Trigonometry & Inverse Trigonometry
Properties of triangles
Grade 11
Question:
<p>In a triangle with sides \(a, b, c\) where \(s - a + s - b + s - c = 15\) (so \(s = 15\)) and the incircle touches side \(BC\) at \(Q\) and side \(CA\) at \(C'\) with \(QC = s - c\). If \(s - a = 3,\; s - b = 5,\; s - c = 7\), find the area of quadrilateral \(QCRI\) (where \(I\) is the incentre and \(R\) is the point of tangency on \(CA\)).</p>
<p>(a) \(5\sqrt{7}\)</p>
<p>(b) \(6\sqrt{7}\)</p>
<p>(c) \(8\sqrt{7}\)</p>
<p>(d) \(7\sqrt{7}\)</p>
Step-by-Step Solution
Key Concept: In a triangle with incircle, tangent segments from vertices are equal: from C, both tangent lengths equal s-c. The quadrilateral QCRI (where Q and R are tangency points on BC and CA respectively, and I is incentre) has area = (s-c)·r, since it's bounded by two tangent segments of length (s-c) and the inradius r acts as the perpendicular distance.
<p><strong>Step 1: Find the sides using s and given conditions.</strong></p><p>Given: s = 15, s-a = 3, s-b = 5, s-c = 7</p><p>Therefore: a = 12, b = 10, c = 8</p><p><strong>Step 2: Calculate area using Heron's formula.</strong></p><p>Area = √[s(s-a)(s-b)(s-c)] = √[15 × 3 × 5 × 7] = √1575 = 15√7</p><p><strong>Step 3: Find the inradius.</strong></p><p>r = Area/s = 15√7/15 = √7</p><p><strong>Step 4: Identify tangency points Q and R.</strong></p><p>Q is the tangency point on BC, R (or C') is the tangency point on CA.</p><p>From property of tangent segments: CQ = CR = s - c = 7</p><p><strong>Step 5: Calculate area of quadrilateral QCRI.</strong></p><p>QCRI is a kite with two sides CQ and CR equal to (s-c) = 7, and I is the incentre at distance r from both BC and CA.</p><p>Area(QCRI) = (1/2) × CQ × CR × sin(∠QCR) is not the direct approach.</p><p>More directly: Area(QCRI) = Area(△QCI) + Area(△RCI) = (1/2)·CQ·r + (1/2)·CR·r = (1/2)·r·(CQ + CR) = (1/2)·√7·(7+7) = (1/2)·√7·14 = 7√7</p><p><strong>Alternate verification:</strong> Area(QCRI) = (s-c)·r = 7·√7 = 7√7</p><p>∴ Answer: D</p>
Correct Answer: D