Probability
Probability
Allen Star Batch
Grade 12
Question:
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
$\frac{2\pi}{9\sqrt{3}}$
$1 - \frac{2\pi}{9\sqrt{3}}$
$\frac{\sqrt{3}\pi}{9}$
$1 - \frac{\sqrt{3}\pi}{9}$
Step-by-Step Solution
Key Concept: Use geometric probability by finding the area where distance to all three corners exceeds 1. Exclude circular sectors of radius 1 centered at each vertex from the total triangle area of 9√3/4, then divide by total area.
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