Question

Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no

colour on it. He wanted to colour it with his crayons. The top is shaped like a cone

surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the

top is 3.5 cm. (Take = Take π = 22/7)​

find the curved surface area of the cone and volume of the hemisphere ​

Answer 1

Answer:

39.5 cm square

Step-by-step explanation:

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Answer 2

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

Here the concept of CSA of Cone and Volume of Hemisphere has been used. We see that we are given the total height of the lattu (top). So the total height will be height of cone in addition with the radius of hemisphere because radius is only its height. Then from this we can find the height of cone. Also since the hemispherical part is surmounted on conical, so radius of both will be equal. After finding these, we can apply values in formula and find the answer.

Let's do it !!

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★ Formula Used :-

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

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

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

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★ Solution :-

Given,

» Total Height of Lattu (top) = 5 cm

» Diameter of top = 3.5 cm

» Radius of top = ½ × 3.5 = 1.75 cm

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~ For Dimensions of different parts of Lattu (Top) ::

Since, hemispherical part is surmounted on Conical Part. Then,

✒ Radius of cone = Radius of hemisphere = 1.75 cm

Also, total Height of the Top is given as -

✒ Total Height of Top = Radius of Hemispherical part + Height of Conical Part

✒ 5 = 1.75 + Height of Conical Part

✒ Height of Conical Part = 5 - 1.75

✒ Height of Conical Part = 3.25 cm

Now, let's find out slant height (L) of Cone. This is given as -

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

$\\\;\;\sf{:\rightarrow\;\;L^{2}\;=\;\bf{(1.75)^{2}\;+\;(3.25)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;L^{2}\;=\;\bf{3.0625\;+\;10.5625}}$

$\\\;\;\sf{:\rightarrow\;\;L^{2}\;=\;\bf{13.625}}$

$\\\;\;\sf{:\rightarrow\;\;L\;=\;\bf{\sqrt{13.625}}}$

$\\\;\;\sf{:\rightarrow\;\;L\;=\;\bf{3.69\;\;cm}}$

$\\\bf{:\rightarrow\;\;Slant\;Height\;of\;Cone,\;L\;=\;\bf{3.7\;\;cm}}$

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~ For the CSA (Curved Surface Area) of Cone ::

Curved Surface Area of cone is given as -

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

$\\\;\;\;\sf{:\Longrightarrow\;\;CSA\;of\;Cone\;=\;\bf{\dfrac{22}{7}\;\times\;1.75\;\times\;3.7}}$

$\\\;\;\;\sf{:\Longrightarrow\;\;CSA\;of\;Cone\;=\;\bf{\dfrac{142.45}{7}}}$

$\\\bf{:\Longrightarrow\;\;CSA\;of\;Cone\;=\;\bf{20.35\;\;cm^{2}}}$

$\\\;\underline{\boxed{\tt{CSA\;\;of\;\;Cone\;=\;\bf{20.35\;\;cm^{2}}}}}$

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~ For the Volume of Hemispherical Part ::

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

$\\\;\;\;\sf{:\mapsto\;\;Volume\;of\;Hemisphere\;=\;\bf{\dfrac{2}{3}\;\times\;\dfrac{22}{7}\;\times\;(1.75)^{3}}}$

$\\\;\;\;\sf{:\mapsto\;\;Volume\;of\;Hemisphere\;=\;\bf{\dfrac{2}{3}\;\times\;\dfrac{22}{7}\;\times\;5.359375}}$

$\\\;\;\;\sf{:\mapsto\;\;Volume\;of\;Hemisphere\;=\;\bf{\dfrac{235.8125}{21}}}$

$\\\bf{:\mapsto\;\;Volume\;of\;Hemisphere\;=\;\bf{11.23\;\;cm^{3}}}$

$\\\;\underline{\boxed{\tt{Volume\;\;of\;\;Hemisphere\;=\;\bf{11.23\;\;cm^{3}}}}}$

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★ More to know :-

Let's now find out Area to be Coloured. This is given as -

$\\\;\tt{\Rightarrow\;\;Area\;\;to\;\;be\;\;Painted\;=\;CSA\;\;of\;\;Cone\;+\;CSA\;\;of\;\;Hemisphere}$

$\\\;\tt{\Rightarrow\;\;Area\;\;to\;\;be\;\;Painted\;=\;\pi rL\;+\;2\pi r^{2}}$

$\\\;\tt{\Rightarrow\;\;Area\;\;to\;\;be\;\;Painted\;=\;20.35\;+\;2\;\times\;\dfrac{22}{7}\;\times\;(1.75)^{2}}$

$\\\;\tt{\Rightarrow\;\;Area\;\;to\;\;be\;\;Painted\;=\;20.35\;+\;\dfrac{134.75}{7}}$

$\\\;\tt{\Rightarrow\;\;Area\;\;to\;\;be\;\;Painted\;=\;20.35\;+\;19.25}$

$\\\;\tt{\Rightarrow\;\;Area\;\;to\;\;be\;\;Painted\;=\;39.55\;\;cm^{2}}$

$\\\;\tt{\bf{\Rightarrow\;\;Area\;\;to\;\;be\;\;Painted\;=\;39.6\;\;cm^{2}}}$

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• More Formulas :-

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

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

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

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

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