Is there a formula to calculate the area of a trapezoid knowing the length of all its sides? If all sides: $a, b, c, d$ are known, is there a formula that can calculate the area of a trapezoid?
I know this formula for calculating the area of a trapezoid from its two bases and its height:

$$S=\frac {a+b}{2}×h$$

And I know a well-known formula for finding the area of a triangle, called Heron's formula:

$$S=\sqrt {p(p-a)(p-b)(p-c)}$$
$$p=\frac{a+b+c}{2}$$

But I could not a formula for finding the area of a trapezoid in the books.

This problem is more subtle than some of the other answers here let on.  A great deal hinges on whether "trapezoid" is defined inclusively (i.e. as a quadrilateral with at least one pair of parallel sides) or exclusively (i.e. as a quadrilateral with exactly one pair of parallel sides).  The former definition is widely considered more mathematically sophisticated, but the latter definition is more traditional, is still extensively used in K-12 education in the United States, and has some advantages.
As the other responses have pointed out, if one defines "trapezoid" inclusively, then any parallelogram is automatically a trapezoid, and as the side-lengths of a parallelogram do not determine its area, it is not possible (even conceptually) that there could be a formula for the area of a trapezoid in terms of its side lengths.
However, if "trapezoid" is defined exclusively, then things are quite different.  Consider a trapezoid with parallel bases of length $a$ and $b$ with $b>a$.  Let $\theta$ and $\phi$ respectively denote the angles formed by the legs $c$ and $d$ with the base $b$.  Then we have the following relationships:
$$c\cos\theta + d\cos\phi = b-a$$
$$c\sin\theta = d\sin\phi$$
These conditions uniquely determine $\theta$ and $\phi$, and therefore among non-parallelogram trapezoids, choosing the lengths of the parallel sides and the lengths of the bases uniquely determines the figure.  In particular we would have $$\cos\theta = \frac{(b-a)^2+c^2-d^2}{2c(b-a)}$$.
The height of the trapezoid would then be $h=c\sin\theta$ (or if you prefer $h=d\sin\phi$, which is equal to it), so the area of the trapezoid can (in principal) be computed.  If you really want to carry it out, you would have
$$\sin\theta = \sqrt{1-\left( \frac{(b-a)^2+c^2-d^2}{2c(b-a)} \right)^2}$$
so the area would be
$$A=\frac{a+b}{2}c\sqrt{1-\left( \frac{(b-a)^2+c^2-d^2}{2c(b-a)} \right)^2}$$
I am not sure if there is a simpler expression, however.
