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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 $A$?

 Jul 24, 2019

Answered here.



Plenty of answers to choose from :D


-\(\pi\) KeyLimePi

 Jul 25, 2019

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.


\(\text{probability} = \frac{\text{area of sector} }{ \text{area of triangle}}\\~\\ \text{probability} = \text{area of sector} \div \text{area of triangle} \\~\\ \text{probability} = \frac{\pi}{6} \div \frac{9\sqrt3}{4} \\~\\ \text{probability} = \frac{\pi}{6} \cdot \frac{4}{9\sqrt3} \\~\\ \text{probability} = \frac{4\pi}{54\sqrt3} \\~\\ \text{probability} = \frac{4\sqrt3\pi}{162} \\~\\ \text{probability} = \frac{2\sqrt3\pi}{81} \)_

 Jul 25, 2019

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