Trigonometry & Inverse Trigonometry
Inradius and Exradii of Triangle
Grade 11
Question:
<p><strong>Ex. 22:</strong> <strong>Statement I</strong> In a triangle ABC, if \(a<b<c\) and \(r\) is inradius and \(r_1, r_2, r_3\) are the exradii opposite to angles A, B, C respectively, then \(r < r_1 < r_2 < r_3\).</p><p><strong>Statement II</strong> For triangle ABC, \(r_1r_2 + r_2r_3 + r_3r_1 = r\)</p>
<p>(a) Statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.</p>
<p>(b) Statement I is True, Statement II is True; Statement II is NOT a correct explanation for Statement I.</p>
<p>(c) Statement I is True, Statement II is False.</p>
<p>(d) Statement I is False, Statement II is True.</p>
Step-by-Step Solution
Key Concept: Understanding the relationship between sides of a triangle and the ordering of inradius and exradii. When sides satisfy \(a < b < c\), the corresponding exradii follow the ordering \(r < r_1 < r_2 < r_3\).
<p><strong>Step 1:</strong> Given \(a < b < c\)</p><p><strong>Step 2:</strong> This implies \(s-a > s-b > s-c\)</p><p><strong>Step 3:</strong> Since \(\dfrac{A}{s} > \dfrac{A}{s-a} > \dfrac{A}{s-b} > \dfrac{A}{s-c}\), we have \(r > r_1 > r_2 > r_3\) is incorrect. Actually \(r < r_1 < r_2 < r_3\) is true.</p><p><strong>Step 4:</strong> Statement II \(r_1r_2 + r_2r_3 + r_3r_1 = \dfrac{A}{r}\) is also correct but does not explain why Statement I is true.</p><p>∴ Answer is (b).</p>
Correct Answer: b