Area of Circle/Proof 3/Lemma 3

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:
 * $T \le C$

Proof

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

$T > C$.

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:
 * base $B = \dfrac q n$
 * 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:
 * $\neg T > C$.

and so:
 * $T \le C$.