Question

The height of a cone is 16cm and its base radius is 12cm. Find the curved surface area and the total surface area of the cone (Use π = 3.14).





Note:-


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Answer 1

★ Concept:-

Here the concept of CSA and TSA of the conce has been used. We see that we are given the height and radius of the cone. So firstly we can find its slant height. After finding this, we can apply required values and thus find the answer.

Let's Do It!!!

★FormulaUsed::

$\begin{gathered}\\\;\boxed{\sf{\purple{L^{2}\;=\;\bf{H^{2}\;+\;r^{2}}}}}\end{gathered}$

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

$\begin{gathered}\\\;\boxed{\sf{\purple{TSA\;of\;Cone\;=\;\bf{\pi rL\;+\;\pi r^{2}}}}}\end{gathered}$

Given,

solution:-

» Height of Cone = H = 16 cm

» Radius of Cone = r = 12 cm

» Value of π = 3.14

~ For the slant Height of the Cone ::

• Let the Slant Height of the cone be L

By relationship of Slant Height, we know that

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

By applying values, we get

$\begin{gathered}\\\;\sf{\rightarrow\;\;L^{2}\;=\;\bf{(16)^{2}\;+\;(12)^{2}}}\end{gathered}$

$\begin{gathered}\\\;\sf{\rightarrow\;\;L^{2}\;=\;\bf{256\;+\;144}}\end{gathered}$

$\begin{gathered}\\\;\sf{\rightarrow\;\;L^{2}\;=\;\bf{400}}\end{gathered}$

$\begin{gathered}\\\;\sf{\rightarrow\;\;L\;=\;\bf{\sqrt{400}}}\end{gathered}$

$\begin{gathered}\\\;\bf{\rightarrow\;\;Slant\;Height,\;L\;=\;\bf{\purple{20\;\;cm}}}\end{gathered}$

~ For the CSA of Cone::

We know that,

$\begin{gathered}\\\;\sf{\Longrightarrow\;\;CSA\;of\;Cone\;=\;\bf{\pi rL}}\end{gathered}$

By applying values, we get

$\begin{gathered}\\\;\sf{\Longrightarrow\;\;CSA\;of\;Cone\;=\;\bf{3.14\:\times\:12\:\times\:20}}\end{gathered}$

$\begin{gathered}\\\;\bf{\Longrightarrow\;\;CSA\;of\;Cone\;=\;\bf{\red{753.6\;\;cm^{2}}}}\end{gathered}$

$\begin{gathered}\\\;\underline{\boxed{\tt{CSA\;\:of\;\:Cone\;=\;\bf{\pink{753.6\;\;cm^{2}}}}}}\end{gathered}$

______________________________

★More to know:-

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

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

$\begin{gathered}\\\;\sf{\leadsto\;\;TSA\;of\;Cylinder\;=\;2\pi rH\;+\;2\pi r^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;CSA\;of\;Cylinder\;=\;2\pi rH}\end{gathered}$

$\begin{gathered}\\\;\sf{\leadsto\;\;Volume\;of\;Hollow\;Cone\;=\;\dfrac{1}{3}\pi (R^{2}\;-\;r^{2})H}\end{gathered}$

Answer 2

Answer:

★ Concept:-

Here the concept of CSA and TSA of the conce has been used. We see that we are given the height and radius of the cone. So firstly we can find its slant height. After finding this, we can apply required values and thus find the answer.

Let's Do It!!!

★FormulaUsed::

\begin{gathered}\begin{gathered}\\\;\boxed{\sf{\purple{L^{2}\;=\;\bf{H^{2}\;+\;r^{2}}}}}\end{gathered}\end{gathered}

L

2

=H

2

+r

2

\begin{gathered}\begin{gathered}\\\;\boxed{\sf{\pink{CSA\;of\;Cone\;=\;\bf{\pi rL}}}}\end{gathered}\end{gathered}

CSAofCone=πrL

\begin{gathered}\begin{gathered}\\\;\boxed{\sf{\purple{TSA\;of\;Cone\;=\;\bf{\pi rL\;+\;\pi r^{2}}}}}\end{gathered}\end{gathered}

