Trigonometry & Inverse Trigonometry
Properties of Triangles
Grade 11
Question:
<p><strong>810.</strong> In \(\triangle ABC\) if inradius \(r = 1\), circumradius \(R = 3\) and semiperimeter \(s = 7\), then find the value of \((a^2 + b^2 + c^2)\), where \(a, b, c\) are the sides of triangle \(ABC\).</p>
Step-by-Step Solution
Key Concept: Use the fundamental relations: Area = rs (connects inradius and semiperimeter), Area = abc/4R (relates circumradius), and the identity a² + b² + c² = s² - 2(rs - r²) derived from (s-a)² + (s-b)² + (s-c)² expansion.
<p><strong>Step 1:</strong> Find the area using inradius formula.</p><p>Area = rs = 1 × 7 = 7</p><p><strong>Step 2:</strong> Verify consistency using circumradius formula: Area = abc/(4R), so abc = 4R × Area = 4(3)(7) = 84.</p><p><strong>Step 3:</strong> Use the identity relating a² + b² + c² to known quantities.</p><p>We know: a + b + c = 2s = 14</p><p>From Area = √[s(s-a)(s-b)(s-c)], we have: 7 = √[7(7-a)(7-b)(7-c)]</p><p>So: 49 = 7(7-a)(7-b)(7-c), giving (7-a)(7-b)(7-c) = 7</p><p><strong>Step 4:</strong> Apply the formula a² + b² + c² = (a+b+c)² - 2(ab+bc+ca).</p><p>We need ab + bc + ca. Using the extended relation:</p><p>(s-a) + (s-b) + (s-c) = 3s - 2s = s = 7</p><p>And (s-a)(s-b) + (s-b)(s-c) + (s-c)(s-a) = s² - r(2s) - r² = 49 - 2(1)(14) - 1 = 49 - 28 - 1 = 20</p><p>This gives: ab + bc + ca = s² - [s² - 2s(s-a-b-c) - (s-a)(s-b) - (s-b)(s-c) - (s-c)(s-a)]</p><p>More directly: ab + bc + ca = s² + r² + 4Rr = 49 + 1 + 4(3)(1) = 49 + 1 + 12 = 62</p><p><strong>Step 5:</strong> Calculate a² + b² + c².</p><p>a² + b² + c² = (a+b+c)² - 2(ab+bc+ca) = 14² - 2(62) = 196 - 124 = 72</p><p>∴ <strong>Answer: 72</strong></p>
Correct Answer: 72