What’s the average length of chords at a given angle in a rectangle? Articles are published on average length of chords on circles, squares, rectangles etc..like [1] where they considered either random chords or chords at all angles in a regular geometry, but I could not find a result for average length of chords which are at a prescribed angle in a rectangle, refer the figure for a clear idea. How would you derive an expression for that?

[1] Kuchel, P. W., & Vaughan, R. J. (1981). Average lengths of chords in a square. Mathematics Magazine, 54(5), 261-269.
 A: For any shape, the area is obtained as the height times the average horizonal chord length.
Hence for a rotated rectangle, the average horizontal chord length is 
$$\frac{WH}{|H\cos\theta|+|W\sin\theta|}.$$
Note that the average value on $\theta$ is $$\frac{2WH}\pi\left|\text{artanh}\left(\frac HD\right)-\text{artanh}\left(\frac WD\right)\right|$$ where $D$ is the diagonal and this is the average chord length regardless the orientation.



For a circle,
$$\frac{\pi r^2}{2r}.$$
For an ellipse,
$$\frac{2\pi ab}{\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}}.$$
For an arbitrary triangle,
$$\frac{\begin{vmatrix}x_1-x_0&y_2-y_0 \\ x_2-x_0&y_2-y_1\end{vmatrix}}{2(\max(y_0,y_1,y_2)-\min(y_0,y_1,y_2))}.$$
If you rotate it, the area remains the same and the ordinates are replaced as
$$y_k\leftrightarrow y_k\cos\theta+x_k\sin\theta.$$
A: We can assume, without loss of generality, that the rectangle is always with its larger side (its length) presented verticaly, along the $y$ axis, its smaller side, its width, being horizontal. Let us call their resp. measures $L$ and $W$ (see Fig. 1 on the left).
When the line with equation $y=ax+b$ for fixed 
$$a = \tan \alpha \ \ (\text{we can assume WLOG that } \ \ 0 < \alpha < \pi/2) \tag{1}$$
and variable $b$ sweeps the rectangle, the length of its "intercepts" (synonym of "chords") as a function of $b$ is a trapezoid-shaped function as represented on the right of the figure.

Fig. 1 : The case where $a=\tan \alpha < \tfrac{L}{W}$ (the other case has a very similar aspect). 
The average intercept length is the mean value of this function.
Instead of using an integral computation, it's worth using a geometric argument. 
With the notations introduced in the figure (in particular with $h:=W\sqrt{1+a^2}$) : 


*

*During the "plateau" phase, which has a "weight" $\color{red}{L-aW}$, the average is plainly $h$.

*During the ascending and descending phases, the average intercept length is $h/2$. The total "weight" of these phases is $\color{red}{2aW}$.
Now, take the barycenter to get :
$$\text{ave. intercept length} \ = \ \dfrac{\color{red}{(L-aW)}h+\color{red}{(2aW)} h/2}{\color{red}{(L-aW)+(2aW)}}=\dfrac{Lh}{L+aW}=\dfrac{LW\sqrt{1+a^2}}{L+aW}$$
which can be written (I am indebted to @David K who has pointed it to me)
$$\text{ave. intercept length} \ = \ \dfrac{LW / \cos\alpha}{L+W\tan\alpha}=\dfrac{LW}{L\cos\alpha+W\sin\alpha}.$$ 
Remarks : 
1) Here is an active link for the first reference given by the OP.
2) For more, see the properties of the so-called "support function" of a convex set.
