An unusual isoperimetric conjecture -- is it known? In a textbook for elementary school mathematics (I happen to coach my grandson) I came across the following exercise:

A pyramid has the base area $60$, the lateral area $120$, and the height $8$.
  Find the surface area and the volume of the pyramid.

The answer is obtained routinely: the surface area is $60+120=180$ and the volume is $60\cdot 8/3=160$. Alas, routine ways of doing exercises are not always correct ways of doing them.
This exercise has something important in common with the problem 6 in V. I. Arnold's
"Problems for children from 5 to 15",
which you might have already encountered somewhere else
on StackExchange Mathematics:

The hypotenuse of a right-angled triangle (in a standard American
  examination) is $10$ inches, the altitude dropped onto it is $6$ inches.
  Find the area of the triangle.
American school students had been coping succesfully with this problem
  over a decade. But the Russian school students arrived from Moscov,
  and none of them was able to solve it as had their American peers
  (giving 30 square inches as the answer). Why?

The answer is simple: the Russian school students actualy could not calculate the area of the triangle for the simple reason that the triangle does not exist.
(Neither could American school students calculate the area, but they never considered this possibility.) For the same reason we cannot calculate the surface area and the volume of the pyramid in the exercise quoted above:
it does not exist -- or, to be precise, I suspect that it does not exist. But, while it is simple to see that the triangle in the Arnold's problem 6 does not exist, proving the
(non)existence of the pyramid
in the exercise above is a far from trivial matter. I did prove that a pyramid with the base area,
the lateral area, and the height as prescribed in the exercise does not exist,
if we require that it is
a right pyramid with a rectangular base. But the exercise is not saying what kind of pyramid it is talking about: the pyramid may be oblique, and its base may be an arbitrary simple polygon with any number of sides. Attentively listening to my mathematical intuition I heard it whispering to me that the following is true (and if it is true, it confirms that 
the pyramid above does not exist):

Conjecture. In the three-dimensional Euclidean space let $V$ be a point
  at a distance $h>0$ from a plane $\Pi$, and let $A>0$. Let $\mathscr{C}(A)$
  be the set of all rectifiable simple closed curves $C$ in the plane $\Pi$
  such that the bounded closed part of the plane $\Pi$ that has the boundary $C$
  has the area $A$. For every $C\in\mathscr{C}(A)$ let $\Lambda(C)$ be
  the area of the surface swept by the line segment $VX$ when the point $X$
  runs around the curve $C$. Then $\Lambda(C) \geq \sqrt{A\,(A+\pi\,h^2)}$
  for every $C\in\mathscr{C}(A)$, where the equality holds iff
  $C$ is the circle in the plane $\Pi$ with the radius $\sqrt{A/\pi}$
  and with the center at the foot of the perpendicular dropped from the
  point $V$ to the plane $\Pi$.

I do not only suspect, I actually know that the conjecture is true, though presently I have no idea
how to go about proving it. Did you see the conjecture, or something equivalent to it, somewhere?
If you did, please let me know.
Added later. Professor Vector has provided an elegant proof of the inequality in the conjecture. Since the conjecture is no longer just a conjecture, it behooves to reformulate it as a proven fact, and while doing it, to cut some crap. We define a generalized cone as the kind of creature implicitly appearing in the conjecture: it has a base that is a bounded closed subset of a plane whose boundary is an arbitrary rectifiable simple closed curve $C$, it has an apex $V$ which is a point not in the base plane (to avoid degeneracy), and it has a lateral which is the union of all line segments $VX$ with $X$
a point in $C$. Also for every generalized cone we denote by $A$ the area of its base, by $h$ its height
(the distance of the apex from the base plane), and by $\Lambda$ the area of its lateral. Then:

Every generalized cone satisfies the inequality
  $$\Lambda \:\geq\: \sqrt{A\,(A+\pi\,h^2)}~,$$
  where the equality holds only for a right circular cone.

Professor Vector's proof of the inequality is restricted to polygonal bases (and hence to pyramids), but it easily extends, by a limiting process, to the general case. The only part of the conjecture that remains unproved is the claim that the equality occurs if and only if the generalized cone is in fact
a right circular cone, but we will manage to somehow survive, at least for a little while, without this particular cherry on top of the cake.
 A: Your intuition hasn't failed you. It's true that such a pyramid can't exist, and for the same reason as in the triangle case: the height of the pyramid would be too big. I can confirm your exact lower bound, with the obvious correction ($h$ instead of $v$): $\Lambda(C)\ge\sqrt{A\,(A+\pi\,h^2)}$.
Proof: We assume that $C$ is a polygon with edges $e_i$ of length $l_i$, so $L=\sum_i\,l_i$ is the length of $C$. Let $V'$ be the projection of $V$ on $\Pi$. To cover the case when $V'$ lies outside of $C$, we give the polygon an orientation, so that the interior of $C$ always lies on the left-hand side of $e_i$ (seen from point $V$). Let $h_i$ be the signed distance of $V'$ from the line containing $e_i$ (negative if $V'$ lies to the right of the line, compare the corresponding section of http://geomalgorithms.com/a01-_area.html), then we have $A=\frac{1}{2}\,\sum_i\,l_i\,h_i$. The distance of $V$ from that line will be $\sqrt{h_i^2+h^2}$, so the area of the triangle formed by $V$ and $e_i$ is $\frac{1}{2}\,l_i\,\sqrt{h_i^2+h^2}$.  Now the function $f(x)=\sqrt{x^2+h^2}$ is convex, so we have Jensen's inequality $\sum_i\,\lambda_i\,f(x_i)\ge f(\sum_i\,\lambda_i\,x_i)$ for all non-negative $\lambda_i$ with $\sum_i\,\lambda_i=1$. Of course, we choose $\lambda_i=l_i/L$ and $x_i=h_i$. Obviously, $\sum_i\,\lambda_i\,x_i=\sum_i\,l_i\,h_i/L=2\,A/L$, so
$$\Lambda(C)=\frac{1}{2}\,L\,\sum_i\,\lambda_i\,f(x_i)\ge\frac{1}{2}\,L\,\sqrt{(2\,A/L)^2+h^2}=\sqrt{A^2+L^2\,h^2/4}.$$
According to the isoperimetric inequality, $L^2\ge 4\,\pi\,A$, and we are done.
