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 Post subject: Geometry or Trig question
PostPosted: Tue, 12 Jun 2012 18:49:42 UTC 
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Can someone help me this afternoon? If I draw eighteen concentric circles, each 71 inches wide, that touch each other to form a large circle, what would the radius of the circle be?


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 Post subject: Re: Geometry or Trig question
PostPosted: Tue, 12 Jun 2012 18:53:19 UTC 
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jimbaumgart wrote:
Can someone help me this afternoon? If I draw eighteen concentric circles, each 71 inches wide, that touch each other to form a large circle, what would the radius of the circle be?


If they are concentric, then they should all be the same circle.

Assuming you mean by "width" the diameter of this circle, then just use the fact that the radius is half of the diameter.

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 Post subject: Re: Geometry or Trig question
PostPosted: Tue, 12 Jun 2012 19:00:24 UTC 
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Topic moved to geometry and trig board.

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 Post subject: Re: Geometry or Trig question
PostPosted: Tue, 12 Jun 2012 20:55:11 UTC 
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Hello, jimbaumgart!

I suspect this is the "pyramid of tires" problem.


Quote:
If I draw eighteen concentric (?) circles, each 71 inches wide,
that touch each other to form a large circle,
what would the radius of the circle be?

Consider two adjacent circles.

Code:
              * * *
          *           *
        *               *
       *                 *

      *         A         *
      *         o         *
      *         |     *   *
                |           *   R
       *        |35.5    *        *
        *       |       *               *
          *     |     *                 10d   *
              * o * - - - - - - - - - - - - - - - - o C
          *     |D    *                       *
        *       |       *               *
       *        |        *        *
                |           *
      *         |     *   *
      *         o         *
      *         B         *

       *                 *
        *               *
          *           *
              * * *

The centers are A and B.
\angle ACB = 20^o,\;\angle ACD = 10^o

In right triangle ADC\!:\;\;\sin10^o \:=\:\dfrac{35.5}{R} \quad\Rightarrow\quad R \:=\:\dfrac{35.5}{\sin10^o}

Therefore: .R \:=\:204.4363522\text{ inches} \:\approx\:17\text{ feet}



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