Trigonometry & Inverse Trigonometry
Properties of Triangles
Grade 11
Question:
<p>Let \(ABC\) be a right angled triangle at \(C\). If the inscribed circle touches the side \(AB\) at \(D\) and \((AD)(BD) = 11\), then find the area of \(\triangle ABC\).</p>
Step-by-Step Solution
Key Concept: In a right angled triangle, the product of segments on the hypotenuse equals the product of semi-perimeter differences, which directly relates to the area.
<p><strong>Step 1:</strong> We have \((AD)(BD) = 11\)</p><p><strong>Step 2:</strong> For a right angled triangle with inscribed circle: \((AD)(BD) = (s-a)(s-b)\)</p><p>Therefore: \((s-a)(s-b) = 11\)</p><p><strong>Step 3:</strong> Using the relation \((2s-2a)(2s-2b) = 44\) and properties of the inscribed circle in a right angled triangle where the inradius \(r = \frac{a+b-c}{2}\)</p><p><strong>Step 4:</strong> For a right angled triangle at \(C\): Area \(= rs = (s-a)(s-b) = 11\)</p><p><strong>Answer:</strong> The area of \(\triangle ABC\) is <strong>11</strong>.</p>
Correct Answer: 11