Annulus Area Question

Status
Not open for further replies.
May 2010
1
0
The given chord length 2X is tangent to the inner circle What is the area of the annulus?"

How can I go about understanding this? There's an image of a circle but not sure how to appropriately solve it. I'd appreciate any feedback. Thanks!
 
Last edited by a moderator:

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, LydiaK67!

The given chord length \(\displaystyle 2x\) is tangent to the inner circle.
What is the area of the annulus?
Code:
                    * * * 
                *           *
            *   x     C     x   *
        A o - - - - * o * - - - - o B
                *     |     *   *
       *      *       |       *      *
             *       r|     *R *
      *               |   *           *
            *         | *       *
      * - - * - - - - o - - - - * - - *
                      O

\(\displaystyle AB\) is a chord of circle \(\displaystyle O\) with radius \(\displaystyle R.\)
It is tangent to the inner circle of radius \(\displaystyle r\) at \(\displaystyle C.\)
. . \(\displaystyle AC \,=\,CB\,=\,x\)


The area of the annulus is: .\(\displaystyle A \;=\;\pi R^2 - \pi r^2 \;=\;\pi(R^2-r^2)\) .[1]


In right triangle \(\displaystyle OCB\!:\;\;r^2 + x^2 \:=\:R^2 \quad\Rightarrow\quad R^2 - r^2 \:=\:x^2\) .[2]


Substitute [2] into [1]: .\(\displaystyle \boxed{A \;=\;\pi x^2}\)

 
Status
Not open for further replies.