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>[<b>Note:</b> 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 a right angle at both Q and R (since the incircle is perpendicular to the sides at contact points). Use the property that CQ = CR = s - c to find that QCRI is a kite with area = (1/2) × d₁ × d₂, where d₁ = QR and d₂ = CI. Alternatively, Area(QCRI) = Area(△CQI) + Area(△CRI) = (1/2)(s-c)·r + (1/2)(s-c)·r = (s-c)·r.
<p><strong>Step 1:</strong> Recall that for tangents from an external point to a circle: tangent segments are equal. From vertex C, the two tangent segments to the incircle are CQ and CR, both equal to <strong>s - c</strong>.</p><p><strong>Step 2:</strong> Given: s - a = 3, s - b = 5, s - c = 7, so CQ = CR = 7. The semiperimeter s = (s-a) + (s-b) + (s-c) + (a+b+c))/2 is found from: a + b + c = 2s, so s - a + s - b + s - c = 3s - (a+b+c) = 3s - 2s = s. Thus s = 3 + 5 + 7 = 15.</p><p><strong>Step 3:</strong> Find the inradius: Area by Heron's formula: a = s - (s-a) = 15 - 3 = 12, b = 15 - 5 = 10, c = 15 - 7 = 8. Area Δ = √[s(s-a)(s-b)(s-c)] = √[15 × 3 × 5 × 7] = √1575 = 15√7. Thus r = Δ/s = 15√7/15 = √7.</p><p><strong>Step 4:</strong> Since IQ ⊥ CQ and IR ⊥ CR (radii to tangent points), the quadrilateral QCRI has two right angles at Q and R. Area(QCRI) = Area(△IQC) + Area(△IRC) = (1/2)·CQ·r + (1/2)·CR·r = (1/2)·7·√7 + (1/2)·7·√7 = 7√7.</p><p>∴ Answer: D</p>
Correct Answer: D