Probability in geometry

[the tree]

There is a large square of a side

s

containing a small circle with a radius

r

, a point is chosen randomly within the square, what is the probability of the point being within the circle?

I'm tempted to think that the awnser is [math]\frac{\pi r^2}{s^2}[/math] but the following question is:

Show that if r increases by 20% then the probability increases by 40%

If [math]r_{1}=1[/math] and [math]r_{2}=1.2[/math] then [math]\pi r^{2}[/math] increases by 44%.

 

Where have I slipped up?

[vaishakh]

If r increases by 20%, then the inintial diameter or 2r was equal to s, the sides of square since theyere tangential to to the circle. But now the circle goes even exterior to the square, then a segment is subtended on all the four sides. Look if there % is 4. Anyway there is some mistake in the problem. Probably if that is four, then we can frame what the question should be, If a random point selected on the circle or square, find the probability that it lies on both.

[the tree]

If r increases by 20%, then the inintial diameter or 2r was equal to s...

What? Why would 2r be equal to s at any time?

From the context I know that the circle is smaller than the square and contained inside it, at all times.

[SilentQ]

I think you're saying that [math]0.2^2=0.04[/math], or 4%. Remember, though, that it is a change of 20%, and thus the multiplication factor is not 0.2, but 1.2. Therefore the factor in the probability formula ([math]\frac{\pi r^2}{s^2}[/math]) is [math]1.2^2=1.44[/math], or a change of 44%, as the Tree says. The textbook is most probably wrong.

[vaishakh]

Sorry? I thought the square is tangential to circle. I am really sorry.