# Area of Circle/Proof 5

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## Theorem

The area $A$ of a circle is given by:

- $A = \pi r^2$

where $r$ is the radius of the circle.

## Proof

Let the circle of radius $r$ be divided into many sectors:

If they are made small enough, they can be approximated to triangles whose heights are all $r$.

Let the bases of these triangles be denoted:

- $b_1, b_2, b_3, \ldots$

From Area of Triangle in Terms of Side and Altitude, their areas are:

- $\dfrac {r b_1} 2, \dfrac {r b_2} 2, \dfrac {r b_3} 2, \ldots$

The area $\mathcal A$ of the circle is given by the sum of the areas of each of these triangles:

\(\ds \mathcal A\) | \(=\) | \(\ds \dfrac {r b_1} 2 + \dfrac {r b_2} 2 + \dfrac {r b_3} 2 + \cdots\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac r 2 \left({b_1 + b_2 + b_3 + \cdots}\right)\) |

But $b_1 + b_2 + b_3 + \cdots$ is the length of the circumference of the circle.

From Perimeter of Circle:

- $b_1 + b_2 + b_3 + \cdots = 2 \pi r$

Hence:

\(\ds \mathcal A\) | \(=\) | \(\ds \dfrac r 2 \left({b_1 + b_2 + b_3 + \cdots}\right)\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac r 2 \left({2 \pi r}\right)\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \pi r^2\) |

It needs to be noted that this proof is intuitive and non-rigorous.

$\blacksquare$

## Historical Note

This was the method used by Johannes Kepler when he was working on his Second Law of Planetary Motion.

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.10$: Kepler ($\text {1571}$ – $\text {1630}$)