Proving that the area of a polygon is less than $\pi/4$ 
The sides and diagonals of a (not necessarily convex) polygon have length at most $1$. Prove that the area of the polygon is less than $\dfrac{\pi}{4}$. 

I was able to prove that the area must be less than $\dfrac{\pi}{2}$ by doing the following

By considering the convex hull, we only need to prove the claim is true for convex polygons. Fixing the diameter as an axis and letting the midpoint of the diameter be the origin. If we let $r$ be the distance of the point on the polygon at an angle $\theta$ to the origin. Then, the area is $$A = \int_0^\pi \frac{r(\theta)^2 + r(\theta + \pi)^2}{2} \, \mathrm{d}\theta \leq \int_0^\pi \frac{1}{2} \, \mathrm{d}\theta = \frac{\pi}{2}$$

But I can't seem to reduce this bound down to $\pi/4$. Any solutions/suggestions are appreciated. 
 A: A key lemma is the same used in the proof of the pizza theorem, namely the Proposition $11$ from Archimedes' Book of Lemmas:

If in a circle $AB$ and $CD$ are perpendicular chords meeting at $O$, 
  $$OA^2+OB^2+OC^2+OD^2=d^2$$
  where $d$ is the diameter of the circle.

This allows to state
$$ A = \int_{0}^{\pi/2}\frac{r(\theta)^2+r(\theta+\pi/2)^2+r(\theta+\pi)^2+r(\theta+3\pi/2)^2}{2}\,d\theta \leq \int_{0}^{\pi/2}\frac{d\theta}{2} = \frac{\pi}{4}.$$
For instance, the area of the Reuleaux triangle with unit diameter is $\frac{\pi-\sqrt{3}}{2}=0.704771\ldots< 0.785398\ldots=\frac{\pi}{4}$.
The Blaschke-Lebesgue theorem is related. 
A: HINT: Show first that the perimeter of your convex polygon is less than $\pi$. 
ADDED: 
For every $\theta \in [0,\pi]$ consider the projection of the polygon (convex!) on a line of direction $\theta$ --  a segment of length denoted by $w_{\theta}$. It is a basic fact of integral geometry ( Crofton formula) that the perimeter $P$ of the polygon equals
$$P=\int_{0}^{\pi} w_{\theta}\, d \theta$$
Now clearly $w_{\theta}$ is at most the diameter of the polygon. Therefore:
$$P \le \pi $$ 
The inequality is in fact strict, since the function $\theta\mapsto w_{\theta}$ cannot be constant for polygons.
Now we can apply some classic facts about polygons with a given perimeter and fixed number of vertices ( the largest area is achieved by regular polygons). Of course that will be smaller than the area of the circle of the same perimeter.
Another approach, that works in all dimensions, is the following. Consider a convex body $K$ of diameter $D$ in dimension $n$.  The problem is to show that its $n$-dimensional volume does not exceed that of an $n$-ball of same diameter.
For this, consider the central symmetrization of $K$
$$\tilde K = \frac{1}{2}K + \frac{1}{2}(-K)$$
(any choice of origin to get the negative gives the same $\tilde K$, up to a translation). It is easy to see that $\tilde K$ is centrally symmetric, and it has the same diameter as $K$. Moreover, by the Brunn-Minkowski theorem for sums of convex bodies, the volume of $\tilde K$ is at least the volume of $K$, with equality if and only if $K$ is centrally symmetric. So now we reduce to proving the result for $\tilde K$. But this is immediate, since $\tilde K$ is contained in a ball of diameter $D$ with same center as $\tilde K$. Therefore the volume of $K$ is at most the volume of a ball of same diameter. We have equality if and only if $K$ is itself a ball.
