<p>How many sides are there in a polygon which has 35 diagonals?</p>
Step-by-Step Solution
Key Concept: A polygon with n sides has n(n-3)/2 diagonals because each vertex connects to (n-3) other vertices (excluding itself and its 2 adjacent vertices). Set this equal to 35 and solve for n.
<p><strong>Step 1:</strong> Recall that the number of diagonals in an n-sided polygon is given by the formula: <br/>D = n(n-3)/2</p><p>This formula arises because: from each vertex, you can draw (n-3) diagonals (to all other vertices except itself and its 2 adjacent vertices). This gives n(n-3) total, but we count each diagonal twice, so we divide by 2.</p><p><strong>Step 2:</strong> Set the formula equal to 35:<br/>n(n-3)/2 = 35<br/>n(n-3) = 70<br/>n² - 3n = 70<br/>n² - 3n - 70 = 0</p><p><strong>Step 3:</strong> Factor the quadratic:<br/>(n - 10)(n + 7) = 0<br/>n = 10 or n = -7</p><p><strong>Step 4:</strong> Since n must be positive (number of sides), n = 10.</p><p><strong>Verification:</strong> A 10-sided polygon has 10(10-3)/2 = 10(7)/2 = 35 diagonals ✓</p><p>∴ Answer: <strong>10</strong></p>
Correct Answer: 10