Surface area of a cone - class-X

Curved surface area of a cone $= \pi rl$, where $r$ is the radius of the cone and $l$ is the slant height.
For a cone,  $l = \sqrt { { h }^{ 2 }+  {r}^{ 2 } } $, where $h$ is the height.

Hence, $ l = \sqrt { { 3 }^{ 2 }+  {4}^{ 2 } } $ 

$ l = \sqrt {25} $

$ l = 5 $ cm

Hence, Curved surface area of this cone, $ =\displaystyle  \frac {22}{7} \times 4 \times 5 = 62 \frac {6}{7}  {cm}^{2} $

Total surface area of a cone $ = \pi r (r + l) $  

Hence, TSA of this cone, $ = \cfrac {22}{7} \times 12 \times (12 + 9) = 792  {m}^{2} $

Curved surface area of a cone $= \pi rl$  

Radius of the cone $ = \cfrac {Diameter}{2} = \cfrac {14}{2}  =  7  cm $

$\displaystyle L.S.A, of, a, cone =\pi rl$
$\displaystyle =\dfrac{22}{7}\times 7\times 13=286cm^{2}$

Let r be the radius of the base of cone.

Total surface area of cone $\displaystyle TSA _{cone}=770cm^{2}$
$\displaystyle l=4r$
$\displaystyle \therefore TSA _{cone}=\pi rl+\pi r^{2}=\pi r\times 4r+\pi r^{2}$
or $\displaystyle 770=5\pi r^{2}$
or $\displaystyle r^{2}=\frac{770}{5\times 22}\times 7=49$
$\displaystyle \therefore r=\sqrt{49}=7$
$\displaystyle \therefore$ Diameter, d=$\displaystyle 2\times 7$=14 cm

Let $\displaystyle \frac{r}{l}=\frac{7}{13}=x$
Curved surface area CSA=286 $\displaystyle cm^{2}$
But CSA=$\displaystyle \pi rl$
$\displaystyle \therefore 286=\frac{22}{7}\times 7x\times 13x$
or $\displaystyle x^{2}=\frac{286}{22\times 13}=1$
or x=1
$\displaystyle \therefore $ r=7 cm
$\displaystyle \therefore $ Diameter d=14 cm

Given r=12 m l=5.6 m B=4 m
Total surface area of the tent=Area of the canvas
i.e. $\displaystyle \pi rl=L\times B$
or $\displaystyle \frac{22}{7}\times 12\times 5.6=L\times 4$
or 211.2=4L
or $\displaystyle L=\frac{211.2}{4}=52.8m$'

Consider the given slant height l and radius$ =r$ of the cone,

 Curved surface area=(Arc length of sector/Circumference of circle)×Area of circle Curved surface area

$=\dfrac{2\pi r}{2\pi l}\times \pi {{l}^{2}}=\pi rl$ 

Given the vertical height of cone is 40 cm and radius is 30 cm 

Given that conical tent of canvas height is 8 cm and diameter of base is 12 m

Given the radius a of height of tent is 12 m and 16 m respt 

Given 84 cm is the height of right circular cone and diameter is 70 cm 

Given that slant height of a right circular cone is 10 cm and its height is 8 cm 

Given height of cone is 24 cm and radius is 7 cm 

Given the vertical height is 4 cm and diameter is 6 cm

Let the radii of two right circular cone is 4x and 5x and siant height is 2y and 3y

Let the radius of the base be r meters Then
$2$$\displaystyle \pi r$$=44$ m
or $\displaystyle r=\frac{44}{2\pi }=\frac{44\times 7}{2\times 22}=7 m$
Let the slant height be l meters Then
$\displaystyle l=\sqrt{h^{2}+r^{2}}=\sqrt{7^{2}+24^{2}}=\sqrt{625}=25m$
Now, Surface area = $\displaystyle \pi rl=\frac{22}{7}\times 7\times 25=550 m^{2}$
$\displaystyle \therefore $ Area of canvas required $= 550$ m$\displaystyle ^{2}$
 Given that width of the canvas is $2$ m therefore
Length of canvas required = $\displaystyle \frac{550}{2}=275m$

Using the forrmula for Curved surface area of the cone$= \pi rs$

Surface area of a cone $=$ $\displaystyle \pi r\left( r+s \right) $ sq.units
Since, surface area of cone $=$  Area of sector $+$ Area of circle.

