<p>In \(\triangle ABC\) if \(AB = AC\) and lengths of the tangents to the incircle from the vertices <em>A</em> and <em>C</em> are 4 and 2 respectively, then identify which of the following statement(s) is(are) <strong>correct</strong>?<br>[Note: Symbols used have usual meaning in \(\triangle ABC\)]</p>
<p>(a) \(AI : BI : CI = \sqrt{3} : 1 : 1\)</p>
<p>(b) Inradius of the triangle \(ABC\) is \(\sqrt{2}\) sq. units</p>
<p>(c) \(R = \dfrac{9}{4}\sqrt{2}\)</p>
<p>(d) \(\Delta R = a^2\)</p>
Step-by-Step Solution
Key Concept: In a triangle, the length of tangent from a vertex to the incircle equals (s - opposite_side), where s is the semi-perimeter. For isosceles triangle AB = AC, use this property along with tangent equality from each vertex to set up equations.
<p><strong>Step 1:</strong> Let tangent lengths from A, B, C be t_A = 4, t_B, t_C = 2 respectively. Since tangents from an external point are equal: AB = AC = b (given), so sides are related to tangent segments.</p><p><strong>Step 2:</strong> For any triangle, tangent length from vertex X = s - (opposite side). Thus: t_A = s - a = 4 and t_C = s - c = 2, where a = BC, c = AB.</p><p><strong>Step 3:</strong> Since AB = AC, we have c = b. From t_A = s - a: s - a = 4 → s = a + 4. From t_C = s - c: s - c = 2 → s = c + 2.</p><p><strong>Step 4:</strong> Therefore a + 4 = c + 2, giving a = c - 2. Since s = 2a + b + c)/2 and a + b + c = 2s, with c = b: 2s = a + 2b. Substituting s = a + 4: 2(a + 4) = a + 2b → a + 8 = 2b → b = (a+8)/2.</p><p><strong>Step 5:</strong> From a = c - 2 = b - 2 and b = (a+8)/2: a = (a+8)/2 - 2 → 2a = a + 8 - 4 → a = 4. Thus b = c = 6, and s = 8.</p><p><strong>Step 6:</strong> Verify: t_A = 8 - 4 = 4 ✓, t_C = 8 - 6 = 2 ✓. Triangle sides are 4, 6, 6.</p><p>∴ Answer: C</p>
Correct Answer: C