# Area of Circle/Proof 3/Lemma 2

< Area of Circle | Proof 3

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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 $T$.

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:

- $T \ge C$

$\blacksquare$