# Area of Circle/Proof 3/Lemma 3

## 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:

$T \le C$

## Proof

Aiming for a contradiction, suppose $T > C$.

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

$P = \dfrac {h q} 2$

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:

$0 < h = r \land 0 < c < q \implies \dfrac {h q} 2 > \dfrac {r c} 2$

$\neg T > C$.
$T \le C$.
$\blacksquare$