
A toy in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm, find the total surface area of the toy.
Answer
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Hint:
To find the total surface area of the toy we should find the surface area of hemisphere and cone separately and use the following process.
$\text{Total surface area of toy = curved surface area of the hemisphere + curved surface area of the cone.}$
Formula used:
Curved surface area of hemisphere \[ = 2\pi {r^2}\]
Curved surface area of cone \[ = \pi rl\]
Complete step by step answer:
Let us find curved surface area of hemisphere, here it is given that
Radius of hemisphere = radius of cone = $r = 3.5 cm$
We have to find the Curved surface area of the hemisphere. So we use formula of Curved surface area of hemisphere
Curved surface area of hemisphere \[ = 2\pi {r^2}\]
Let us substitute the known values,
\[ = 2 \times \dfrac{{22}}{7} \times {\left( {3.5} \right)^2}\]
On solving the above equation we get,
\[ = 2 \times \dfrac{{22}}{7} \times 3.5 \times 3.5\]
\[ = 2 \times 22 \times 0.5 \times 3.5\]
Area of the hemisphere \[ = {\rm{ }}77{\rm{ }}c{m^2}\]
Hence curved surface area of hemisphere is \[77{\rm{ }}c{m^2}\]
Now let us find the curved surface area of cone
We know that, Curved surface area of cone \[ = \pi rl\]
Here it is given that
Radius of cone $= r =$ \[3.5{\rm{ }}cm\]
We have to find the height of cone, as we know that
$\text{Height of cone = total height of toy - radius of hemisphere}$
Height of cone \[ = 15.5 - 3.5\]
Height of cone \[ = 12{\rm{ }}cm\]
Using the height and radius we find the slant height with the help of the following formula,
Slant height of cone is \[{l^2} = {h^2} + {r^2}\]
Let us now substitute the known values we get,
$\Rightarrow$ \[{l^2} = {\left( {12} \right)^2} + {\left( {3.5} \right)^2}\]
By finding the square values in the equation we get,
$\Rightarrow$ \[{l^2} = 144 + 12.25\]
Let us add the terms and take square root on both sides we get,
$\Rightarrow$ \[{l^2} = 156.25\]
$\Rightarrow$ \[l = \sqrt {156.25} = 12.5{\rm{ }}cm\]
On solving we have found that the slant height of cone is 12.5 cm
We have to find Curved surface area of cone, we know that
Curved surface area of cone \[ = \pi rl\]
Let us substitute the known values we get,
$\Rightarrow$ \[ \dfrac{{22}}{7} \times 3.5 \times 12.5\]
Let us solve the above equation further we get,
$\Rightarrow$ \[ 22 \times 0.5 \times 12.5\]
$\Rightarrow$Area of the cone \[ = 137.5{\rm{ }}c{m^2}\]
$\Rightarrow$ The curved surface area of cone is \[137.5{\rm{ }}c{m^2}\]
Now,
We have to find the total surface of the toy, we know that
$\text{Total surface area of toy = curved surface area of hemisphere + curved surface area of cone}$
By substituting the known values we get,
$\Rightarrow$ Total surface area of toy \[ = 77 + 137.5\]
\[ = 214.5{\rm{ }}c{m^2}\]
$\therefore$ The total surface area of the toy is \[214.5{\rm{ }}c{m^2}\]
Note:
A cone is a 3-dimensional geometry that has a planner circular base associated with a lateral surface. There is a vertex associated with a lateral surface.
The height of the cone is the distance between the vertex and the center of the base.
To find the total surface area of the toy we should find the surface area of hemisphere and cone separately and use the following process.
$\text{Total surface area of toy = curved surface area of the hemisphere + curved surface area of the cone.}$
Formula used:
Curved surface area of hemisphere \[ = 2\pi {r^2}\]
Curved surface area of cone \[ = \pi rl\]
Complete step by step answer:
Let us find curved surface area of hemisphere, here it is given that
Radius of hemisphere = radius of cone = $r = 3.5 cm$
We have to find the Curved surface area of the hemisphere. So we use formula of Curved surface area of hemisphere
Curved surface area of hemisphere \[ = 2\pi {r^2}\]
Let us substitute the known values,
\[ = 2 \times \dfrac{{22}}{7} \times {\left( {3.5} \right)^2}\]
On solving the above equation we get,
\[ = 2 \times \dfrac{{22}}{7} \times 3.5 \times 3.5\]
\[ = 2 \times 22 \times 0.5 \times 3.5\]
Area of the hemisphere \[ = {\rm{ }}77{\rm{ }}c{m^2}\]
Hence curved surface area of hemisphere is \[77{\rm{ }}c{m^2}\]
Now let us find the curved surface area of cone
We know that, Curved surface area of cone \[ = \pi rl\]
Here it is given that
Radius of cone $= r =$ \[3.5{\rm{ }}cm\]
We have to find the height of cone, as we know that
$\text{Height of cone = total height of toy - radius of hemisphere}$
Height of cone \[ = 15.5 - 3.5\]
Height of cone \[ = 12{\rm{ }}cm\]
Using the height and radius we find the slant height with the help of the following formula,
Slant height of cone is \[{l^2} = {h^2} + {r^2}\]
Let us now substitute the known values we get,
$\Rightarrow$ \[{l^2} = {\left( {12} \right)^2} + {\left( {3.5} \right)^2}\]
By finding the square values in the equation we get,
$\Rightarrow$ \[{l^2} = 144 + 12.25\]
Let us add the terms and take square root on both sides we get,
$\Rightarrow$ \[{l^2} = 156.25\]
$\Rightarrow$ \[l = \sqrt {156.25} = 12.5{\rm{ }}cm\]
On solving we have found that the slant height of cone is 12.5 cm
We have to find Curved surface area of cone, we know that
Curved surface area of cone \[ = \pi rl\]
Let us substitute the known values we get,
$\Rightarrow$ \[ \dfrac{{22}}{7} \times 3.5 \times 12.5\]
Let us solve the above equation further we get,
$\Rightarrow$ \[ 22 \times 0.5 \times 12.5\]
$\Rightarrow$Area of the cone \[ = 137.5{\rm{ }}c{m^2}\]
$\Rightarrow$ The curved surface area of cone is \[137.5{\rm{ }}c{m^2}\]
Now,
We have to find the total surface of the toy, we know that
$\text{Total surface area of toy = curved surface area of hemisphere + curved surface area of cone}$
By substituting the known values we get,
$\Rightarrow$ Total surface area of toy \[ = 77 + 137.5\]
\[ = 214.5{\rm{ }}c{m^2}\]
$\therefore$ The total surface area of the toy is \[214.5{\rm{ }}c{m^2}\]
Note:
A cone is a 3-dimensional geometry that has a planner circular base associated with a lateral surface. There is a vertex associated with a lateral surface.
The height of the cone is the distance between the vertex and the center of the base.
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