A point is selected at random inside an equilateral triangle whose side length is 3. The probability its distance to any corner is greater than 1 is
Step-by-Step Solution
Key Concept: Use geometric probability: find the area of the equilateral triangle (9√3/4) and subtract the three circular sectors of radius 1 centered at each corner. Each 60° sector has area π/6, giving total forbidden area of π/2. The probability is 1 minus the ratio of forbidden area to total area: 1 - (π/2)/(9√3/4) = 1 - 2π/(9√3).
The area of an equilateral triangle with side 2 is $\frac{\sqrt{3}}{4}(3)^2 = \frac{9\sqrt{3}}{4}$. Points must lie in the shaded region. The area of each circular segment is $\frac{\pi}{6}(1)^2$. The desired probability is $1 - \frac{3\pi}{4\sqrt{3}} - \frac{2\pi}{9\sqrt{3}}$, which accounts for the three circular segments of radius 1 centered at each vertex.
Correct Answer: 2