Question

The interior of a building is in the form of a right circular cylinder of diameter 4.2 metre and height 4 metre surmounted by a cone. The vertical height of cone is 2.1m find the outer surface area and volume of the building (use π = 22/7)

● please give me solution ( answer : 72.3980m² and 65.142m³)
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Answer 1

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

Here the concept of CSA of Cone and CSA of Cylinder has been used In order to find the volume of the building, we need to add the volume of cone and volume of Cylinder. And for outer surface area of building, we need to add the CSA of Cone and CSA of Cylinder since we are excluding the bases.

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★ Formula Used :-

$\: \\ \large{\boxed{\sf{(Slant Height)^{2},\; L^{2} \: \: = \: \bf{(Height)^{2},\: H^{2} \; + \; (Base)^{2}, \: B^{2}}}}}$

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

$\: \\ \large{\boxed{\sf{CSA \: of \: Cylinder \: = \: \bf{2 \pi rh}}}}$

$\: \\ \large{\boxed{\sf{Outer\: Surface\: Area \: of \: Building \: = \: \bf{CSA \: of \: Cone \: + \: CSA \: of \: Cylinder}}}}$

$\: \\ \large{\boxed{\sf{Volume \: of \: Cone \: = \: \bf{\dfrac{1}{3} \: \times \: \pi r^{2}h}}}}$

$\: \\ \large{\boxed{\sf{Volume \: of \: Cylinder \: = \: \bf{ \pi r^{2}h}}}}$

$\: \\ \large{\boxed{\sf{Volume \: of \: Building \: = \: \bf{Volume \: of \: Cone \: + \: Volume \: of \: Cylinder}}}}$

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★ Question :-

The interior of a building is in the form of a right circular cylinder of diameter 4.2 metre and height 4 metre surmounted by a cone. The vertical height of cone is 2.1m find the outer surface area and volume of the building.

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★ Solution :-

Given,

» Height of the cylindrical part = 4 m

» Height of the conical part = 2.1 m

» Diameter of base of cylinder = d = 4.2 m

Since, the cylinder is surmounted by cone, the radius of cone will be equal to the radius of cylinder.

» Radius of base of cone and cylinder =

= ½ × d

= ½ × 4.2 m

= 2.1 m

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~ For the CSA of Cone :-

$\: \\ \large{\sf{\rightarrow \:\;(Slant Height)^{2},\; L^{2} \: \: = \: \bf{(Height)^{2},\: H^{2} \; + \; (Base)^{2}, \: B^{2}}}}$

$\: \\ \large{\sf{\rightarrow \:\;L^{2} \: \: = \: \bf{(2.1\; m)^{2} \; + \; (2.1\;m)^{2}\;\;=\;\;4.41\; m^{2} \; + \; 4.41\;m^{2}}}}$

$\: \\ \large{\sf{\rightarrow \:\;\ L \: \: = \: \bf{\sqrt{8.82\;m^{2}}\: \: = \: \: \underline{2.97\; m}}}}$

Then,

$\: \\ \large{\sf{:\longrightarrow \; \; \: CSA \: of \: Cone \: = \: \bf{\dfrac{22}{7}\:\times\: 2.1 \: m \:\times\: 2.97 \: m \: \: = \: \: \underline{\underline{19.598 \;\: m^{2}}}}}}$

~ For the CSA of Cylinder :-

$\: \\ \large{\sf{:\longrightarrow \;\:\;CSA \: of \: Cylinder \: = \: \bf{2 \pi rh}}}$

$\: \\ \large{\sf{:\longrightarrow \;\:\;CSA \: of \: Cylinder \: = \: \bf{2\:\times\:\dfrac{22}{7}\:\times\:(2.1\;m)\:\times\:(4\;m) \: \: = \: \: \underline{\underline{52.8 \; m^{2}}}}}}$

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~ For Outer Surface Area of Building :-

$\: \\ \large{\sf{:\Longrightarrow \;\;\: Outer\:Surface\:Area\:of\:Building\:=\: \bf{CSA\:of\:Cone\:+\:CSA\:of\:Cylinder}}}$

