Permutations & Combinations
Polygons and Diagonals
Grade 11
Question:
<p>Consider the convex polygon, which has 35 diagonals. Match the following lists and choose the correct code.</p><table border='1'><tr><th>List I</th><th>List II</th></tr><tr><td>a. Number of triangles joining the vertices of the polygon</td><td>p. 210</td></tr><tr><td>b. Number of points of intersections of diagonal which lies inside the polygon</td><td>q. 120</td></tr><tr><td>c. Number of triangles in which exactly one side is common with that of polygon</td><td>r. 10</td></tr><tr><td>d. Number of triangles in which exactly two sides are common with that of polygon</td><td>s. 60</td></tr></table><p><strong>Codes:</strong></p><table border='1'><tr><th></th><th>a</th><th>b</th><th>c</th><th>d</th></tr><tr><td>(1)</td><td>q</td><td>p</td><td>s</td><td>r</td></tr><tr><td>(2)</td><td>r</td><td>q</td><td>p</td><td>s</td></tr><tr><td>(3)</td><td>s</td><td>r</td><td>q</td><td>p</td></tr><tr><td>(4)</td><td>p</td><td>r</td><td>s</td><td>q</td></tr></table>
<p>(1) a-q, b-p, c-s, d-r</p>
<p>(2) a-r, b-q, c-p, d-s</p>
<p>(3) a-s, b-r, c-q, d-p</p>
<p>(4) a-p, b-r, c-s, d-q</p>
Step-by-Step Solution
Key Concept: First find the number of sides (n) using the diagonal formula: n(n-3)/2 = 35, giving n=10. Then systematically count triangles by their relationship to polygon sides: total C(n,3), subtract those sharing sides, and use intersection properties of diagonals.
<p><strong>Step 1: Find n (number of sides)</strong></p><p>Diagonals = n(n-3)/2 = 35</p><p>n² - 3n - 70 = 0 → n = 10</p><p><strong>Step 2: Count total triangles</strong></p><p>Total triangles from 10 vertices: C(10,3) = 120</p><p><strong>Step 3: Triangles with exactly 2 common sides (two adjacent sides of polygon)</strong></p><p>These are formed by choosing 2 adjacent vertices and one more vertex. Count = n = 10 (one for each vertex as the common vertex)</p><p><strong>Step 4: Triangles with exactly 1 common side</strong></p><p>Choose one side (n ways) and a vertex not adjacent to it: (n-3) ways</p><p>Count = n(n-3) = 10(7) = 70 ✗ [This needs adjustment]</p><p>Correct: C(n,2) - n = 45 - 10 = 35... Actually: n(n-4) = 10(6) = 60</p><p><strong>Step 5: Triangles with no common side</strong></p><p>Total - (2 sides) - (1 side) = 120 - 10 - 60 = 50 ✗</p><p>Correct count: C(10,3) - 10 - 60 = 120 - 70 = 50... → Actually 0 common sides = 120 - 10 - 60 = 50</p><p><strong>Step 6: Diagonal intersections inside polygon</strong></p><p>Each 4 vertices form exactly one intersection point inside: C(10,4) = 210</p><p>a → 120 → q<br/>b → 210 → p<br/>c → 60 → s<br/>d → 10 → r</p><p>∴ Code: qpsr → Answer: (1)</p>
Correct Answer: A