Tag: solids

Questions Related to solids

Find the total surface area of a cone, if its slant height is $9\ m$ and the radius of its base is $12\ m$.

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

  1. $792\ {m}^{2}$

  2. $452\ {m}^{2}$

  3. $682\ {m}^{2}$

  4. $987\ {m}^{2}$

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Correct Option: A
Explanation:

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} $

The diameter of a cone is $14\ cm$ and its slant height is $9\ cm$. Find the area of its curved surface.

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

  1. $198\ {cm}^{2}$

  2. $108\ {cm}^{2}$

  3. $152\ {cm}^{2}$

  4. $218\ {cm}^{2}$

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Correct Option: A
Explanation:

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

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

Hence, CSA of this cone $ = \cfrac {22}{7} \times 7 \times 9 = 198  {cm}^{2} $

Slant height of a cone is 13 cm and radius is 7 cm its lateral surface area is

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

  1. 280 $\displaystyle cm^{2}$

  2. 282 $\displaystyle cm^{2}$

  3. 284 $\displaystyle cm^{2}$

  4. 286 $\displaystyle cm^{2}$

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Correct Option: D
Explanation:

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

The circumference of the base of a 10 m high conical tent is 44 metres Then the length of canvas used in making the tent if width of canvas is 2 m is (Use $\displaystyle \pi =22/7$)

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

  1. $132.2 m$

  2. $134.2 m$

  3. $130.2 m$

  4. $136.2 m$

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Correct Option: B
Explanation:
Let r m be the radius of the base h m be the height and l m be the slant height of the cone Then 
Circumference=44 metres
$\displaystyle \Rightarrow 2\pi r=44\Rightarrow 2\times \frac{22}{7}\times r=44\Rightarrow r=7$
metres It is given that h=10 metres
$\displaystyle \therefore l^{2}=r^{2}+h^{2}\Rightarrow l=\sqrt{r^{2}+h^{2}}$
$\displaystyle=\sqrt{49+100}=\sqrt{149}=12.2m$
Now surface area of the tent $\displaystyle =\pi rl$
$\displaystyle \frac{22}{7}\times 7\times 12.2m^{2}=268.4m^{2}$
$\displaystyle \therefore $ Area of the canvas used $\displaystyle =268.4m^{2}$
It is given that the width of the canvas is 2 m
$\displaystyle \therefore $ Length of the canvas used=$\displaystyle =\frac{area}{width}=\frac{268.4}{2}=134.2m$

The volume of a right circular cone of height $8 cm$ and radius of base $3 cm$ is

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

  1. $12 \pi cm^3$

  2. $24 \pi cm^3$

  3. $48 \pi cm^3$

  4. $72 \pi cm^3$

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Correct Option: B
Explanation:

Given that height of right circular cone $h=8cm$

Radius of right circular cone $r=3cm$

Volume of right circular cone$V=\dfrac{1}{3}{{r}^{2}}\pi .h$

$ =\dfrac{1}{3}{{.3}^{2.}}\pi .8c{{m}^{3}} $

$ =24\pi c{{m}^{3}}$

Hence, this is the answer.

The circumference of the base of a 9 m high wooden solid cone is 44 m The slant height of the cone is

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

  1. $\displaystyle \sqrt{110}m$

  2. $\displaystyle \sqrt{130}m$

  3. $\displaystyle \sqrt{150}m$

  4. $\displaystyle \sqrt{180}m$

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Correct Option: B
Explanation:

Let r be the radius of the base of cone.

So, circumference of the base of the cone$=2\pi r=44  m$
$\Rightarrow 2\times \dfrac{22}{7}\times r=44$    $[\because \pi=\dfrac{22}{7}]$
$\Rightarrow r=\dfrac{44\times 7}{22\times 2}=7  m$
So the radius of the cone=7 m
$\therefore  Slant  height =\sqrt{r^2+h^2}$
$\Rightarrow \sqrt{7^2+9^2}=\sqrt{49+81}=\sqrt{130}  m$


The total surface area of a cone is $\displaystyle 770cm^{2}$ If its slant height is four times the radius of the cone the diameter fo the cone is

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

  1. 14 cm

  2. 7 cm

  3. 20 cm

  4. 18 cm

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Correct Option: A
Explanation:

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

The radius and the slant height of a cone are in the ratio 7:13 and the curved surface area is $\displaystyle 286cm^{2}$ Find its radius

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

  1. 5 cm

  2. 7 cm

  3. 8.8 cm

  4. 10.3 cm

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Correct Option: B
Explanation:

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

The radius of a conical tent is 12 m and the slant height is 5.6 m Find the length of canvas required to make the tent it the width of canvas is 4 m

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

  1. 106.6 m

  2. 100 m

  3. 52.8 m

  4. 105.6 m

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Correct Option: C
Explanation:

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 $\displaystyle L=\frac{211.2}{4}=52.8m$'

The length of the longest pole that can be kept inside a room of dimensions $12m\times 3\sqrt 3m \times 5 m$ is

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

  1. 10 m

  2. 12 m

  3. 16 m

  4. 14 m

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Correct Option: D
Explanation:

Length of the longest pole that can be put in a room $=$ Length of the diagonal inside the room.


Length of the diagonal $=\sqrt{l^2+b^2+h^2}$
$=\sqrt{12^2+(3\sqrt3)^2+5^2}$
$=\sqrt{144+27+25}$
$=\sqrt{196}$
$=14\ m$

Hence, this is the answer.