Trigonometry & Inverse Trigonometry
Properties of Triangle
Grade 11
Question:
<p>If a right angled \(\Delta ABC\) of maximum area is inscribed within a circle of radius R, then (\(\Delta\) represents area of triangle ABC and \(r, r_1, r_2, r_3\) represent inradius and exradii, and s is the semi perimeter of \(\Delta ABC\)) then</p>
<p>(a) \(\Delta = R^2\)</p>
<p>(b) \(\dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3} = \dfrac{\sqrt{2}+1}{R}\)</p>
<p>(c) \(r = (\sqrt{2}-1)R\)</p>
<p>(d) \(s = (1+\sqrt{2})R\)</p>
Step-by-Step Solution
Key Concept: A right-angled triangle inscribed in a circle has its hypotenuse as the diameter (Thales' theorem). Maximum area occurs when the triangle is isosceles right-angled, giving hypotenuse = 2R and legs = R√2 each.
<p><strong>Step 1: Setup</strong> For a right-angled triangle inscribed in a circle of radius R, the hypotenuse is a diameter (Thales' theorem), so hypotenuse c = 2R.</p><p><strong>Step 2: Maximize Area</strong> Let legs be a and b. Then a² + b² = 4R². Area Δ = ½ab. By AM-GM: (a² + b²)/2 ≥ ab, so ab ≤ 2R². Maximum when a = b = R√2. Thus Δ_max = R².</p><p><strong>Step 3: Calculate Parameters</strong> For isosceles right triangle with legs R√2 and hypotenuse 2R: s = (R√2 + R√2 + 2R)/2 = R(√2 + 1); r = Δ/s = R²/[R(√2 + 1)] = R(√2 - 1); c - side = 2R, so r₃ = Δ/(s-c) = R²/[R(√2-1)] = R(√2+1); a - side = R√2, so r₁ = Δ/(s-a) = R²/[R(1)] = R; similarly r₂ = R.</p><p><strong>Step 4: Verify Relations</strong> Key relations: Δ = r·s = R²✓; r₃ = s = R(√2+1)✓; r₁ = r₂ = R (legs equal)✓; r₁·r₂·r₃ = R·R·R(√2+1) = R³(√2+1)✓</p><p>∴ Answer: ABCD (All statements about maximum area right triangle are correct)</p>
Correct Answer: ABCD