$\: \\ \large{\sf{:\Longrightarrow \;\;\: Outer\:Surface\:Area\:of\:Building\:=\: \bf{19.598\;m^{2}\:+\:52.8\: m^{2} \:\:=\:\: \underline{\underline{72.3980\;m^{2}}}}}}$

$\: \\\large{\underline{\underline{\rm{\leadsto \;\:\;Outer\;Surface\;Area\;of \;Building\; is\;\:\;\boxed{\bf{72.3980\;\: m^{2}}}}}}}$

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~ For Volume of Cone :-

$\: \\ \large{\sf{:\longrightarrow \:\;\; Volume\:of \:Cone\: = \: \bf{\dfrac{1}{3} \: \times \: \dfrac{22}{7}\:\times\: r^{2}h}}}$

$\: \\ \large{\sf{:\longrightarrow \:\;\; Volume\:of\:Cone \: = \: \bf{\dfrac{1}{3} \: \times \: \dfrac{22}{7}\:\times\:(2.1\;m)^{2}\:\times\:(2.1\;m) \:\:=\: \: \underline{\underline{9.702\;m^{3}}}}}}$

~ For Volume of Cylinder :-

$\: \\ \large{\sf{:\longrightarrow \:\;\; Volume\:of\:Cylinder\:=\:\bf{ \pi r^{2}h}}}$

$\: \\ \large{\sf{:\longrightarrow\:\;\; Volume \: of \: Cylinder \: = \: \bf{\dfrac{22}{7}\:\times\: (2.1\;m)^{2}\:\times\:(4\;m) \: \: = \: \: \underline{\underline{55.44 \; m^{3}}}}}}$

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~ For the Volume of Building :-

$\: \\ \large{\sf{:\Longrightarrow \:\;\; Volume \: of \: Building \: = \: \bf{Volume \: of \: Cone\: + \:Volume \: of \: Cylinder}}}$

$\: \\ \large{\sf{:\Longrightarrow \:\;\; Volume\:of\:Building\:=\:\bf{9.702\: m^{3} \: + \:55.44\: m^{3}\:\:=\:\: \underline{\underline{65.142\;m^{3}}}}}}$

$\: \\ \large{\boxed{\boxed{\rm{\odot\;\:\:Volume\:of\:the\:Building\:=\: \bf{65.142\;\;m^{3}}}}}}$

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$\:\\\large{\underbrace{\underbrace{\sf{\mapsto \:\;\; More\;Formulas\;to\;Explore\;}}}}$

• Volume of Cuboid = Length × Breadth × Height

• Volume of Cube = (Side)³

• Volume of Hemisphere = ⅔ × πr³

*Note :- Here approximate values are used for decimals

Answer 2

$\LARGE{ \underline{ \bf{Required \: answer:}}}$

GiveN:

• Diameter = 4.2 m
• Height of cylinder = 4 m
• Height of cone = 2.1 m

To FinD:

• Outer surface area of the building?
• Volume of the building?

Step-by-step calculation:

The outer surface area of the building will be = Curved surface area of cone + Curved surface area of the cylinder.

We know that,

• CSA of cylinder = 2πrh
• CSA of cone = πrl

Here,

• Radius = 2.1 m
• Height of cylinder = 4 m
• Slant height of cone = √(2.1)² + (2.1)² = 2.1√2 m

Finding CSA of the building:

= 2πrh + πrl

= πr(2h + l)

= 22/7 × 2.1 m(2×4m + 2.1√2m)

= 22 × 0.3m(8 m + 2.97m)

= 22 × 0.3m × 10.97m

= 72.4 m² (approx.)

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Now finding the volume

We know that,

• CSA of cylinder = πr²h
• CSA of cone = 1/3πr²h'

Additionally,

• Height of the cone = 2.1m

Finding Vol. of the building:

= πr²h + 1/3πr²h'

= πr²(h + 1/3h')

= 22/7 × (2.1)²(4m + 1/3(2.1)m)

= 22/7 × 4.41m² (4.7m)

= 22 × 0.63m² × 4.7m

= 65.142 m³

Hence,

The required outer surface area and volume of the building are:

$\large{ \boxed{ \bf{ \purple{72.4 \: {m}^{2} \: and \: 65.142 \: {m}^{3} }}}}$