Question

$\huge \bf Question$

● A joker's cap is in the form of a right circular cone of base radius 7cm and height 24cm. Find the area of the sheet required to make 10 such caps.

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Answer 1

Answer:

Given:-

• A right circular cone in shape of cap.
• Radius of cap is 7 cm
• Height of cap is 24 cm

find area of sheet of 10 caps.

Solution. ↓ ↓ ↓ ↓

Salient height. (l)

=_/(24)^2+(7)^2

=_/576+49

=_/625

=25 cm

Area required for making cap

=22/7*r*l

=22/7*7*25

=550 cm^2

Area of sheet required to make 10 such caps

=10*Area of sheet of one cap

=10*550

=5500 cm^2

↑ ↑ ↑

Here is your answer

Answer 2

★ Concept :-

Here the concept of CSA of Cone has been used. We see that the shape of the solid is a Conical Cap. We know that a Conical Cap is hollow from below so that it can be worn out. This means we have to calculate its Curved Surface Area to find the area of sheet required to make one cap. Firstly we will find the Slant Height of the conical cap and then its CSA. After that we can multiply this area required with 10 to find the area of sheet required to make 10 such caps.

Let's do it !!

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★ Formula Used :-

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

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

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★ Solution :-

Given,

» Radius of base of cone = r = 7 cm

» Height of the cone = H = 24 cm

» Number of caps = 10

» Shape of Cap = Conical

• Let the slant height of cone be 'h'

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~ For the Slant Height of the Cap ::

We know that,

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

By applying values, we get

$\;\sf{\rightarrow\;\;L^{2}\;=\;\bf{(24)^{2}\;+\;(7)^{2}}}$

$\;\sf{\rightarrow\;\;L^{2}\;=\;\bf{576\;+\;49}}$

$\;\sf{\rightarrow\;\;L^{2}\;=\;\bf{625}}$

$\;\sf{\rightarrow\;\;L\;=\;\bf{\sqrt{625}}}$

$\;\bf{\rightarrow\;\;L\;=\;\bf{\red{25\;\;cm}}}$

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~ For the CSA of one Cap ::

We know that,

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

By applying values, we get

$\;\sf{\rightarrow\;\;CSA\;of\;Cone\;=\;\bf{\dfrac{22}{7}\:\times\:7\:\times\:25}}$

$\;\sf{\rightarrow\;\;CSA\;of\;Cone\;=\;\bf{22\:\times\:25}}$

$\;\bf{\rightarrow\;\;CSA\;of\;Cone\;=\;\bf{\green{550\;\;cm^{2}}}}$

This is the Area of sheet Required to make one cap.

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~ Area of sheet required to make 10 such caps ::

Since it is given that the caps are same type. Thus their area will be same. So,

$\;\bf{\Longrightarrow\;\;CSA\;Area\;of\;10\;Caps\;=\;\bf{10\:\times\:Area\;of\;one\;cap}}$

By applying values, we get

$\;\bf{\Longrightarrow\;\;CSA\;Area\;of\;10\;Caps\;=\;\bf{10\:\times\:550}}$

$\;\bf{\Longrightarrow\;\;CSA\;Area\;of\;10\;Caps\;=\;\bf{\blue{5500\;\;cm^{2}}}}$

This is ths Area of sheet required for 10 such caps.

$\;\underline{\boxed{\tt{Area\;of\;sheet\;required\;for\;10\;caps\;=\;\bf{\purple{5500\;\;cm^{2}}}}}}$

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★ More to know :-

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

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

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

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

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