What is the average area of all triangles that can be inscribed in a unit circle? There is only one circle that passes through any three given points. Hence by suitable scaling, we can inscribe every triangle inside a unit circle of radius $1$. We define distinct triangles as triangles which have different sides regardless of of the order. Hence a triangle with sides $(a,b,c)$ and a triangle with sides $(b,c,a)$ are not distinct.
Question 1: What is the average area of all distinct triangles that can be inscribed in a unit circle?
Question 2: What is the average perimeter of all distinct triangles that can be inscribed in a unit circle?
Motivation: This question was motivated by this related question where it was proved that the average perimeter of all right triangles inscribed in a semi-circle of unit diameter is $1+4/\pi$.
 A: Comment:
As can be seen in figure the area and perimeter of right angled triangle in half circle is maximum when the height is maximum , i.e. $h=r$ where h is height of triangle and r is the radius of circle. The area and perimeter are minimum when $h → 0$. So we can write:
$a/2 = r= 1$
$S_{max}=(\frac{(\sqrt 2)^2}{2})=1$ 
$S_{min}= 0$
$S_{ave.}=\frac{1+2\times 0}{3}=\frac{1}{3}$
and perimeter:
$P_{max}=2+2\sqrt2$
$P_{min}=0$
$P_{ave}=\frac{2(1+\sqrt2)+2\times 0}{3} ≈ 2.27 ≈ 1+\frac{4}{\pi}$
Now suppose we want to find the average area and perimeter of isosceles triangles which can be inscribed in circle with radius unit. In triangle $OHC_1$ we have:
$(\frac{a}{2})^2+(dr)^2=r^2=1$
$S_{AB_1C_1}=\frac{a}{2}\times (r+dr)=\frac{a}{2}(1+dr)$
Eliminating $dr$ and letting $\frac{a}{2}=x$ we get:
$S=x(1+\sqrt{1-x^2})$
⇒$S'=\frac{\sqrt{1-x^2}+1-2x^2}{\sqrt{1-x^2}}=0$
⇒$x=0$ and $x=\frac{\sqrt 3}{2}$
⇒$a= \sqrt 3$
⇒$S_{max}=\frac{3}{2}\times\frac{\sqrt3}{2}=\frac{3\sqrt3}{4}$
That is S is maximum when $a=b=c=\sqrt 3$, i.e. when the triangle is equilateral.The minimum value is when $h →0$ or $a →0$ , so the average of area can be:
$S_{ave.}=\frac{\frac{3\sqrt3}{4}+2\times 0}{3}=\frac{\sqrt 3}{4}$
Similarly you can find the average of perimeter:
$P=a+2\sqrt{2+2\sqrt{1-(a/2)^2}}$
Now take derivative and so on.  
