Just click on the green flag to run an experiment (Monte Carlo) that picks two points on the circumference of the circle, draws the chord between the two points, and compares the length of the chord to the length of the side of the equilateral triangle. There is a theoretical solution but several runs of the experiment should provide you enough information to make an informed conjecture as to the value of the theoretical probability. The next related project, Bertrand's Paradox 2, uses the same circle and equilateral triangle but a different method for picking the chord. Are the probabilities the same? Also, the Probability Experiment that compares the length of the chord to the radius of the circle can be seen at the following link. http://scratch.mit.edu/projects/19924829/
I used the parametric equation for a circle(x = radius x cos [angle], y = radius x sin [angle]) to draw a circle of radius 100. Then, on the background, I used the vector option to size a vector circle to the project-drawn circle. I then deleted the script that drew the circle. This shortened the code so that just the chord had to be repeatedly drawn. The code shows how to compute the length of the chord (using the Pythagorean theorem) knowing the (x,y) coordinates of the endpoints of the chord.