TSAofCone=πrL+πr

2

Given,

solution:-

» Height of Cone = H = 16 cm

» Radius of Cone = r = 12 cm

» Value of π = 3.14

~ For the slant Height of the Cone ::

Let the Slant Height of the cone be L

By relationship of Slant Height, we know that

\begin{gathered}\begin{gathered}\\\;\sf{\rightarrow\;\;L^{2}\;=\;\bf{H^{2}\;+\;r^{2}}}\end{gathered}\end{gathered}

→L

2

=H

2

+r

2

By applying values, we get

\begin{gathered}\begin{gathered}\\\;\sf{\rightarrow\;\;L^{2}\;=\;\bf{(16)^{2}\;+\;(12)^{2}}}\end{gathered}\end{gathered}

→L

2

=(16)

2

+(12)

2

\begin{gathered}\begin{gathered}\\\;\sf{\rightarrow\;\;L^{2}\;=\;\bf{256\;+\;144}}\end{gathered}\end{gathered}

→L

2

=256+144

\begin{gathered}\begin{gathered}\\\;\sf{\rightarrow\;\;L^{2}\;=\;\bf{400}}\end{gathered}\end{gathered}

→L

2

=400

\begin{gathered}\begin{gathered}\\\;\sf{\rightarrow\;\;L\;=\;\bf{\sqrt{400}}}\end{gathered}\end{gathered}

→L=

400

\begin{gathered}\begin{gathered}\\\;\bf{\rightarrow\;\;Slant\;Height,\;L\;=\;\bf{\purple{20\;\;cm}}}\end{gathered}\end{gathered}

→SlantHeight,L=20cm

~ For the CSA of Cone::

We know that,

\begin{gathered}\begin{gathered}\\\;\sf{\Longrightarrow\;\;CSA\;of\;Cone\;=\;\bf{\pi rL}}\end{gathered}\end{gathered}

⟹CSAofCone=πrL

By applying values, we get

\begin{gathered}\begin{gathered}\\\;\sf{\Longrightarrow\;\;CSA\;of\;Cone\;=\;\bf{3.14\:\times\:12\:\times\:20}}\end{gathered}\end{gathered}

⟹CSAofCone=3.14×12×20

\begin{gathered}\begin{gathered}\\\;\bf{\Longrightarrow\;\;CSA\;of\;Cone\;=\;\bf{\red{753.6\;\;cm^{2}}}}\end{gathered}\end{gathered}

⟹CSAofCone=753.6cm

2

\begin{gathered}\begin{gathered}\\\;\underline{\boxed{\tt{CSA\;\:of\;\:Cone\;=\;\bf{\pink{753.6\;\;cm^{2}}}}}}\end{gathered}\end{gathered}

CSAofCone=753.6cm

2

______________________________

★More to know:-

\begin{gathered}\begin{gathered}\\\;\sf{\leadsto\;\;Volume\;of\;Cone\;=\;\dfrac{1}{3}\pi r^{2}H}\end{gathered}\end{gathered}

⇝VolumeofCone=

3

1

πr

2

H

\begin{gathered}\begin{gathered}\\\;\sf{\leadsto\;\;Volume\;of\;Cylinder\;=\;\pi r^{2}H}\end{gathered}\end{gathered}

⇝VolumeofCylinder=πr

2

H

\begin{gathered}\begin{gathered}\\\;\sf{\leadsto\;\;TSA\;of\;Cylinder\;=\;2\pi rH\;+\;2\pi r^{2}}\end{gathered}\end{gathered}

⇝TSAofCylinder=2πrH+2πr

2

\begin{gathered}\begin{gathered}\\\;\sf{\leadsto\;\;CSA\;of\;Cylinder\;=\;2\pi rH}\end{gathered}\end{gathered}

⇝CSAofCylinder=2πrH

\begin{gathered}\begin{gathered}\\\;\sf{\leadsto\;\;Volume\;of\;Hollow\;Cone\;=\;\dfrac{1}{3}\pi (R^{2}\;-\;r^{2})H}\end{gathered}\end{gathered}

⇝VolumeofHollowCone=

3

1

π(R

2

−r

2

)H