Trigonometry & Inverse Trigonometry
Properties of Triangle
Grade 11
Question:
<p>In a triangle ABC, perpendiculars drawn from vertices A, B, C meet the opposite sides BC, CA, AB at D, E, F respectively, triangle DEF is completed. The perimeter of triangle DEF is greater than or equal to \(3\sqrt{3}r\), where r is in-radius of triangle ABC. Also \(r = \sqrt{3}\) and perimeter of triangle ABC is 18. Then</p>
<p>(a) triangle ABC is right angled</p>
<p>(b) triangle ABC is equilateral</p>
<p>(c) area of triangle ABC is \(9\sqrt{3}\)</p>
<p>(d) ratio of area of triangle ABC to triangle DEF is 4 : 1</p>
Step-by-Step Solution
Key Concept: The orthic triangle DEF (formed by feet of altitudes) has perimeter ≥ 3√3r with equality when ABC is equilateral. Use the constraint that perimeter of ABC = 18 and r = √3 to determine the triangle's nature.
<p><strong>Step 1: Use inradius formula</strong></p><p>Given: perimeter = 18, so semi-perimeter s = 9</p><p>r = Area/s → √3 = Area/9 → Area = 9√3</p><p><strong>Step 2: Check if triangle is equilateral</strong></p><p>For equilateral triangle with side a: s = 3a/2 = 9 → a = 6</p><p>Area of equilateral triangle = (√3/4)a² = (√3/4)(36) = 9√3 ✓</p><p>Inradius of equilateral triangle = a/(2√3) = 6/(2√3) = √3 ✓</p><p><strong>Step 3: Verify orthic triangle perimeter</strong></p><p>For equilateral triangle ABC with side 6, the orthic triangle DEF is also equilateral with side = a/2 = 3</p><p>Perimeter of DEF = 9</p><p>Check constraint: 3√3r = 3√3(√3) = 9 ✓</p><p>Equality holds, confirming ABC is equilateral</p><p><strong>Step 4: Determine valid options</strong></p><p>Since ABC is equilateral with side 6:</p><p>• All angles are 60°</p><p>• All altitudes are equal: h = (√3/2)(6) = 3√3</p><p>• Triangle is acute-angled</p><p>• DEF is equilateral with side 3 and perimeter 9</p><p>∴ Answer: BCD</p>
Correct Answer: BCD