Area of Circle/Proof 3/Lemma 2

Lemma for Area of Circle: Proof 3

 * Area-of-Circle-Proof-3.png

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

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

Then:
 * $T \ge C$

Proof

 * Area-of-Circle-Proof-3-Lemma-2.png

$T < C$.

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 regular polygon
 * $h$ is the height of any given triangular part of it
 * $P$ is the area.

Onthe 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:
 * $\neg T < C$

and so:
 * $T \ge C$