Question

find the total curved surface area of a cone whose height is 12 cm and base radius is 5 cm​

Answer 1

Answer:

The curved surface area of the cone will

be equal to 204.28 square cm

Step-by-step explanation:

Given, height of cone = 12cm

Base radius = 5 cm

By Pythagoras theorem,

Slant height of cone

|= V[(12)^2 +(5)^2} =

| = 13cm

Now, curved surface area of cone

= Trl

=(22/7)x5x13

=1430/7

=204.28 square cm

Answer 2

Correct Question-:

• Find the curved surface area of a cone whose height is 12 cm and base radius is 5 cm .

AnswEr -:

• $\boxed{\sf{\purple{ Curved\:Surface \:Are\:or \: (CSA ) \:of \:cone\:= 204.1 cm²}}}$

Explanation-:

Given-:

• Radius of a cone is 5 cm .
• Height of a cone is 12 cm .

To Find,

• Curved Surface Area of a cone .

Solution-:

☆ The Slant Height or (l) of a cone -:

• $\boxed{\sf{\purple{\:(Slant \: Height)²\: =\:(Radius)²+\: (Height)²\: }}}$

Here ,

• Radius of a cone is 5 cm or (r) .
• Height of a cone is 12 cm or (h) .
• Slant Height of cone is ? or (l) .

Now ,

$\large{\sf{\Rightarrow{\pink{ (l)²\: = (12)²+(5)²}}}}$

$\large{\sf{\Rightarrow{\pink{ (l)²\: = 144+25}}}}$

$\large{\sf{\Rightarrow{\pink{ (l)²\: = 169}}}}$

$\large{\sf{\Rightarrow{\pink{ l\: = \sqrt 169}}}}$

$\large{\sf{\Rightarrow{\pink{ l\: = 13 cm}}}}$

Therefore ,

$\boxed{\sf{\purple{ \:Slant\:Height\:or \: (l ) \:of \:cone\:= 13 cm }}}$ ........... [1]

Now ,

• $\boxed{\sf{\purple{\:Curved\:Surface \:Area\:or\:CSA\:of \: cone\: =\:\:  π  \times \: r \times \: l }}}$

Here ,

•  π  = $\sf{\frac{22}{7}}$ or 3.14
• r = Radius of a cone = 5 cm .
• l = Slant height of a cone = 13 cm

Now ,

$\large{\sf{\Rightarrow{\pink{ CSA \: = \frac{22}{7} \times 5 \times 13 }}}}$

$\large{\sf{\Rightarrow{\pink{ CSA \: = \frac{22}{7} \times 65 }}}}$

$\large{\sf{\Rightarrow{\pink{ CSA \: = 3.14 \times 65 }}}}$

$\large{\sf{\Rightarrow{\pink{ CSA \: = 204.1 cm ² \: [Approx \: Value ] }}}}$

Hence ,

$\boxed{\sf{\purple{ Curved\:Surface \:Are\:or \: (CSA ) \:of \:cone\:= 204.1 cm²\:[Approx \: Value\:]}}}$

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