Question

A ᴛᴇɴᴛ ᴏғ ᴀ ᴄɪʀᴄᴜꜱ ɪꜱ ꜱᴜᴄʜ ᴛʜᴀᴛ ɪᴛ'ꜱ ʟᴏᴡᴇʀ ᴘᴀʀᴛ ɪꜱ ᴄʏʟɪɴᴅʀɪᴄᴀʟ ᴀɴᴅ ᴜᴘᴘᴇʀ ᴘᴀʀᴛ ɪꜱ ᴄᴏɴɪᴄᴀʟ. Tʜᴇ ᴅɪᴀᴍᴇᴛᴇʀ ᴏғ ᴛʜᴇ ʙᴀꜱᴇ ᴏғ ᴛʜᴇ ᴛᴇɴᴛ ɪꜱ 48 ᴍ ᴀɴᴅ ᴛʜᴇ ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ᴄʏʟɪɴᴅʀɪᴄᴀʟ ᴘᴀʀᴛ ɪꜱ 15 ᴍ. Tᴏᴛᴀʟ ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ᴛᴇɴᴛ ɪꜱ 33 ᴍ. Fɪɴᴅ ᴀʀᴇᴀ ᴏғ ᴄᴀɴᴠᴀꜱ ʀᴇϙᴜɪʀᴇᴅ ᴛᴏ ᴍᴀᴋᴇ ᴛʜᴇ ᴛᴇɴᴛ. Aʟꜱᴏ ғɪɴᴅ ᴠᴏʟᴜᴍᴇ ᴏғ ᴀɪʀ ɪɴ ᴛʜᴇ ᴛᴇɴᴛ.

♡_Tɦαɳҡ µɦ_♡
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Answer 1

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$\huge{ \underline{ \sf{ \green{Answer....}}}}$

♣️ GiveN:

• Tent with lower part cylindrical and upper part conical.
• Height of the cylinder = 15 m
• Total height of the tent = 33 m
• Diameter of the cylindrical portion = 48 m

♣️ To FinD:

• Area of the canvas needed....?!
• Volume of air inside the tent.....?!

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$\huge{ \underline{ \sf{ \green{Explanation \: of \: Q.}}}}$

The question is asking for finding the area of canvas of tent and the volume of air inside the tent. These required certain formulae to be used are:-

♠️ Formulae to be used:

$\large{ \ast \: { \red{ \sf{ CSA \: of \: cone= \pi \: r \: l}}}} \\ \\ \large{ \ast \: \: { \red{ \sf{volume \: of \: cone = \frac{1}{3}\pi \: {r}^{2}h}}}} \\ \\ \large{ \ast \: \: { \red{ \sf{CSA\: of \: cylinder = 2\pi rh}}}} \\ \\ \large{ \ast \: \: { \red{ \sf{volume \: of \: cylinder = \pi {r}^{2} h}}}}$

Where, l = slant height, h= height and r=radius

By using these formulae, we are going to solve the questions. ✏In the attachment, the rough figure of tent is provided. Pls refer to that.

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$\huge{ \underline{ \sf{ \green{Solution....}}}}$

The total cloth or canvas used is the curved surface area of the cone and the curved surface area of cylinder. The bottom part of cylinder is not canvased and the top part is not there.

So, the canvas required,

$\large{ \sf{ \dashrightarrow \boxed{ \sf{ \purple{\: CSA \: of \: cone + CSA \: of \: cylinder}}}}}$

We have height of cylinder = 15 m

Total height of tent = 33 m

Then, height of the cone = 33 m - 15 m = 18 m

Radius of the cone = Radius of cylinder = 48/2m = 24 m

As, CSA of cone required slant height

Slant height of cone,

$\large{ \sf{ \dashrightarrow \sqrt{ {r}^{2} + {h}^{2} } \: m}} \\ \\ \large{ \sf{ \dashrightarrow \: \sqrt{ {24}^{2} + {18}^{2} } \: m}} \\ \\ \large{ \sf{ \dashrightarrow \: \sqrt{576 + 324} \: m}} \\ \\ \large{ \sf{ \dashrightarrow \: \sqrt{900} \: m}} \\ \\ \large{ \sf{ \dashrightarrow \: \boxed{ \sf{ \purple{30 \: m}}}}}$

So, slant height of cone = 30 m

Now, we need to find the area of canvas required by using the required formulae,

By using formula,

$\large{ \sf{ \dashrightarrow \: \pi rl + 2\pi rh_{cylinder}}} \\ \\ \large{ \sf{ \dashrightarrow \: \pi r(l + 2h_{cylinder})}} \\ \\ \large{ \sf{ \dashrightarrow \: \frac{22}{7} \times 24(30 + 2 \times 15) \: {m}^{2} }} \\ \\ \large{ \sf{ \dashrightarrow \: \frac{22}{7} \times 24 \times 60 \: {m}^{2}}} \\ \\ \large{ \sf{ \dashrightarrow \: \frac{31680}{7} \: {m}^{2} }} \\ \\ \large{ \sf{ \dashrightarrow \: \boxed{ \sf{ \red{4525.71 \: {m}^{2} }}}}} \\ \\ \large{ \therefore{ \underline{ \sf{ \purple{Area \: of \: canvas = 4525.71 \: {m}^{2}}}}}}$

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So, volume of air inside tent,

$\large{ \sf{ \dashrightarrow \boxed{ \sf{ \purple{\: Volume \: of \: cone + Volume\: of \: cylinder}}}}}$

We have all our required values, putting the values in formulae and finding the volume of the air in tent.

By using formula,

$\large{ \sf{ \dashrightarrow \: \frac{1}{3}\pi {r}^{2} h_{cone} + \pi {r}^{2} h_{cylinder}}} \\ \\ \large{ \sf{ \dashrightarrow \: \pi {r}^{2}( \frac{1}{3} h_{cone} + h_{cylinder})}} \\ \\ \large{ \sf{ \dashrightarrow \: \frac{22}{7} \times 24 ^{2}( \frac{1}{3} \times 18 + 15) \: {m}^{3} }} \\ \\ \large{ \sf{ \dashrightarrow \: \frac{22}{7} \times 24 \times 24 \times 21 \: {m}^{3} }} \\ \\ \large{ \sf{ \dashrightarrow \: \cancel{\frac{22 \times 24 \times 24 \times 21}{7}} \: {m}^{3} }} \\ \\ \large{ \sf{ \dashrightarrow \: \boxed{ \sf{ \red{38016 \: {m}^{3} }}}}} \\ \\ \large{ \therefore{ \sf{ \underline{ \purple{Volume \: of \: air \: = 38016 \: {m}^{3} }}}}}$

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