Question

A cylinder of same height and radius is placed on the top of a hemisphere. find the curved surface area of the sphere if the length of shape formed be 7cm.​

Answer 1

Answer :-

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of CSA of Hemisphere ans Equality in Radius of the figure has been used. We see that its given in question that Radius of Cylinder is equal to Height of Cylinder. Also since cylinder is placed on hemisphere. So, Radius of Hemisphere will be equal to the Radius of Cylinder. Also Height of hemisphere will be equal to the Radius of Hemisphere since its a curved solid. By equating this we can our answer.

Let's do it !!

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★ Equations Used :-

$\\\;\boxed{\sf{Height\;of\;Cylinder\;+\;Height\;of\;Hemisphere\;=\;\bf{Height\;of\;Solid}}}$

$\\\;\boxed{\sf{CSA\;of\;Hemisphere\;=\;\bf{2\pi r^{2}}}}$

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

$\\\;\boxed{\sf{CSA\;of\;Shape\;=\;\bf{CSA\;of\;Cylinder\;+\;CSA\;of\;Hemisphere}}}$

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★ Correct Question :-

A cylinder of same height and radius is placed on the top of a hemisphere. Find the curved surface area of the hemisphere and curved surface area of shape, if the length of shape formed be 7cm.

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★ Solution :-

Given,

» Radius of Cylinder = Height of Cylinder

» Radius of Cylinder = Radius of Hemisphere

» Total length of shape = 7 cm

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~ For the measurements of figure :-

• Let the radius of the Cylinder be r

• Let the height of the Cylinder be h

So, h = r (given)

Then,

• Radius of Hemisphere = r (given)

• Height of Hemisphere = r (given)

Also, its given, length of shape = 7 cm

$\\\;\;\sf{:\rightarrow\;\;Height\;of\;Cylinder\;+\;Height\;of\;Hemisphere\;=\;\bf{Height\;of\;Solid}}$

$\\\;\;\sf{:\rightarrow\;\;h\;+\;r\;=\;\bf{7}}$

$\\\;\;\sf{:\rightarrow\;\;r\;+\;r\;=\;\bf{7}}$

$\\\;\;\sf{:\rightarrow\;\;2r\;=\;\bf{7}}$

$\\\;\;\bf{:\rightarrow\;\;r\;=\;h\;=\;\bf{\dfrac{7}{2}\;cm}}$

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~ For the CSA of Hemisphere :-

$\\\;\;\;\;\sf{:\Rightarrow\;\;CSA\;of\;Hemisphere\;=\;\bf{2\pi r^{2}}}$

$\\\;\;\;\;\sf{:\Rightarrow\;\;CSA\;of\;Hemisphere\;=\;\bf{2\;\times\;\dfrac{22}{7}\;\times\;(\;\dfrac{7}{2})^{2}}}$

$\\\;\;\;\;\sf{:\Rightarrow\;\;CSA\;of\;Hemisphere\;=\;\bf{\cancel{2}\;\times\;\dfrac{\cancel{22}}{\cancel{7}}\;\times\;(\;\dfrac{\cancel{49}}{\cancel{4}})}}$

$\\\;\;\;\;\sf{:\Rightarrow\;\;CSA\;of\;Hemisphere\;=\;\bf{11\;\times\;7}}$

$\\\;\;\;\;\sf{:\Rightarrow\;\;CSA\;of\;Hemisphere\;=\;\bf{77\;\;cm^{2}}}$

$\\\;\large{\underline{\underline{\rm{CSA\;of\;hemisphere\;is\;\;\boxed{\bf{77\;\;cm^{2}}}}}}}$

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~ For CSA of Shape :-

$\\\;\;\;\;\sf{:\mapsto\;\;CSA\;of\;Shape\;=\;\bf{CSA\;of\;Cylinder\;+\;CSA\;of\;Hemisphere}}$

We already know the formula of CSA of Cylinder. By applying that value, we get,

$\\\;\;\;\;\sf{:\mapsto\;\;CSA\;of\;Shape\;=\;\bf{(2\;\times\;\dfrac{22}{7}\;\times\;(\dfrac{7}{2})\;\times\;(\dfrac{7}{2}))\;+\;77}}$

$\\\;\;\;\;\sf{:\mapsto\;\;CSA\;of\;Shape\;=\;\bf{(11\;\times\;7)\;+\;77}}$

$\\\;\;\;\;\sf{:\mapsto\;\;CSA\;of\;Shape\;=\;\bf{77\;+\;77}}$

$\\\;\;\;\;\sf{:\mapsto\;\;CSA\;of\;Shape\;=\;\bf{154\;\;cm^{2}}}$

$\\\;\large{\underline{\underline{\rm{CSA\;of\;shape\;is\;\;\boxed{\bf{154\;\;cm^{2}}}}}}}$

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★ More Formulas to know :-

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

$\\\;\sf{\leadsto\;\;TSA\;of\;Hemisphere\;=\;3\pi r^{2}}$

$\\\;\sf{\leadsto\;\;TSA\;of\;Sphere\;=\;CSA\;=\;4\pi r^{2}}$

$\\\;\sf{\leadsto\;\;CSA\;of\;Cone\;=\;\pi rL}$

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

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

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

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

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