What is the expected perimeter of the quadrilateral implied by choosing points uniformly on each side of a unit square? On each side of a square with unit length sides choose a point uniformly. Connect these points to form a quadrilateral. What is the expected perimeter of this quadrilateral? 
I am able to simulate and also solve the integral to get the expected area of $\frac12$.
$$1-\frac12\int_{d=0}^{1}\int_{c=0}^{1}\int_{b=0}^{1}\int_{a=0}^{1}a(1-b)+b(1-c)+c(1-d)+d(1-a) da db dc dd=\frac12$$
The corresponding integral for the perimeter does not appear to have a straightforward solution but simulation shows the expected value to be somewhere around $3.06$.
 A: Your vertices lie on $(a,0),(1,b),(c,1),(0,d)$ hence by the Pythagorean Theorem, the perimeter is
$$
P(a,b,c,d) = \sqrt{(a-1)^2+b^2} + \sqrt{(b-1)^2+(c-1)^2} + \sqrt{c^2+(d-1)^2} + \sqrt{a^2+d^2}
$$
and you need to compute $\mathbb{E}[P]$ over a joint distribution of 4 independent uniforms, i.e.
$$
\int_{[0,1]^4} P(a,b,c,d) da\ db\ dc\ dd \approx 3.0676,
$$
with the approximation provided by Wolfram Alpha.
To be more precise, note that each of the roots basically yield the same result (since $1-U$ and $U$ have the same distribution if $U \sim \mathcal{U}(0,1)$). So we get
$$
\mathbb{E}[P]
 = 4\int_0^1 \int_0^1 \sqrt{a^2 + d^2}\ da\ dd
 = \frac{4}{3} \left[ \sqrt2 + \sinh^{-1}(1) \right]
 = \frac{4}{3} \left[ \sqrt2 + \ln \left(1 + \sqrt2 \right) \right]
$$

With that approach, area becomes simple as well. Notice that the area of your quadrilateral is exactly the area of the unit square less areas of 4 right triangles, each of which has legs sides, which are uniformly distributed and hence has an expected area of 
$$
\int_0^1 \int_0^1 \frac{ad}{2} \ da\ dd
 = \frac12 \left(\int_0^1 a\ da\right) \left(\int_0^1 d\ dd\right)
 = \frac18, 
$$
so the total area is $A = 1 - \frac48= \frac12$ as you computed.
A: The expected length of a side is $$\int_0^1 \int_0^1 \sqrt{x^2+y^2}\ dx \ dy$$
I was able to get Alpha do do the two integrals, the $x$ one giving
$$\frac 14 \left(2 (\sqrt{y^2 + 1} + y^2 \log(\sqrt{y^2 + 1} + 1)) - y^2 \log(y^2)\right)$$
and the second giving 
$$\frac 13 (\sqrt2 + \sinh^{-1}(1))≈0.76520$$
The expected perimeter is $4$ times this, or about $3.0608$
