Question

The radius of an official NBA basketball is about \[4.7{\text{ inches}}\] . What is the volume of an NBA basketball?

Answer 1

Hint:To solve this problem, we use the concepts of volume and area of a sphere. A sphere is a 3-Dimensional structure, which has a radius \['r'\] which is equal to the distance between the centre of the sphere and any point on the sphere. And twice the radius is called diameter. So, \[d = 2r\] .

Complete step by step solution:
We know that a sphere is a 3-Dimensional object which has a radius represented as \[r\] . And, its base shape i.e., its 2-Dimensional shape is a circle.

And we also know that, the ratio of the circumference of a circle to its diameter is always constant. And we call that constant as ‘pi’ and represent it as \[\pi \] . Its value is \[\pi = \dfrac{{22}}{7} = 3.14\] .
In a 2D object, we can find area, whereas in a 3D object, we can find surface area and volume.
So, for a circle of radius \[r\] , the area is equal to \[\pi {r^2}\] square units.
And for a sphere, the surface area is equal to \[4\pi {r^2}\] square units and volume is equal to \[\dfrac{4}{3}\pi {r^3}\] cubic units.
Here, in the question, the units are inches, so we get volume in cubic inches.
So, the radius of basketball is \[r = 4.7{\text{ inches}}\] .
So, volume is \[V = \dfrac{4}{3}\pi {(4.7)^3}{\text{ cubic inches}}\]
\[ \Rightarrow V = \dfrac{4}{3}\pi (4.7)(4.7)(4.7){\text{ cubic inches}}\]
\[ \Rightarrow V = \dfrac{4}{3}\pi (103.823){\text{ cubic inches}}\]
Now, substitute the value of pi and calculate the volume.
\[ \Rightarrow V = \dfrac{4}{3}(3.14)(103.823){\text{ cubic inches}}\]
\[ \Rightarrow V = \dfrac{4}{3}(326.00422){\text{ cubic inches}}\]
\[\therefore V = 434.67{\text{ cubic inches}}\]
This is the volume of the required NBA basketball.

Note:
The value that we got after calculations is rounded off to two decimal places after the decimal point.
The surface area which is mentioned above is about the total surface area. However, there is only “total surface area” present for a sphere, but for a hemi-sphere, there is curved surface area and total surface area.
So, for a hemi-sphere, total surface area is equal to \[3\pi {r^2}\] and curved surface area is equal to \[2\pi {r^2}\] .