How do you find the volume of a sphere with a diameter of 25 cm?
Answer
556.5k+ views
Hint: Here in this question, we have to find the volume of the sphere of given diameter of sphere is 25 cm. this can be find by using the formula of Volume of sphere is \[V = \dfrac{4}{3}\pi {r^3}\], where r is the radius of the sphere this can be find by using the diameter, as we know radius is the half of diameter. On simplification on formula we get the required solution.
Complete step by step answer:
The volume of the sphere is the capacity it has. The shape of the sphere is round and three -dimensional. It has three axes such as x-axis, y-axis and z-axis which defines its shape. All the things like football and basketball are examples of the sphere which have volume.
The volume here depends on the diameter of radius of the sphere since if we take the cross-section of the sphere, it is a circle. The surface area of a sphere is the area or region of its outer surface. To calculate the sphere volume, whose radius is ‘r’ we have the below formula:
Volume of a sphere = \[\dfrac{4}{3}\pi {r^3}\].
Consider the given question:
The sphere has a diameter of 25 cm.
As we know the radius is half of the diameter. Then the radius of the given sphere is
\[ \Rightarrow \,\,r = \dfrac{d}{2}\]
\[ \Rightarrow \,\,r = \dfrac{{25}}{2}\]
\[ \Rightarrow \,\,r = 12.5\] cm
And the value of \[\pi \] is \[\,\pi = \dfrac{{22}}{7}\].
Consider the formula of volume of sphere
\[ \Rightarrow \,\,V = \dfrac{4}{3}\pi {r^3}\]
On substituting the values, we get
\[ \Rightarrow \,\,V = \dfrac{4}{3}\pi {r^3}\]
\[ \Rightarrow \,\,V = \dfrac{4}{3} \times \dfrac{{22}}{7} \times {\left( {12.5} \right)^3}\]`
On simplification, we get
\[ \Rightarrow \,\,V = \dfrac{{88}}{{21}} \times 1,953.125\]
\[ \Rightarrow \,\,V = \dfrac{{171,875}}{{21}}\]
\[ \Rightarrow \,\,V = \dfrac{{171,875}}{{21}}\]
\[ \Rightarrow \,\,V = 8,184.52\,\,c{m^3}\]
Hence, the volume of the sphere is \[8,184.52\,\,c{m^3}\].
Note: While Determining the volume of the sphere we use the formula \[V = \dfrac{4}{3}\pi {r^3}\], here in the question they have given the value of diameter and we know that the \[d = 2r\] and hence we determine the radius and then we find the value of volume of the sphere and we use the simple arithmetic operations while simplifying.
Complete step by step answer:
The volume of the sphere is the capacity it has. The shape of the sphere is round and three -dimensional. It has three axes such as x-axis, y-axis and z-axis which defines its shape. All the things like football and basketball are examples of the sphere which have volume.
The volume here depends on the diameter of radius of the sphere since if we take the cross-section of the sphere, it is a circle. The surface area of a sphere is the area or region of its outer surface. To calculate the sphere volume, whose radius is ‘r’ we have the below formula:
Volume of a sphere = \[\dfrac{4}{3}\pi {r^3}\].
Consider the given question:
The sphere has a diameter of 25 cm.
As we know the radius is half of the diameter. Then the radius of the given sphere is
\[ \Rightarrow \,\,r = \dfrac{d}{2}\]
\[ \Rightarrow \,\,r = \dfrac{{25}}{2}\]
\[ \Rightarrow \,\,r = 12.5\] cm
And the value of \[\pi \] is \[\,\pi = \dfrac{{22}}{7}\].
Consider the formula of volume of sphere
\[ \Rightarrow \,\,V = \dfrac{4}{3}\pi {r^3}\]
On substituting the values, we get
\[ \Rightarrow \,\,V = \dfrac{4}{3}\pi {r^3}\]
\[ \Rightarrow \,\,V = \dfrac{4}{3} \times \dfrac{{22}}{7} \times {\left( {12.5} \right)^3}\]`
On simplification, we get
\[ \Rightarrow \,\,V = \dfrac{{88}}{{21}} \times 1,953.125\]
\[ \Rightarrow \,\,V = \dfrac{{171,875}}{{21}}\]
\[ \Rightarrow \,\,V = \dfrac{{171,875}}{{21}}\]
\[ \Rightarrow \,\,V = 8,184.52\,\,c{m^3}\]
Hence, the volume of the sphere is \[8,184.52\,\,c{m^3}\].
Note: While Determining the volume of the sphere we use the formula \[V = \dfrac{4}{3}\pi {r^3}\], here in the question they have given the value of diameter and we know that the \[d = 2r\] and hence we determine the radius and then we find the value of volume of the sphere and we use the simple arithmetic operations while simplifying.
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