# Thread: Sector area and probability questions

1. ## Sector area and probability questions

Hello. For the first question refer to this:

http://img693.imageshack.us/img693/8362/46035151.jpg - the outer numbers are degrees

What I want to do is find the area in the middle there (where the two and arrows are). Not quite sure how to go about it though. I was thinking divide the sections into triangle but I'd need a radius. Anyway the answer is $\displaystyle 4pi + 6\sqrt{3} over 3$.

Second question: "A dart is thrown at a board 12m long and 5m wide. Attached to the board are 30 balloons, each with radius 10cm. Assuming each balloon lies entirely on the board, find the probability that a dart that hits the board will also hit a balloon."

I'm not even sure where to start here.

The answer is pi/200 is about 0.016.

Basically I just want to see how each was worked out. Thanks.

2. Hello, Phil!

Find the area of the middle region.

Answer: .$\displaystyle \dfrac{4\pi + 6\sqrt{3}}{3}$
Code:
              * * *
*           *
A ♥ - - - + - - - ♥ B
*  *     |     *  *
*   |1  *
*       * | *       *
*         ♥O        *
*       * | *       *
*   |1  *
*  *     |     *  *
* - - - + - - - *
*           *
* * *

Note all the $\displaystyle 60^o$ angles.

We find that the radius is 2.
The area of the circle is: .$\displaystyle \pi r^2 \:=\:\pi(2^2) \:=\:4\pi$

The area of sector $\displaystyle OAB$ is: .$\displaystyle \frac{1}{3}(4\pi)$

The area of $\displaystyle \Delta OAB\:=\:\frac{1}{2}(2)(2)\sin120^o \;=\;\sqrt{3}$

The area of the segment is: .$\displaystyle \dfrac{4\pi}{3} - \sqrt{3}$

The area of the middle region is: .$\displaystyle 4\pi - 2\left(\dfrac{4\pi}{3} - \sqrt{3}\right)$

. . $\displaystyle =\;\;4\pi - \dfrac{8\pi}{3} + 2\sqrt{3} \;\;=\;\;\dfrac{4\pi}{3} + 2\sqrt{3} \;\;=\;\;\dfrac{4\pi + 6\sqrt{3}}{3}$

3. Thank you - was very helpful.