Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm.
Answer
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Hint: The largest circular cone that can be cut out of the cube will have height equal to cube edge and base diameter of cone will be equal to edge of the cube.
Largest right circular cone will have the same height. Let this height be h.
Diameter of the base of the cone is equal to the edge of the cube since it touches all the edges of the cube side.
Let the radius be r.
$ \Rightarrow h = 9cm\;\& \;r = \dfrac{{diameter}}{2} = \dfrac{9}{2} = 4.5cm$
We know that volume of a cone with radius r and height h is
Volume $ = \dfrac{1}{3}\pi {r^2}h$
Substituting r and h values in the above formula, we get
$ \Rightarrow Volume = \dfrac{1}{3}\pi {\left( {4.5} \right)^2}9 = 190.85c{m^3}$
$\therefore $ The volume of the largest right circular cone formed from the given cube is $190.85c{m^3}$
Note:
Here the base of the cone will be the circle inscribed in the face of the cube and its height will be equal to the edge length of the cube. We need to visualize the given problem in geometrical structures to solve the problem easily. When we represent the given problem in graphical form it looks similar to the below figure.
Largest right circular cone will have the same height. Let this height be h.
Diameter of the base of the cone is equal to the edge of the cube since it touches all the edges of the cube side.
Let the radius be r.
$ \Rightarrow h = 9cm\;\& \;r = \dfrac{{diameter}}{2} = \dfrac{9}{2} = 4.5cm$
We know that volume of a cone with radius r and height h is
Volume $ = \dfrac{1}{3}\pi {r^2}h$
Substituting r and h values in the above formula, we get
$ \Rightarrow Volume = \dfrac{1}{3}\pi {\left( {4.5} \right)^2}9 = 190.85c{m^3}$
$\therefore $ The volume of the largest right circular cone formed from the given cube is $190.85c{m^3}$
Note:
Here the base of the cone will be the circle inscribed in the face of the cube and its height will be equal to the edge length of the cube. We need to visualize the given problem in geometrical structures to solve the problem easily. When we represent the given problem in graphical form it looks similar to the below figure.
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