Surface area of cone $=$ CSA of cone + area of circular base

We have,

We have,

The formula of curved surface area of a right circular cone $ =\pi \times r\times l $

Given Height $h=5$ and Slant height $l=13$

Base area = pi * r^2 = 616. r^2 = 616 / (22/7) = 196. r = 14. Given height h = 48, slant height l = sqrt(r^2 + h^2) = sqrt(14^2 + 48^2) = sqrt(196 + 2304) = sqrt(2500) = 50.

Given the radius of the base of a right circular cone $(r)=10$ cm, and its slant height $(l)=3.5$ cm.

Given the radius of the base of a right circular cone $(r)=2$ cm, and its slant height $(l)=3.5$ cm.

Let $r$ and $l$ be base radius and slant height of cone.

For a cone, $l = \sqrt { { h }^{ 2 }+  {r}^{ 2 } } $ where h is the height. 
Hence, $l = \sqrt { { 28 }^{ 2 }+  {21}^{ 2 } } $

$ l = 35 $ cm 

Area of the canvas $ = $ Curved surface area of the conical tent
Since the canvas is rectangular in shape, its area is $=$ length $\times $ width
Curved surface area of a cone $= \pi rl$, where $r$ is the radius of the cone and $l$ is the slant height.
For a cone, $ l= \sqrt { { h }^{ 2 }+  {r}^{ 2 } } $, where $l$ is the slant height.

Hence, $ l = \sqrt { { 24 }^{ 2 }+  {7}^{ 2 } } $ 

$\Rightarrow  l = \sqrt {625} $

$ \Rightarrow l = 25 $ cm

Hence, length $ \times 5 =\displaystyle \frac { 22 }{ 7 } \times 7\times 25 $

Let the radius of the cone be $ 4a$ and slant height $ = 7a$ 

Curved surface area of a cone $= \pi rl$, where $r$ is the radius of the cone and $l$ is the slant height.
Hence, curved surface area of this cone, $ = \dfrac {22}{7} \times 56 \times l = 12320 $ 

$ \therefore l = 70  cm $

For a cone, $ l = \sqrt { { h }^{ 2 }+  {r}^{ 2 } } $, where $h$ is the height. 

Hence, $ 70 = \sqrt { { h }^{ 2 }+  {56}^{ 2 } } $ 

$\Rightarrow  4900 = { h }^{ 2 } + 3136 $

$\Rightarrow  { h }^{ 2 } = 1764 $

$ \Rightarrow  h = 42 $ cm

Given,

Area of sheet required to make a cap is the Curved surface area of a cone which is $= \pi rl$, where $r$ is the radius of the cone and $l$ is the slant height.

For a cone, $l = \sqrt { { h }^{ 2 }+  {r}^{ 2 } } $, where $h$ is the height. 

Hence, $ l = \sqrt { { 24 }^{ 2 }+  {7}^{ 2 } } $ 

$ \therefore l = 25 $ cm

Curved surface area of a cone $= \pi rl$, where $r$ is the radius of the cone and $l$ is the slant height.
Radius of the tomb $ = \dfrac {\text{Diameter}}{2} = 7  m $
Hence, CSA of this conical tomb , $ = \dfrac {22}{7} \times 7 \times 25 = 550$  sq. m.
Cost of white washing the tomb at Rs. $210$ per $100 {m}^{2} = \dfrac {210}{100} \times 550 =  $ Rs. $  1155 $

Given, $h = 14$ m and floor area $= 346.5$ $m^{ 2 }$
$\Rightarrow \pi { r }^{ 2 }=346.5\ \Rightarrow { r }^{ 2 }=346.5\times \dfrac { 7 }{ 22 } =110.25\ \Rightarrow r=10.5$
$\therefore$ $l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { { \left( 10.5 \right)  }^{ 2 }+{ \left( 14 \right)  }^{ 2 } } =\sqrt { 110.25+196 } $
$=\sqrt { 306.25 } =17.5$ m
$\therefore$ curved surface area $=\pi rl$
$=\cfrac { 22 }{ 7 } \times 10.5\times 17.5=\cfrac { 4042.5 }{ 7 } { m }^{ 2 }$
Width of cloth $=1.1$ m
$\therefore$ length of cloth required $=\cfrac { \cfrac { 4042.5 }{ 7 }  }{ 1.1 } =\cfrac {4042.5 }{ 7.7 } = 525$ m
Hence, $525$ m length of canvas is required to build the conical tent.

