Triangle $ABC$ is equilateral with side length of 3. A point $X$ is randomly chosen within $\triangle ABC$. What is the probability that $X$ is no more than 1 unit away from vertex $A$?
Within the circle section around A, X is not more than 1 unit from A.
The probability sought is equal to the ratio of the area of the circular segment to the rest of the area of the triangle.
The side of the triangle is a. r = 1
\(probability=\frac{\pi r^2}{6}:\\(\frac{a^2\cdot\sqrt{3}}{4}-\frac{\pi r^2}{6})\)
\(probability=\frac{\pi }{6}:(\frac{a^2\cdot\sqrt{3}}{4}-\frac{\pi}{6})\) [ \(\times\frac{6}{\pi}\)
\(probability=1:(\frac{a^2\cdot 6 \cdot\sqrt{3}}{4\pi}-1)\)
\(prohability=1:(\frac{a^2\cdot 3 \cdot\sqrt{3}}{2\pi}-1)\)
\(prohablity=1:(0.827a^2-1)\)
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