<p>Number of quadrilaterals that can be made using the vertices of the polygon of sides 'n' if exactly two adjacent sides of the quadrilateral are common to the sides of the n-gon, is</p>
Step-by-Step Solution
Key Concept: Identify the configuration where two consecutive sides of the polygon form two sides of the quadrilateral, then count valid placements of the fourth vertex.
<p><strong>Solution:</strong> From the passage, $n = 10$.</p><p>For exactly 2 adjacent sides of the quadrilateral common with the 10-gon:</p><p>Choose 2 adjacent sides of the 10-gon sharing a vertex, say A-B-C: 10 ways</p><p>We need one more vertex D such that:</p><p>- CD is not a side of the 10-gon</p><p>- DA is not a side of the 10-gon</p><p>Available vertices = $10 - 5 = 5$ (excluding A, B, C and their adjacent vertices)</p><p>From these 5 vertices, choose 1: 5 ways</p><p>But we must ensure no three consecutive vertices, which adds constraints.</p><p>Total = $10 \times 7 = 70$</p><p>∴ Answer is (c) 70</p>
Correct Answer: c