Area of Circle/Proof 3

Proof

 * [[File:Area Equal.jpg]]

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 2: $T \ge C$

 * [[File:Area Smaller.jpg]]

$T < C$.

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 a contradiction is obtained.

Hence $\neg T < C$.

Hence $T \ge C$.

Lemma 3: $T \le C$

 * [[File:Area Greater.jpg]]

$T > C$.

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

From Area of Polygon by Inradius and Perimeter:


 * $P = \dfrac {h q} 2$

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 a contradiction is obtained.

Hence $\neg T > C$.

Hence $T \le C$.

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)