Question

the interior of a building is in the form of a right circular cylinder of diameter 4 .2 m and height 4m surmounted by a cone the verticle height of cone is 2.1 m find the outer surface area of the building ​

Answer 1

Answer :-

$\;\\\large{\underbrace{\underline{\sf{Let's\;analyse\;the\;question\;first\;:-}}}}$

Here the concept of CSA of Cone and CSA of Cylinder has been used. We are given the dimensions of different solids. Now they are attached to each other by their circular base. We need to find their outer surface area. We can do this by adding the CSA of Cone and CSA of Cylinder since from outer view the bases of these aren't visible.

Let's do it !!

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★ Formula Used :-

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

$\:\\\large{\boxed{\sf{L^{2}\;=\;\bf{r^{2}\;+\;h^{2}}}}}$

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

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

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★ Solution :-

Given,

» Diameter of Cylindrical Part = 4.2 m

» Radius of Cylindrical Part = ½ × 4.2 = 2.1 m

» Height of Cylindrical Part = 4 m

» Height of Conical Part = 2.1 m

Since both the cylindrical and conical parts are attached with bases, their radius will be equal.

» Radius of Conical Part = 2.1 m

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~ For the Outer Surface Area of Cylinder :-

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

$\\\:\;\;\;\;\large{\sf{:\rightarrow\;\;\:CSA\;of\;Cylinder\;=\;\bf{2\:\times\:\dfrac{22}{7}\:\times\:2.1\:\times\:4}}}$

$\\\:\;\;\;\;\large{\sf{:\rightarrow\;\;\:CSA\;of\;Cylinder\;=\;\bf{2\:\times\:22\:\times\:0.3\:\times\:4\;\:=\;\:\underline{\underline{52.8\;\;m^{2}}}}}}$

$\\\:\large{\boxed{\boxed{\tt{Outer\;Surface\;Area\;of\;Cylindrical\;Part\;=\;\bf{52.8\;\;m^{2}}}}}}$

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~ For the Slant Height of the Conical Part :-

Let the slant height of the Conical Part be L.

Then,

$\\\:\;\;\;\;\;\large{\sf{:\Rightarrow\;\;\; L^{2}\;=\;\bf{r^{2}\;+\;h^{2}}}}$

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

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

$\\\:\;\;\;\;\;\large{\sf{:\Rightarrow\;\;\; (L)^{2}\;=\;\bf{8.82}}}$

$\\\:\;\;\;\;\;\large{\sf{:\Rightarrow\;\;\; (L)\;=\;\bf{\sqrt{8.82}\;\:=\;\:\underline{\underline{2.97\;\;m}}}}}$

$\:\\\large{\boxed{\boxed{\tt{Slant\;Height\;of\;Conical\;Part\;=\;\bf{2.97\;\;m}}}}}$

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~ For the Outer Surface Area of Cone :-

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

$\\\:\;\;\;\;\large{\sf{:\rightarrow\;\;\:CSA\;of\;Cone\;=\;\bf{\dfrac{22}{7}\:\times\:2.1\:\times\:2.97}}}$

$\\\:\;\;\;\;\large{\sf{:\rightarrow\;\;\:CSA\;of\;Cone\;=\;\bf{22\:\times\:0.3\:\times\:2.97\;\:=\;\:\underline{\underline{19.602\;\;m^{2}}}}}}$

$\:\\\large{\boxed{\boxed{\tt{Outer\;Surface\;Area\;of\;Conical\;Part\;=\;\bf{19.602\;\;m^{2}}}}}}$

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~ For the Outer Surface Area of Building :-

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

$\\\:\;\;\;\;\large{\sf{:\Longrightarrow\;\;\:Outer\;Surface\;Area\;of\;Building\;=\;\bf{52.8\;+\;19.602\;\:=\;\:\underline{\underline{72.402\;\;m^{2}}}}}}$

$\\\;\large{\underline{\underline{\rm{\;\mapsto\;\;Thus,\;outer\;surface\;area\;of\;building\;is\;\;\boxed{\bf{72.402\;\;m^{2}}}}}}}$

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★ More Formulas to Know :-

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

$\\\:\sf{\leadsto\;\;\;Volume\;of\;Cone\;=\;\dfrac{1}{3}\:\times\:\pi r^{2}h}$

$\\\:\sf{\leadsto\;\;\;Volume\;of\;Hemisphere\;=\;\dfrac{2}{3}\:\times\:\pi r^{3}}$

$\\\:\sf{\leadsto\;\;\;Volume\;of\;Hemisphere\;=\;\dfrac{4}{3}\:\times\:\pi r^{3}}$

$\\\:\sf{\leadsto\;\;\;TSA\;of\;Cylinder\;=\;2\pi r^{2} \;+\;2\pi rh}$

$\\\:\sf{\leadsto\;\;\;TSA\;of\;Cone\;=\;\pi r^{2} \;+\;\pi r}$

$\\\:\sf{\leadsto\;\;\;Volume\;of\;Cube\;=\;(Side)^{3}}$

$\\\:\sf{\leadsto\;\;\;Volume\;of\;Cuboid\;=\;Length\:\times\:Breadth\:\times\;Height}$

Answer 2

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★ For the Outer Surface Area of Cylinder :-

→CSA of Cylinder=2πrh

→CSA of Cylinder=2× 22/7 ×2.1×4

→CSA of Cylinder= 2×22×0.3×4= 52.8m²

Outer Surface area of cylindrical part=52.8m²

For the Slant Height of the Conical Part :-

Let the slant height of the Conical Part be L.

Then,

⇒L² =r² +h²

⇒(L)² =(2.1)² +(2.1)²

⇒(L)² =4.41+4.41

⇒(L)² =8.82

⇒(L)= 8.82 = 2.97m

Slant height of conical part=2.97m

For the Outer Surface Area of Cone :-

→ CSA of Cone=πrl

→ CSA of Cone= 22/7 ×2.1×2.97

→ CSA of Cone =22×0.3×2.97= 19.602m²

Outer surface area of conical part =19.602m²

For the Outer Surface Area of Building :-

⟹Outer Surface Area of Building=CSA of Cone+CSA of Cylinder

⟹ Outer Surface Area of Building= 52.8+19.602= 72.402m²

↦Thus, outer surface area of building is 72.402m². Ans.