Trapezoid's area 
A trapezoid was inscribed into a semicirle of radius R. The side of
  the trapezoid is slanting alpha against the base which is the diameter
  of the semicirlce. Compute the area of the trapezoid.

So the base is 2R. The bad thing is: that's all I know. How should I move on? Don't know what to do with that angle.
 A: Hint:  both base angles are $\alpha$ to make the top parallel to the base.  Draw a radius to the upper corners and you have three isosceles triangles.
A: There is a reasonable brute force approach, using the standard formula for the area of a trapezoid.  We need a picture. Instead I will label the vertices, and rely on you to draw the picture. Please note that your picture is an essential part of the calculation below.
Let the vertices of the trapezoid be $A$, $B$, $C$, $D$.  These labels go counterclockwise, and $A$, $B$ are the endpoints of the diameter of the semicircle, with $A$ on the left.  We assume that by "the side of the trapezoid is slanting $\alpha$ against the base" you mean that $\angle DAB=\alpha$.  Note that for the geometry to work, we need $\alpha \ge 45^\circ$. If $\alpha=45^\circ$, our "trapezoid" degenerates to a triangle.
Note that $\triangle ADB$ is right-angled at $D$.  It follows that $AD=2R\cos\alpha$. 
Drop a perpendicular from $D$ to the diameter $AB$, meeting $AB$ at say $P$. Then by looking at $\triangle APD$, we can see that $DP=AD \sin\alpha=2R\sin\alpha\cos\alpha$. Progress, we now know the height of the trapezoid.
Now we sort of need to know the shorter one of the two parallel sides. Note that $AP=AD\cos\alpha=2R\cos^2\alpha$.
It follows that the shorter parallel side has length $2R-2AP=2R(1-2\cos^2\alpha)$. 
So the average of the two parallel sides is $2R(1-\cos^2\alpha)$, which  simplifies to $2R\sin^2\alpha$.  
Multiply the average of the two parallel sides by the height. We get 
$4R^2\sin^3\alpha\cos\alpha$.  
Comment: We really didn't need to calculate the shorter parallel side, since it is obvious that our trapezoid has the same area as the rectangle of base $PB$ and height $DP$, and once we know that $AP=2R\cos^2\alpha$, we know that $PB=2R-2R\cos^2\alpha$.  And we didn't need the formula for the area of a trapezoid.
My preference would be to find the answer when the radius is $1$, then scale by the linear factor $R$ at the end, which scales the area by the factor $R^2$.
