Question

The circumference of a circle is 50.24 centimeters. How do you find the area of the circle?

Answer 1

Hint: We have been given the circumference of a circle and using this information, we have to calculate the area of the same circle. Thus, we shall first apply the formula of the circumference of a circle and calculate the radius using it. Then, we shall find the area of the circle using the radius calculated.

Complete step by step solution:
The most essential yet common quantity on which the various measurements like circumference and radius of a circle depends upon is the radius of the circle.
We know that the circumference, C of a circle is expressed as:
$C=2\pi r$ ……………………. (1)
Where, r is the radius of the circle.
Also, the area of a circle is calculated using the formula:
$A=\pi {{r}^{2}}$ …………………….. (2)
Where, r is again the radius of the circle.
Thus, we shall first find out the radius of the circle given to us whose circumference is $50.24cm$.
We will substitute the value of circumference of the circle in equation (1).
$\Rightarrow 50.24=2\pi r$
Dividing both sides by $2\pi $, we get
 $\begin{align}
  & \Rightarrow \dfrac{50.24}{2\pi }=\dfrac{2\pi r}{2\pi } \\
 & \Rightarrow r=\dfrac{50.24}{2\pi } \\
\end{align}$
Putting $\pi =3.14$ in the above equation, we get
$\Rightarrow r=\dfrac{50.24}{2\times 3.14}$
$\Rightarrow r=\dfrac{50.24}{6.28}$
$\Rightarrow r=8cm$
Now, we shall use this calculated value of the radius in equation (2) to find the area of the circle.
$\begin{align}
  & \Rightarrow A=\pi {{\left( 8 \right)}^{2}} \\
 & \Rightarrow A=\pi \left( 8\times 8 \right) \\
\end{align}$
Since, we know that $8\times 8=64$,
$\Rightarrow A=64\pi $
Again substituting the value of $\pi $, we get
$\Rightarrow A=200.96c{{m}^{2}}$
Therefore, the area of the circle is $200.96c{{m}^{2}}$.

Note:
We must perform calculations carefully and must be well-versed with the values of a few basic constants like $\pi $. The radius and center of a circle are its fundamental quantities which define a particular circle and differentiate it from the circles.