What is the probability that a randomly drawn chord will be longer than the radius of the circle?
Prove it.
We only need to consider at most half of the circle since the other half will not affect the solution due to symmetry. But we could break it up to consider the halfchord only instead of the full chord.
It's hard to do it without a picture. Draw one radius of the circle. We need to consider how far down this radius where the halfchord is r/2. Draw a radius to the point where this cord meets the circle and you'd get a right triangle with hypotenus r and one leg r/2. The other leg is length √(3)*r/2.
If you "travel" less than √(3)*r/2 down the radius, then the semichord will be longer than r/2, which means the chord will be longer than r. Hence the probability is (√(3)*r/2)/r = √(3)/2.

Posted by np_rt
on 20031009 15:00:05 