Area of Circle/Proof 3/Lemma 2

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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 $T$.


$T \ge C$



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:

$P = \dfrac {h q} 2$


$q$ 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.


$\neg T < C$

and so:

$T \ge C$