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.
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
$$E(X)=\frac{4r}{\pi}.$$
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