Trigonometry & Inverse Trigonometry
Properties of triangles
Grade 11
Question:
<p>In a triangle <em>ABC</em>, let \(\angle C = \pi/2\). If <em>r</em> is the inradius and <em>R</em> is the circumradius of the triangle <em>ABC</em>, then 2(<em>r</em> + <em>R</em>) equals</p>
<p>\(b + c\)</p>
<p>\(a + b\)</p>
<p>\(a + b + c\)</p>
<p>\(c + a\)</p>
Step-by-Step Solution
Key Concept: In a right triangle with angle C = π/2, the circumradius R = c/2 (half the hypotenuse) and inradius r = (a+b-c)/2. Adding these and using the relationship between sides gives a direct formula in terms of the legs.
<p><strong>Step 1:</strong> For a right triangle with right angle at C, the hypotenuse is c = AB. The circumradius of a right triangle equals half the hypotenuse: <strong>R = c/2</strong></p><p><strong>Step 2:</strong> The inradius of a right triangle with legs a, b and hypotenuse c is: <strong>r = (a + b - c)/2</strong></p><p><strong>Step 3:</strong> Calculate r + R:<br>r + R = (a + b - c)/2 + c/2 = (a + b - c + c)/2 = (a + b)/2</p><p><strong>Step 4:</strong> Therefore:<br>2(r + R) = 2 · (a + b)/2 = <strong>a + b</strong></p><p>∴ Answer: <strong>B</strong> (where B represents a + b, the sum of the two legs)</p>
Correct Answer: B