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### Point in a Circle

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Date: 04/29/97 at 23:21:47
From: zolnierz
Subject: Probability with a circle

Given a circle with two chords (six inches in length) running across
the top and the bottom and enclosing an area with a height of 6
inches, find the probability that a point chosen at random is in the
region between the chords.

I have no idea how to go about solving this problem.  The height given
does not go all the way through the circle so I don't know how to find
the radius which would give me the area of the circle.  Also, the
region between the chords is not a square, which is confusing me even
more.

```

```
Date: 05/01/97 at 16:23:05
From: Doctor Anthony
Subject: Re: Probability with a circle

As in all problems, it is always best to try to draw a picture first
so you can see what's going on.  Here's an attempt (sorry it looks
more like a hexagon than a circle!):

_____________
/             \
/___________3___\
/        |       /\
|       3|  r /    |
|        |/        |
|        |         |
\        |         /
\_______6________/
\              /
\____________/

But what do we want to do with this diagram?  We must decide what it
means to find the "probability that a point chosen at random is in the
region between the chords." A point chosen at random can either be
inside the chords or outside the chords. The greater the area between
the chords, the more likely it is that the random point will be there,
right? This means that we can find the probability of the point being
between the chords by finding the area of the whole circle and the
area between the chords. Our sample space, the denominator of the
probability fraction, is the area of the entire circle.  Our favorable
outcome, the numerator of the probability fraction, is the area of the
region between the two chords.

In order to find the area of the circle, we need to find the circle's
radius. To do this, we make note of the triangle formed by a side
which is one half the length of one of the chords, a side which is one
half of the distance between the chords, and the radius of the circle
(labelled r above). This triangle has two sides of length 3 and a
hypoteneuse (the radius of the circle) of unknown length. Using the
Pythagorean theorem, the length of the hypoteneuseis equal to
sqrt(3^2 + 3^2), which simplifies to 3*sqrt2.  Thus the radius of the
circle has length 3*sqrt(2).

Since we know the radius of the circle, we can find its area, which is
equal to pi(r^2). This means that the area of the whole circle is
equal to: pi(3*sqrt2)^2 = pi(18) = 56.55 square units.

It is now relatively easy to find the area of the circle outside the
chords if we make use of the fact that the chord subtends a right
angle at the center of the circle. This means that to find the area
outside the chord, we make use of the fact that pi(r^2/4) is the area
of this sector of the circle that is enclosed by the right angle that
is formed at the center. Then the area outside the chord is equal to
the area of the sector minus the area of the triangle that has the
chord as its base and two radii as its other sides. The area of the
triangle is (1/2)(3.sqrt(2))^2 = 9 square units.

So the area outside the chord is: pi(3*sqrt(2))^2/4 - 9 = 5.13

We double this to get the area outside both chords: 10.27

The area between the chords is: pi(r^2) - 10.27 = 46.27

Thus we have the numerator and denominator of our probability
fraction, so the probability of a point lying between the chords is:

46.27/56.55 = 0.82

This means that a random point has an 82 percent chance of being in
the region between the chords.

-Doctors Anthony and Rachel,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Probability

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