Sum and average length of chords From a specific point A on a circle's circumference, am drawing various chords. 
a) what is the sum of length of such chords and 
b) what is the average length of such chords?
Assuming a radius r, I'm trying to solve a) in the below manner:
One specific chord length = 2rCos(Theta)
To get the sum, I'm integrating over -pi/2 to pi/2 ( 2rCos(Theta) dTheta )
But i don't think this is right. Please point me in the correct direction.
Thanks.
 A: There are many possible answers, depending on the probability distribution of the parameter that defines the other endpoint of the chord. That is a standard issue in problems of geometric probability.
We give one interpretation, and derive the answer under that interpretation. Other interpretations will yield a different answer.
Imagine that we select the other endpoint $B$ by choosing $\theta$ with uniform distribution in the interval $[0,2\pi]$, and letting $B$ be the point obtained by rotating the point $A$ counterclockwise through the angle $\theta$.  
Then the length $X$ of the chord $AB$ is, by basic trigonometry, $2r\sin(\theta/2)$. It follows that 
$$E(X)=\int_0^{2\pi} 2r\sin(\theta/2)\frac{1}{2\pi}\,d\theta.$$
Integrate. We get
$$E(X)=\frac{4r}{\pi}.$$
A: If you imagine many chords of a circle arranged in a stack, with appropriate lengths, it can be made into a circle. so, the total of all the chords will give us the area of the circle. This can be easily proved and imagined. For example, take a rectangle. Have you ever wondered why the area is l x b?? It is because we are adding the length breadth number of times. Hence applying the same logic to a circle, we can understand that if we add the lengths of all the chords in a circle we would get the circle's area..pi*R^2.
The total number of chords would be equal to 2*R as the length of the circle is 2*R.
Hence the average chord length = (pi*R^2)/(2*R).
=> pi*r/2
