Trigonometry & Inverse Trigonometry
Properties of Triangles
Grade 11
Question:
<p>In \(\triangle ABC\), if incircle touches the sides \(AB\), \(BC\) and \(CA\) at \(P\), \(Q\) and \(R\) respectively and \(s - a = 3\), \(s - b = 5\) and \(s - c = 7\), then area of the quadrilateral \(QCRI\) is, where \(I\) is incentre of \(\triangle ABC\):<br>[Note: Symbols used have usual meaning in \(\triangle ABC\).]</p>
<p>(a) \(\sqrt{7}\)</p>
<p>(b) \(5\sqrt{7}\)</p>
<p>(c) \(3\sqrt{7}\)</p>
<p>(d) \(7\sqrt{7}\)</p>
Step-by-Step Solution
Key Concept: The quadrilateral QCRI has two right angles at Q and R (since the incircle is tangent to BC and CA). Use the property that tangent segments from a vertex are equal: CQ = CR = s - c, and the inradius r = Area/s to find the area.
<p><strong>Step 1:</strong> Find the semiperimeter and side lengths.</p><p>Given: s - a = 3, s - b = 5, s - c = 7</p><p>Adding: 3s - (a + b + c) = 15 → 3s - 2s = 15 → <strong>s = 15</strong></p><p>Therefore: a = 12, b = 10, c = 8</p><p><strong>Step 2:</strong> Recall tangent segment properties.</p><p>From vertex C: CQ = CR = s - c = 7 (equal tangent segments from external point)</p><p><strong>Step 3:</strong> Find the inradius using Area formula.</p><p>By Heron's formula: Area = √[15(15-12)(15-10)(15-8)] = √[15 × 3 × 5 × 7] = √1575 = 15√7</p><p>Inradius: r = Area/s = 15√7/15 = √7</p><p><strong>Step 4:</strong> Calculate area of quadrilateral QCRI.</p><p>Since IQ ⊥ BC and IR ⊥ CA (radii to tangent points are perpendicular):</p><p>Quadrilateral QCRI has right angles at Q and R.</p><p>Area(QCRI) = Area(△CIQ) + Area(△CIR) = (1/2)·CQ·IQ + (1/2)·CR·IR</p><p>= (1/2)·7·√7 + (1/2)·7·√7 = (7√7/2) + (7√7/2) = <strong>7√7</strong></p><p>∴ Answer: D</p>
Correct Answer: D