The base of a cone is circular. 

The amount of canvas used to make the tent $ = $ Curved surface area of cylindrical part $ + $ Curved surface area of the conical part.
Curved Surface Area of a Cylinder of Radius "$R$" and height "$h$" $ = 2\pi Rh$
Radius of the cylindrical part $ = \dfrac {Diameter}{2} = \dfrac {9}{2} $ m 
Curved surface area of a cone $= \pi rl$, where $r$ is the radius of the cone and $l$ is the slant height.

Radius of the conical part $ = \dfrac {Diameter}{2} = \dfrac {9}{2}  m $

Height of the conical part $ = 10.8 - 4.8 = 6 $ m 

For a cone, $  l= \sqrt { { h }^{ 2 }+  {r}^{ 2 } } $, where $h$ is the height. 

Hence, $l = \sqrt { { 6 }^{ 2 }+  {4.5}^{ 2 } } $

$\therefore  l = 7.5  $ m 

Hence, area of canvas $ = 2 \times \pi \times 4.5 \times 4.8 + \pi \times 4.5 \times 7.5$

$ = 76.95 \pi  $

$= 241.84  {m}^{2} $

Curved surface area of a cone $= \pi rl$

For a cone, $l = \sqrt { {h }^{ 2 }+  {r}^{ 2 } } $

Hence, $l = \sqrt { { 3 }^{2 }+  {4}^{ 2 } } $

Radius $= 7m$,  height $= 24m$

Let radius be $4x$ and height be $3x$.

Curved surface area of a cone $= \pi rl$  where r is the radius of
the cone and l is the slant height.
Hence, CSA of this cone, $ = \frac {22}{7} \times r \times 13 = 286 $
$ =>r = 7  m $

Area of base $ = \pi {r}^{2} = \frac {22}{7} \times 7 \times 7 = 154  m^2 $

We know that the total surface area S of a right circular cylinder of radius r and slant height l is given by 
$\displaystyle S=\pi r^{2}+\pi rl=\pi r\left ( r+l \right )$
Here, $\displaystyle r=12m, and, l=9m$
$\displaystyle \therefore S=\left { \frac{22}{7}\times 12\times \left ( 12+9 \right ) \right }^{2}=792m^{2}$

A cone is a three-dimensional geometric shape consisting of all line segments joining a single point to every point of a two-dimensional figure.

Given the height of cone 24 m and base radius is 24 m

Given, radius $=20$ cm and height $=0.$ cm

Given, surface area of cone $=20$ $m^2$ and radius $7$ m

Surface area of cone is $A=πr(r+l)$

First need to find the value of slant height(s) of a cone, using Pythagoras theorem, since the cross section is a right triangle.
$\displaystyle { s }^{ 2 }={ h }^{ 2 }+{ r }^{ 2 }$
$\displaystyle { s }^{ 2 }={ 3 }^{ 2 }+{ 2 }^{ 2 }$
$\displaystyle { s }^{ 2 }=9+4$
$\displaystyle s=\sqrt { \left( 13 \right)  } $
$\displaystyle s=3.6$ cm
Then, surface area of a cone $\displaystyle =\pi rs+\pi { r }^{ 2 }$
$\displaystyle =3.14\times 2\times 3.6+3.14\times 2\times 2$
$\displaystyle =22.608+12.56$
$\displaystyle= 35.168{ cm }^{ 2 }$

