9]

⇒ l = √[21.25]

⇒ l = 4.61 m

Now, tarpaulin is CSA of tent.

⇒ CSA of cone = πrl

⇒ CSA of tent = 22/7 * 3.5 * 4.61

⇒ CSA of tent = 50.71 m²

Now,

⇒ Area of tarpaulin materials = CSA of tent

⇒ Length * 1 = 50.71

⇒ Length = 50.71/1

⇒ Length = 50.71 m

Therefore,

50.71 m of tarpaulin is required for tent.