(a) $ \boldsymbol{ r \ = \ 2 \pi C } $
(b) $ \boldsymbol{ r \ = \ \dfrac{ C – \pi }{ 2 } } $
(c) $ \boldsymbol{ r \ = \ \dfrac{ C }{ 2 \pi } } $
(d) $ \boldsymbol{ r \ = \ C – 2 \pi } $
This question aims to develop an understanding of the algebraic simplification of the equation for the circumference of a circle using basic arithmetic operations.
The circumference of a circle is the length of its outer periphery. It is mathematically defined by the following formula:
\[ \boldsymbol{ C \ = \ 2 \pi r } \]
Where $ C $ represents the circumference and $ r $ represents the radius of the subject circle. Now this formula can be directly used to calculate the circumference given the radius of the circle, however, if we were to evaluate the value of $ r $ given the circumference, then we may have to modify it a little bit. This rearrangement process is called the algebraic simplification process which is further explained in the following solution.
Expert Answer
Given the formula of the circumference of the circle:
\[ C \ = \ 2 \pi r \]
Dividing both sides by $ 2 $:
\[ \dfrac{ C }{ 2 } \ = \ \dfrac{ 2 \pi r }{ 2 } \]
\[ \Rightarrow \dfrac{ C }{ 2 } \ = \ \pi r \]
Dividing both sides by $ \pi $:
\[ \dfrac{ C }{ 2 \pi } \ = \ \dfrac{ \pi r }{ \pi } \]
\[ \Rightarrow \dfrac{ C }{ 2 \pi } \ = \ r \]
Exchanging sides:
\[ r \ = \ \dfrac{ C }{ 2 \pi } \]
Which is the required expression. If we compare it with the given options, we can see that option (c) is the right answer.
Numerical Result
\[ r \ = \ \dfrac{ C }{ 2 \pi } \]
Example
The area of a circle is given by the following formula:
\[ A \ = \ \pi r^{ 2 } \]
Find the value of $ r $.
Dividing the above equation by $ \pi $:
\[ \dfrac{ A }{ \pi } \ = \ \dfrac{ \pi r^{ 2 } }{ \pi } \]
\[ \Rightarrow \dfrac{ A }{ \pi } \ = \ r^{ 2 } \]
Taking square root on both sides:
\[ \sqrt{ \dfrac{ A }{ \pi } } \ = \ \sqrt{ r^{ 2 } } \]
Since $ \sqrt{ r^{ 2 } } \ = \ \pm r $, above equation becomes:
\[ \sqrt{ \dfrac{ A }{ \pi } } \ = \ \pm r \]
Exchanging sides:
\[ r \ = \ \pm \sqrt{ \dfrac{ A }{ \pi } } \]