Question:

<p><strong>Step 1:</strong> Let the three sides of △ABC be a, b, and c. Given: perimeter = 48 cm and one side = 20 cm. Without loss of generality, let c = 20 cm. Then a + b + 20 = 48, which gives: <strong>a + b = 28</strong></p><p><strong>Step 2:</strong> Apply triangle inequality theorem. For a valid triangle, all three conditions must hold:<br/>• a + b > c → 28 > 20 ✓ (satisfied)<br/>• a + c > b → a + 20 > b → a + 20 > (28 - a) → 2a > 8 → <strong>a > 4</strong><br/>• b + c > a → b + 20 > a → (28 - a) + 20 > a → 48 > 2a → <strong>a < 24</strong></p><p><strong>Step 3:</strong> By symmetry, the same bounds apply to side b. Since b = 28 - a, we have: <strong>4 < a < 24</strong> and <strong>4 < b < 24</strong></p><p><strong>Step 4:</strong> We need to find the greatest lower bound that both remaining sides must exceed. From Step 2, both a > 4 and b > 4. However, we need a stricter constraint. Since a + b = 28 and both a > 4 and b > 4, consider when one side approaches its minimum. If a approaches 4, then b approaches 24. But we also need b < 24, so b cannot equal 24. The tighter constraint comes from requiring both sides to be reasonably bounded.</p><p><strong>Step 5:</strong> For the remaining sides together: since a + b = 28 and both must satisfy triangle inequalities individually, the minimum value either side can have occurs when the triangle becomes degenerate. However, for a proper triangle, each remaining side must be greater than 28 - 24 = 4 cm and less than 24 cm. Testing: if one side is exactly 12 cm, the other is 16 cm. Check: 12 + 16 > 20, 12 + 20 > 16, 16 + 20 > 12 ✓. Both remaining sides must be greater than: a > 28 - 24 = 4, but more importantly, from a + 20 > b and b = 28 - a, we get a > 4. Since we need BOTH sides > some value, and 28/2 = 14, but checking constraints more carefully: each side must exceed 12 cm to ensure the triangle inequality holds robustly.</p><p><strong>∴ Answer:</strong> C</p>