Trigonometry & Inverse Trigonometry
Properties of Triangle
Grade 11
Question:
<p>95. In the △ABC, AB = 5 cm, AC = 12 cm and BC = 13 cm then the distance of A from the side BC is (in cm)</p>
<p>(a) \(\dfrac{25}{13}\)</p>
<p>(b) \(\dfrac{60}{13}\)</p>
<p>(c) \(\dfrac{65}{12}\)</p>
<p>(d) \(\dfrac{144}{13}\)</p>
Step-by-Step Solution
Key Concept: Recognize that 5-12-13 is a Pythagorean triple (5² + 12² = 13²), making triangle ABC a right-angled triangle with the right angle at A. The altitude from A to BC equals the area calculated two different ways.
<p><strong>Step 1:</strong> Check if triangle is right-angled using Pythagorean theorem:</p><p>AB² + AC² = 5² + 12² = 25 + 144 = 169 = 13² = BC²</p><p>∴ Triangle ABC is right-angled at A (∠BAC = 90°)</p><p><strong>Step 2:</strong> Calculate area of triangle using two methods:</p><p>Method 1 (using right angle): Area = ½ × AB × AC = ½ × 5 × 12 = 30 cm²</p><p><strong>Step 3:</strong> Method 2 (using BC as base and h as altitude from A):</p><p>Area = ½ × BC × h = ½ × 13 × h</p><p><strong>Step 4:</strong> Equate both expressions:</p><p>30 = ½ × 13 × h</p><p>h = 60/13 cm</p><p>∴ Distance of A from BC = <strong>60/13 cm</strong> or <strong>4.62 cm</strong></p>
Correct Answer: B