# concentric circle problem

• Apr 22nd 2013, 03:31 PM
aaronrpoole
concentric circle problem
I have no idea how to solve this. I love geometry, but I've found that I'm not very naturally good at it. please help!

The circles in the figure I have drawn out are concentric. The chord AB is tangent to the inner circle and has a length of 12 cm. What is the Area of of the non-shaded region? (A of big triangle - A of small triangle)
Attachment 28090
• Apr 22nd 2013, 03:46 PM
Plato
Re: concentric circle problem
Quote:

Originally Posted by aaronrpoole
The circles in the figure I have drawn out are concentric. The chord AB is tangent to the inner circle and has a length of 12 cm. What is the Area of of the non-shaded region? (A of big triangle - A of small triangle)
Attachment 28090

As posed, I fear there is no unique answer to this question.
Look at this webpage.

I think that you need to know the radius of one of those two circles. Or some other term from that webpage.
• Apr 22nd 2013, 05:35 PM
Soroban
Re: concentric circle problem
Hello, aaronrpoole!

This is a classic problem . . . with a surprising punchline.

Quote:

The circles in the figure I have drawn are concentric.
The chord AB is tangent to the inner circle and has a length of 12 cm.
What is the area of of the non-shaded region? (Area of big circle - Area of small circle)
Code:

              * * *           *          *         *      C  6  *     A *- - - - ♥ - - - -♥ B             *  |  *  o       *    *  r|  o* R  *       *    *  ♥  *    *       *    *  O  *    *             *    *       *        *        *         *              *           *          *               * * *

$\displaystyle O$ is the center of the circles.
$\displaystyle C$ is the midpoint of chord $\displaystyle AB\!:\;CB = 6$
Let $\displaystyle R = OB$, the radius of the large circle.
Let $\displaystyle r = OC$, the radius of the small circle.

From right triangle $\displaystyle BOC\!:\;r^2 + 6^2 \,=\,R^2 \quad\Rightarrow\quad R^2-r^2 \,=\,36$ .[1]

The area of the large circle is: $\displaystyle \pi R^2$
The area of the small circle is: $\displaystyle \pi r^2$

The area of the ring is: $\displaystyle A \:=\:\pi R^2 - \pi r^2 \:=\:\pi(R^2-r^2)$

Substitute [1]: .$\displaystyle A \:=\:\pi(36) \:=\:36\pi$

Surprise! .We didn't need to know the two radii.

The small circle could be a golfball or the Earth.
The area of the ring is constant!
• Apr 22nd 2013, 11:41 PM
first123
Re: concentric circle problem
While this subject can be very touchy for most people, my opinion is that there has to be a middle or common ground that we all can find. I do appreciate that youve added relevant and intelligent commentary here though. Thank you!