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HINT- Proceed the solution of this question, first visualising the remaining part that will be left after a hemispherical depression is cut out from one face of the cubical wooden Block then using mensuration formulae we can find the area of remaining solid.
Complete step-by-step answer:
Consider the diagram shown below.
It is given that a hemisphere of radius \[\dfrac{l}{2}\] (diameter = l ) is cut out from the top face of the cubical wooden block.
Here, the diameter of the hemisphere is equal to the edge of the cube.
So, diameter = side of cube = l
Here, the base of the hemisphere would not be included in the total surface area of the wooden cube.
Therefore, surface area of the remaining solid
= surface area of the cubical box who’s each edge is of length l − Area of the top of the hemispherical part + curved surface area of the hemispherical part
Area of cube
Here, side = l
Area of cube = $6 \times {({\text{side}})^2} = 6 \times {(l)^2}$
Curved surface area of hemisphere
Diameter of hemisphere = l
Hence, radius = r = $\dfrac{{{\text{Diameter}}}}{2} = \dfrac{l}{2}$
We know that
Curved surface area of hemisphere = $2\pi {{\text{r}}^2}$
On putting $r = \dfrac{l}{2}$
Curved surface area of hemisphere = $2\pi {\left( {\dfrac{l}{2}} \right)^2} = 2\pi \times \dfrac{{{l^2}}}{4} = \pi \times \dfrac{{{l^2}}}{2}$
Base area of hemisphere
Base area of hemisphere is a circle with radius = radius of hemisphere = r $ = \dfrac{l}{2}$
Base area of hemisphere = $\pi {{\text{r}}^2}$
$ \Rightarrow \pi {\left( {\dfrac{l}{2}} \right)^2}$
$ \Rightarrow \pi \times \dfrac{{{l^2}}}{4}$
Base area of hemisphere $ = \pi \times \dfrac{{{l^2}}}{4}$
Now, surface area of the remaining solid
= surface area of the cubical box who’s each edge is of length l − Area of the top of the hemispherical part + curved surface area of the hemispherical part
$ \Rightarrow 6 \times {(l)^2}$- $\pi \times \dfrac{{{l^2}}}{4}$ + $\pi \times \dfrac{{{l^2}}}{2}$
On further solving
$ \Rightarrow 6 \times {(l)^2}$ + $\pi \times \dfrac{{{l^2}}}{4}$
On taking ${l^2}$ as a common
\[ \Rightarrow {(l)^2}\left( {6 + \dfrac{\pi }{4}} \right) \Rightarrow {(l)^2}\left( {\dfrac{{24 + \pi }}{4}} \right) \Rightarrow \dfrac{{{{(l)}^2}}}{4}\left( {24 + \pi } \right)\]
\[ \Rightarrow \dfrac{{{{(l)}^2}}}{4}\left( {24 + \pi } \right)\] sq. units
Hence, surface area of the remaining solid = \[\dfrac{{{{(l)}^2}}}{4}\left( {24 + \pi } \right)\]
Note- In this particular question, sometimes it is difficult to visualise the red portion area (shown in below figure) and simply tells that surface area of the remaining solid will sum of
= surface area of 5 faces of the cubical box + curved surface area of the hemispherical part
But it’s not true.
Hence we can also write surface area of the remaining solid
= surface area of 5 faces of the cubical box who’s each edge is of length l + surface area of upper face − Area of the top of the hemispherical part which is a circle + curved surface area of the hemispherical part
By solving like this, we will get the same answer.
Complete step-by-step answer:
Consider the diagram shown below.
It is given that a hemisphere of radius \[\dfrac{l}{2}\] (diameter = l ) is cut out from the top face of the cubical wooden block.
Here, the diameter of the hemisphere is equal to the edge of the cube.
So, diameter = side of cube = l
Here, the base of the hemisphere would not be included in the total surface area of the wooden cube.
Therefore, surface area of the remaining solid
= surface area of the cubical box who’s each edge is of length l − Area of the top of the hemispherical part + curved surface area of the hemispherical part
Area of cube
Here, side = l
Area of cube = $6 \times {({\text{side}})^2} = 6 \times {(l)^2}$
Curved surface area of hemisphere
Diameter of hemisphere = l
Hence, radius = r = $\dfrac{{{\text{Diameter}}}}{2} = \dfrac{l}{2}$
We know that
Curved surface area of hemisphere = $2\pi {{\text{r}}^2}$
On putting $r = \dfrac{l}{2}$
Curved surface area of hemisphere = $2\pi {\left( {\dfrac{l}{2}} \right)^2} = 2\pi \times \dfrac{{{l^2}}}{4} = \pi \times \dfrac{{{l^2}}}{2}$
Base area of hemisphere
Base area of hemisphere is a circle with radius = radius of hemisphere = r $ = \dfrac{l}{2}$
Base area of hemisphere = $\pi {{\text{r}}^2}$
$ \Rightarrow \pi {\left( {\dfrac{l}{2}} \right)^2}$
$ \Rightarrow \pi \times \dfrac{{{l^2}}}{4}$
Base area of hemisphere $ = \pi \times \dfrac{{{l^2}}}{4}$
Now, surface area of the remaining solid
= surface area of the cubical box who’s each edge is of length l − Area of the top of the hemispherical part + curved surface area of the hemispherical part
$ \Rightarrow 6 \times {(l)^2}$- $\pi \times \dfrac{{{l^2}}}{4}$ + $\pi \times \dfrac{{{l^2}}}{2}$
On further solving
$ \Rightarrow 6 \times {(l)^2}$ + $\pi \times \dfrac{{{l^2}}}{4}$
On taking ${l^2}$ as a common
\[ \Rightarrow {(l)^2}\left( {6 + \dfrac{\pi }{4}} \right) \Rightarrow {(l)^2}\left( {\dfrac{{24 + \pi }}{4}} \right) \Rightarrow \dfrac{{{{(l)}^2}}}{4}\left( {24 + \pi } \right)\]
\[ \Rightarrow \dfrac{{{{(l)}^2}}}{4}\left( {24 + \pi } \right)\] sq. units
Hence, surface area of the remaining solid = \[\dfrac{{{{(l)}^2}}}{4}\left( {24 + \pi } \right)\]
Note- In this particular question, sometimes it is difficult to visualise the red portion area (shown in below figure) and simply tells that surface area of the remaining solid will sum of
= surface area of 5 faces of the cubical box + curved surface area of the hemispherical part
But it’s not true.
Hence we can also write surface area of the remaining solid
= surface area of 5 faces of the cubical box who’s each edge is of length l + surface area of upper face − Area of the top of the hemispherical part which is a circle + curved surface area of the hemispherical part
By solving like this, we will get the same answer.
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