<p>Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle formed by these vertices is equilateral, is</p>
<p>(a) \(\frac{1}{2}\)</p>
<p>(b) \(\frac{1}{3}\)</p>
<p>(c) \(\frac{1}{10}\)</p>
<p>(d) \(\frac{1}{20}\)</p>
Step-by-Step Solution
Key Concept: In a regular hexagon, an equilateral triangle can only be formed by selecting alternate vertices (every second vertex). We need to count such favorable triangles and divide by the total number of ways to choose 3 vertices from 6.
<p><strong>Step 1: Label the hexagon vertices.</strong> Label the vertices of a regular hexagon as 1, 2, 3, 4, 5, 6 going around clockwise.</p><p><strong>Step 2: Find the total number of ways to choose 3 vertices from 6.</strong></p><p>Total ways = C(6,3) = 6!/(3!×3!) = 20</p><p><strong>Step 3: Identify which triangles are equilateral.</strong> In a regular hexagon, all vertices are equidistant from the center. For a triangle to be equilateral, the three chosen vertices must be equally spaced around the hexagon, meaning they must be alternate vertices (every second vertex).</p><p><strong>Step 4: Count equilateral triangles.</strong> There are exactly 2 ways to choose alternate vertices:</p><p>• Triangle 1: vertices {1, 3, 5}</p><p>• Triangle 2: vertices {2, 4, 6}</p><p>These are the ONLY two equilateral triangles possible, as they form 120° angles at the center.</p><p><strong>Step 5: Calculate the probability.</strong></p><p>Probability = (Number of equilateral triangles)/(Total number of triangles)</p><p>Probability = 2/20 = 1/10</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C