Area of Circle/Proof 3/Lemma 2

Lemma for Area of Circle: Proof 3

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

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 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:

$T \ge C$

$\blacksquare$