Given, radius $=0.2$ mm and height $=1.2$ mm 
Then find the value of slant height(s) of a conical water tank, using Pythagoras theorem, since the cross section is a right triangle.
$\displaystyle { s }^{ 2 }={ h }^{ 2 }+{ r }^{ 2 }$
$\displaystyle { s }^{ 2 }={ 1.2 }^{ 2 }+{ 0.2 }^{ 2 }$
$\displaystyle { s }^{ 2 }=1.44+0.4$
$\displaystyle s=\sqrt { \left( 1.84 \right)  } $
$\displaystyle s=1.35$ mm
So, surface area of a cone $\displaystyle =\pi rs+\pi { r }^{ 2 }$
$\displaystyle =3.14\times 0.2\times 1.35+3.14\times 0.2\times 0.2$
$\displaystyle =0.8478+0.1256$
$\displaystyle =0.9734{\  mm }^{ 2 }$

Surface area of a cone $= $ Curved surface area of a cone $+$ Area of circle
So, SA $\displaystyle =\pi rs+\pi { r }^{ 2 }$
$\displaystyle =100+3.14\times 200\times 200$
$\displaystyle =100+125600$
$\displaystyle =125700{ cm }^{ 2 }$

Surface area of cone is $A=πr(r+l)$

Formula:

$\displaystyle Diameter=\frac { radius }{ 2 } $
$\displaystyle radius=\dfrac{1}{2}ft$
Surface area of a cone $\displaystyle =\pi rs+\pi { r }^{ 2 }$
$\displaystyle 3.14\times { 1 }/{ 2 }\times 12+3.14\times { 1 }/{ 2 }\times { 1 }/{ 2 }$
$\displaystyle 18.84+0.785$
$\displaystyle 19.625{\  ft }^{ 2 }$

Find the value of slant height(s) of a ice-cream cone, using Pythagoras theorem, since the cross section is a right triangle.
$\displaystyle { s }^{ 2 }={ h }^{ 2 }+{ r }^{ 2 }$
$\displaystyle { s }^{ 2 }={ 20 }^{ 2 }+{ 21 }^{ 2 }$
$\displaystyle { s }^{ 2 }=400+441$
$\displaystyle s=\sqrt { 841 } $
$\displaystyle s=29$ m
So, surface area of a cone $\displaystyle =\pi rs+\pi { r }^{ 2 }$
$\displaystyle =3.14\times 21\times 29+3.14\times 21\times 21$
$\displaystyle =1912.26+1384.74$
$\displaystyle =3297{ m }^{ 2 }$

The value of slant height(s) of a cone, using Pythagoras theorem, since the cross section is a right triangle is

Formula:

Total surface area of cone  = Area of base + Area of curved surface
$\displaystyle =\quad \pi { r }^{ 2 }+\pi rs$

Given, Lateral surface area $=6 \pi$ and slant height $=6$

Radius $=5.6$ cm, vertical height $= h$, slant height $=$ $l$
Curved Surface Area of cone $=\pi r l = 158.4 cm^2$
$\Rightarrow \dfrac{22}{7} \times 5.6 \times l = 158.4$
$\Rightarrow l = \dfrac{158.4 \times 7}{22 \times 5.6} = \dfrac{18}{2} = 9$ cm
We know $l^2 = r^2 + h^2$
Thus $h^2 = l^2 - r^2 $

$l^2=(3.5)^2+12^2=156.25$
$l=12.5$
Curved surface area$=\pi rl$
$=471.42m^2$

Given the radius of cylinder and cone are the same because there base areas are same.
Curved surface of cylinder $=$ curved surface area of the cone
$\therefore 2 \pi r h = \pi r l$
$\therefore l = 2h $
Given, height of cylinder $= 3$ cm
Therefore, slant height of cone $= 6$ cm

Given,
Radius of cone $= 7$ cm
Height of cone $= 24$ cm
Curved surface area = $\pi r \sqrt{(r^2 + l^2)}$
= $\pi \times 7 (\sqrt{(7^2 + (24)^2)}$
= $\pi \times 70\times 25$
= $550 cm^2$

Let the curved surface areas of two cones be $C _1$ and $C _2$

To find the amount of canvas required to make a conical tent, we need to calculate the lateral surface area of the tent.
Lateral surface area of a cone of radius $ r $, height $ h = \pi r\sqrt{{h}^{2} + {r}^{2}} $
As the total cost to make the tent is Rs $10048 $ at the rate of Rs $ 40 $ per sq m, total area LSA $ = \dfrac {10048}{40} = 251.2 $ sq m