# Area of Circle/Proof 3

## 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: $T = \pi r^2$

 $\ds T$ $=$ $\ds \frac {r c} 2$ Area of Triangle in Terms of Side and Altitude $\ds$ $=$ $\ds \frac {r 2 \pi r} 2$ Perimeter of Circle $\ds$ $=$ $\ds \pi r^2$

$\Box$

### Lemma 2: $T \ge C$ Aiming for a contradiction, suppose $T < C$.

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

For any given regular polygon:

$P = \dfrac {h q} 2$

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:

$P > T \implies \dfrac {h q} 2 > \dfrac {r c} 2$

On the other hand:

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

Hence $\neg T < C$.

Hence $T \ge C$.

$\Box$

### Lemma 3: $T \le C$ Aiming for a contradiction, suppose $T > C$.

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

$P = \dfrac {h q} 2$

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 $A = \dfrac {h q} {2 n}$ and with $n$ triangles we get $P = \dfrac {h q} 2$

On one hand:

$P < T \implies \dfrac {h q} 2 < \dfrac {r c} 2$

On the other hand:

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

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.