Question

The surface area of a football is $616m{{m}^{2}}$ . Calculate its radius.
$\begin{align}
  & a)4mm \\
 & b)7mm \\
 & c)6mm \\
 & d)5mm \\
\end{align}$

Answer 1

Hint: Now we know that the sphere is a curved surface whose surface area is given by the formula $4\pi {{r}^{2}}$ where r is the radius of the sphere. Now we know the surface area, hence we will use the formula to find the radius of the sphere. For calculation we will assume the value of $\pi $ to be $\dfrac{22}{7}$ .

Complete step by step solution:
Now let us first understand the concept of sphere.
A sphere is a solid round shaped 3 dimensional structure.
The sphere can be defined as a set of all points which are at equal distance from a particular point. Now this fixed point is called the centre of the sphere. The distance between the center and the surface is called radius of sphere.
Now a sphere is a curved symmetrical surface with no points or flat surface. Hence the only surface it has is curved surface area. An example of a sphere is a ball.
Now for a sphere with radius r, we have the area of sphere is $\dfrac{4}{3}\pi {{r}^{3}}$ and the Surface area of sphere is $4\pi {{r}^{2}}$ .
Now consider the given example.
We are given that the surface area of a ball is $616m{{m}^{2}}$
Now we know that for a sphere with radius r the surface area is given by $4\pi {{r}^{2}}$ .
Hence we have
\[\begin{align}
  & \Rightarrow 4\pi {{r}^{2}}=616 \\
 & \Rightarrow {{r}^{2}}=\dfrac{616}{4\pi } \\
 & \Rightarrow {{r}^{2}}=\dfrac{154}{\pi } \\
\end{align}\]
Now let us take the approximate value of $\pi $ as $\dfrac{22}{7}$ hence we get,
$\Rightarrow {{r}^{2}}=\dfrac{154\times 7}{22}=49$
Hence taking the square root we get the radius r = 7mm.
Hence the radius of the sphere is 7mm.
So, the correct answer is “Option b”.

Note: Now note that the area of sphere is $4\pi {{r}^{2}}$ and the volume of sphere is $\dfrac{4}{3}\pi {{r}^{3}}$ . Not to be confused between the two. Also note that the diameter of the sphere is 2r. Hence we can also write the formula in terms of diameter as $\pi {{d}^{2}}$ and $\dfrac{1}{6}\pi {{d}^{3}}$ respectively.