<p>Number of points with integral co-ordinates that lie inside a triangle whose co-ordinates are \((0, 0)\), \((0, 21)\) and \((21, 0)\):</p>
Step-by-Step Solution
Key Concept: Apply Pick's Theorem: \(A = I + \frac{B}{2} - 1\) to find the number of interior lattice points from the area and boundary points.
<p><strong>Solution:</strong> The triangle has vertices at \((0, 0)\), \((0, 21)\), and \((21, 0)\).</p><p>The equation of the hypotenuse is: \(x + y = 21\)</p><p><strong>Using Pick's Theorem:</strong> \(A = I + \frac{B}{2} - 1\), where \(A\) is area, \(I\) is interior lattice points, and \(B\) is boundary lattice points.</p><p><strong>Area of triangle:</strong> \(A = \frac{1}{2} \cdot 21 \cdot 21 = \frac{441}{2} = 220.5\)</p><p><strong>Boundary points:</strong></p><ul><li>On \((0,0)\) to \((0,21)\): 22 points</li><li>On \((0,21)\) to \((21,0)\): Points on \(x+y=21\) with \(0 \le x \le 21\): 22 points</li><li>On \((21,0)\) to \((0,0)\): 22 points</li><li>Subtract vertices counted thrice: \(B = 22 + 22 + 22 - 3 = 63\)</li></ul><p>From Pick's theorem: \(220.5 = I + \frac{63}{2} - 1 = I + 31.5 - 1\)</p><p>\(I = 220.5 - 30.5 = 190\)</p>
Correct Answer: B