# Area of Circle/Proof 3

## Contents

## Theorem

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

- $A = \pi r^2$

where $r$ is the radius of the circle.

## Proof

Refer to the figure.

Construct a circle with radius r and circumference $c$, where its area is denoted by $C$.

Construct a triangle with height r and base $c$, where its area is denoted by $T$.

### Lemma 1: $T \mathop = \pi r^2$

\(\displaystyle T\) | \(=\) | \(\displaystyle \frac{rc}2\) | $\quad$ Area of Triangle in Terms of Side and Altitude | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac{r 2 \pi r} 2\) | $\quad$ Perimeter of Circle | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \pi r^2\) | $\quad$ | $\quad$ |

$\Box$

### Lemma 2: $T \ge C$

This will be proven by contradiction.

Assume $T < C$.

It should be possible to construct a regular polygon with area $P$, where $T < P < C$.

For any given regular polygon:

- $P = \dfrac {hq} 2$

where:

- $q$ is the perimeter of the polygon
- $h$ is the height of any given triangular part of it
- $P$ is the area.

On one hand:

- $P > T \implies \dfrac {hq} 2 > \dfrac {rc} 2$

On the other hand:

- $0 < h < r \land 0 < q < c \implies \dfrac {hq} 2 < \dfrac {rc} 2$

Hence a contradiction is obtained.

Hence $\neg T < C$.

Hence $T \ge C$.

$\Box$

### Lemma 3: $T \le C$

This will be proven by contradiction.

Assume $T > C$.

It should be possible to construct a regular polygon with area $P$, where $C < P < T$.

From Area of Polygon by Inradius and Perimeter:

- $P = \dfrac {hq} 2$

where:

- $q$ is the perimeter of the regular polygon
- $h$ is the inradius of the regular polygon
- $P$ is the area.

as each triangle has the base $B = \dfrac {q} n$ and area $A = \dfrac {hq}{2n}$ and with $n$ triangles we get $P = \dfrac {hq} 2$

On one hand:

- $P < T \implies \dfrac {hq} 2 < \dfrac {rc} 2$

On the other hand:

- $0 < h = r \land 0 < c < q \implies \dfrac {hq} 2 > \dfrac {rc} 2$

Hence a contradiction is obtained.

Hence $\neg T > C$.

Hence $T \le C$.

$\Box$

### Final Proof

- $T \ge C$ (from Lemma 2)
- $T \le C$ (from Lemma 3)
- $\therefore T \mathop = C$
- $\therefore C \mathop = T \mathop = \pi r^2$ (from Lemma 1)

$\blacksquare$

## Historical Note

The area of a circle was determined by Archimedes in his *Measurement of a Circle*.

## Sources

- 1939: R.P. Gillespie:
*Integration*(2nd ed.) ... (previous) ... (next): Chapter $\text I$: $\S 1$. Area of a Circle - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.5$: Archimedes (ca. $287$ – $212$ B.C.)