Average length of chord of circle - definite integral

Given a circle C of radius a, and a diameter AB of C. Chords are drawn perpendicular to AB, intercepting equal arcs along the circumference of C. Find the limit of the average of the lengths of these chords, as the number of chords tends to infinity.

I can find the average lengths of the chords that intercept equal segments along AB: $$L_{ave}= 2*\frac{1}{a}\int_{0}^{a}\sqrt{a^2-x^2}\quad dx = 2*\frac{1}{a}* \frac{1}{4}\pi a^2 = \frac{\pi a}{2}$$

This is straightforward since $$dx$$ is invariable.

However, when solving the problem for chords intercepting equal arc segments, then $$dx$$ varies across the diameter of the circle. I was trying to set it up as a function of $$sin$$ or $$cos$$, but then I have both x and an angle as variables. Any hints on how to approach this?

• Make it a trig function of $n$, the number of chords. – Gerry Myerson Oct 10 '18 at 11:56
• Fun fact: For spheres these two notions (planar cuts with equal distance along the diameter, and planar cuts with equal surface area) coincide. – Arthur Oct 10 '18 at 11:59

Without loss of generality assume the circle is of radius 1 and divided by $$n$$ chords into $$2n$$ arcs. The required limit is then the Riemann sum $$\lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}2\sin\frac{(2k+1)\pi}{2n}$$ which is thus equivalent to the integral $$\int_0^12\sin\pi x\,dx=\frac4\pi$$ Therefore, for a circle of radius $$a$$, the average length is $$\frac{4a}\pi$$.