Tag: surface area of cone

Questions Related to surface area of cone

Mark the correct alternative of the following.
If the base radius and the height of a right circular cone are increased by $20\%$, then the percentage increase in volume is approximately.

maths solids surface area of a cone surface area of cone the surface area of a cone

  1. $60$

  2. $68$

  3. $73$

  4. $78$

Show answer
Correct Option: C
Explanation:
Let original radius of base and height are $R$ and $H$ respectively.
Original height $=H$

New radius $=\dfrac{120}{100}R=\dfrac{6}{5}R$

And new height $=\dfrac{6}{5}H$

Original volume of cone $V _1=\dfrac{1}{3}\pi R^2 H$

New volume of cone $V _2=\dfrac{1}{3}\pi \left(\dfrac{6}{5}R\right)^2\times \dfrac{6}{5}H$

                                          $=\dfrac{216}{125}V _1$

Now, increased in volume $=\dfrac{216}{125}V _1-V _1=\dfrac{91}{125}V _1$

$\therefore$  Percentage increased in volume $=\left(\dfrac{91}{125}V _1\times\dfrac{1}{V _1}\right)\times 100\%$

                                                              $=72.81\%\approx 73\%$

Mark the correct alternative of the following.
The total surface area of a cone of radius $\dfrac{r}{2}$ and length $2l$, is?

maths solids surface area of a cone surface area of cone the surface area of a cone

  1. $2\pi r(l+r)$

  2. $\pi r\left(1+\dfrac{r}{4}\right)$

  3. $\pi r(l+r)$

  4. $2\pi rl$

Show answer
Correct Option: B
Explanation:

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

Total surface area $=\pi r (l+r)$
Here, it is given that, 
The base radius is $\dfrac{r}{2}$ and that the slant height is $2l.$
Substituting these values in the above equation we have,
Total surface area $=\pi\left(\dfrac{r}{2}\right)\left(2l+\dfrac{r}{2}\right)$

                               $=\pi r\left(l+\dfrac{r}{4}\right)$

If the radius of the base and the height of a right circular cone are respectively $21$ cm and $28$ cm, then the curved surface area of the cone is $\displaystyle \left(\pi\, =\, \frac{22}{7}\right)$

maths solids surface area of a cone surface area of cone the surface area of a cone

  1. $3696\, cm^{2}$

  2. $2310\, cm^{2}$

  3. $2550\, cm^{2}$

  4. $2410\, cm^{2}$

Show answer
Correct Option: B
Explanation:

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

$ l = 35 $ cm 

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 21 \times 35 = 2310  {cm}^{2} $

A conical tent with base-radius $7$ m and height $24$ m is made from $5$ m wide canvas. The length of the canvas used is $\displaystyle \left( \pi\, =\, \frac{22}{7}\right)$

maths solids surface area of a cone surface area of cone the surface area of a cone

  1. $100$ m

  2. $105$ m

  3. $110$ m

  4. $115$ m

Show answer
Correct Option: C
Explanation:

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 $

$ \therefore $ length $= 110 $ m

If the radius and slant height of a cone are in the ratio $4 : 7$ and its curved surface area is $792 cm^{2}$, then its radius is $\displaystyle \left(\pi\, =\, \frac{22}{7}\right)$.

maths solids surface area of a cone surface area of cone the surface area of a cone

  1. $10$ cm

  2. $8$ cm

  3. $12$ cm

  4. $9$ cm

Show answer
Correct Option: C
Explanation:

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

Given, curved surface area $=792$ $cm^2$
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 $ = \displaystyle \frac {22}{7} \times 4a \times 7a = 792 $
$ \therefore  a = 3 $ cm 

Hence, radius $ = 4a  = 12 $ cm 

If the curved surface area of a right circular cone is $12,320 cm^{2}$ and its base radius is $56$ cm, then its height is $\displaystyle \left(\pi\, =\, \frac{22}{7}\right)$

maths solids surface area of a cone surface area of cone the surface area of a cone

  1. $42$ cm

  2. $36$ cm

  3. $48$ cm

  4. $50$ cm

Show answer
Correct Option: A
Explanation:

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

What length of tarpaulin $3$ m wide will be required to make conical tent of height $8$ m and base radius $6$ m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately $20$ cm.

 (Use $\pi=3.14$)

maths solids surface area of a cone surface area of cone the surface area of a cone

  1. $63$ m

  2. $85$ m

  3. $74$ m

  4. $92$ m

Show answer
Correct Option: A
Explanation:

Given,

$h=8m$, $r=6m$

$\therefore l=\sqrt {r^2+h^2}$

 $=\sqrt {6^2+8^2}$

 $=\sqrt {36+64}$

 $=\sqrt {100}$

 $=10$ m

$\therefore$ width surface area $=$ $\pi(r)(l)$

                                    $=3.14(6)(10)$

                                    $=1883.4m^2$

Width of tarpaulin $= 3$ m

$\therefore$ length of tarpaulin $= \displaystyle \frac{188.4}{3}$
                                   $=62.8$ m

Extra length of material required $=20$ cm
                                                       $=0.2$ m

$\therefore$ actual length of tarpaulin required $= 62.8$ m $+0.2$ m
                                                              $=63$ m

A joker's cap is in the form of a right circular cone of base radius $7$ cm and height $24$ cm. Find the area of the sheet required to make $100$ such caps.

maths solids surface area of a cone surface area of cone the surface area of a cone

  1. $55000{cm}^{2}$

  2. $48724{cm}^{2}$

  3. $30000{cm}^{2}$

  4. $11256{cm}^{2}$

Show answer
Correct Option: A
Explanation:

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

Hence, area of sheet required to make one cap $

= \dfrac {22}{7} \times 7 \times 25 = 550 $ sq.cm  
Area of sheet required to make $100$ caps $ 100\times 550=55000$ sq.cm

The slant height and base diameter of a conical tomb are $25$ m and $14$ m respectively. Find the cost of white washing its curved surface at the rate of Rs. $210$ per $100{m}^{2}$.

maths solids surface area of a cone surface area of cone the surface area of a cone

  1. Rs. $5627$

  2. Rs. $4156$

  3. Rs. $1155$

  4. Rs. $964$

Show answer
Correct Option: C
Explanation:

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 $

The length of canvas $1.1$ m wide required to build a conical tent of height $14$ m and the floor area $346.5{m}^{2}$, is

maths solids surface area of a cone surface area of cone the surface area of a cone

  1. $65$ m

  2. $525$ m

  3. $490$ m

  4. $860$ m

Show answer
Correct Option: B
Explanation:

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.