Hello, Outragexl10!
Question 10 .(Essay  5 points)
A circle is given with an inscribed equilateral triangle and with sides 24 cm.
What is the probability of selecting a point at random inside the circular region
and outside the triangle. .Leave your answer in exact form. The probability is: .$\displaystyle \frac{\text{Area of circle}  \text{Area of triangle}}{\text{Area of circle}} $
The area of an equilateral triangle with side $\displaystyle S$ is: .$\displaystyle A \;=\;\frac{\sqrt{3}}{4}S^2$
We have $\displaystyle S = 24$, hence: .$\displaystyle A_{\Delta} \;=\;\frac{\sqrt{3}}{4}(24^2) \;=\;144\sqrt{3}$ cm²
Given inscribed equilateral triangle $\displaystyle ABC$ with side 24,
. . $\displaystyle \angle BOC = 120^o$ and $\displaystyle OA = OB = OC = r$ Code:
A
* o *
*  *
*  *
* r *

*  *
* *O *
* / \ *
r/ \r
* / \ *
B o        o C
* 24 *
* * *
There are many ways to determine the radius.
. . I'll use the Law of Cosines on $\displaystyle \Delta OBC$.
$\displaystyle r^2 + r^2  2\cdot r\cdot r\cos120^o \:=\:24^2$
. . $\displaystyle 2r^2  2r^2\left(\frac{1}{2}\right) \:=\:576$
. . $\displaystyle 3r^2 \,=\,576\quad\Rightarrow\quad r^2 \,=\,192\quad\Rightarrow\quad r \,= \,8\sqrt{3}$
Hence, the area of the circle is: .$\displaystyle A_{\circ} \;=\;\pi r^2\;=\;\pi(8\sqrt{3})^2\;=\;192\pi$ cm²
The area in the circle but outside the triangle is: .$\displaystyle 192\pi  144\sqrt{3}$
Therefore, the probability is: .$\displaystyle \frac{192\pi  144\sqrt{3}}{192\pi} \;=\;1  \frac{3\sqrt{3}}{4\pi}
$