Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Topic: Chords of Concentric Circles
Replies: 2   Last Post: Mar 23, 1993 4:05 PM

 Messages: [ Previous | Next ]
 Rouben Rostamian Posts: 16 Registered: 12/6/04
Re: Chords of Concentric Circles [WARNING! SPOILER!]
Posted: Mar 23, 1993 3:32 PM

In article <1olds5\$m16@forum.swarthmore.edu> Bill Schneider <Bill.Schneider@f32.n272.z1.fidonet.org> writes:

> There are two concentric circles such that the radius of the smaller
> plus the radius of the larger is 36. A chord of the larger circle is
> trisected by the smaller; that is, it must pass through the smaller
> circle. This chord is length 36 also. What is the radius of the large
> circle?

Let a and b denote the radii of the small and large circles, respectively.
Let L denote the length of the chord of the larger circle, and let l denote
the length of its sub-segment that forms a chord of the smaller circle.
Finally, let h denote the distance of the chord from the common center of
the circles.

By the Pythagorean theorem: (L/2)^2 + h^2 = b^2 and (l/2)^2 + h^2 = a^2.
Subtract these to obtain: L^2 - l^2 = 4*(b^2 - a^2). Therefore:
L^2 - l^2 = 4*(b-a)*(b+a).

Now, plug in L=36, l=L/3=36/3, and b+a=36, to obtain b-a=8.
Solve the system { b+a=36, b-a=8 } to arrrive at: a=14, b=22. Q.E.D.

--
Rouben Rostamian

Date Subject Author
3/22/93 Bill Schneider
3/23/93 Rouben Rostamian
3/23/93 Gregory Russell