Triangle $ABC$ is equilateral with side length 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 $A$?
Answered here.
https://web2.0calc.com/questions/help_3820
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-\(\pi\) KeyLimePi
+9466 Here's equilateral triangle ABC and a circle with radius 1 centered on point A:
The intersecion of the triangle and the circle is the highlighted sector.
probability that a randomly chosen point lands in highlighted sector = area of sector / area of triangle
So we just have to find the area of the sector and the area of the triangle.
Let's find the area of the sector:
\(\frac{\text{area of sector}}{\text{area of circle}}\ =\ \frac{\text{measure of central angle}}{360^\circ} \\~\\ \frac{\text{area of sector}}{\pi\cdot1^2}\ =\ \frac{60^\circ}{360^\circ} \\~\\ \text{area of sector}\ =\ \frac{60^\circ}{360^\circ}\cdot\pi\cdot1^2 \\~\\ \text{area of sector}\ =\ \frac{\pi}{6} \)
Now let's find the area of the triangle:
\(\text{area of triangle}\ =\ \frac12\cdot\text{base}\cdot\text{height}\\~\\ \text{area of triangle}\ =\ \frac12\cdot3\cdot\frac{3\sqrt3}{2}\\~\\ \text{area of triangle}\ =\ \frac{9\sqrt3}{4}\)
Now we can find the probability in question.
\(\text{probability} = \frac{\text{area of sector} }{ \text{area of triangle}}\\~\\ \text{probability} = \text{area of sector} \div \text{area of triangle} \\~\\ \text{probability} = \frac{\pi}{6} \div \frac{9\sqrt3}{4} \\~\\ \text{probability} = \frac{\pi}{6} \cdot \frac{4}{9\sqrt3} \\~\\ \text{probability} = \frac{4\pi}{54\sqrt3} \\~\\ \text{probability} = \frac{4\sqrt3\pi}{162} \\~\\ \text{probability} = \frac{2\sqrt3\pi}{81} \)_
