Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle.
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Last updated at Oct. 24, 2019 by Teachoo
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Question 35 (OR 2nd Question) Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle. Let R be the radius of Circle and h be height of triangle 2r be the base of triangle Let AD be the height, it is perpendicular to BC β΄ OD be perpendicular to chord BC Since perpendicular from chord bisects the chord BD = π΅πΆ/2 = r Now, Ξ OBD is right angle triangle OB2 = OD2 + BD2 π ^2=π^2+(ββπ )^2 π ^2=π^2+β^2+π ^2β2βπ 0=π^2+β^2β2βπ 2βπ ββ^2=π^2 π^2 = 2βπ ββ^2 Now, we need to maximize Area of Triangle Area of Triangle = 1/2 Γ Base Γ Height A = 1/2 Γ 2r Γ h A = r Γ h Now, we know the value of r2 In value of r, we will get square root Which will be difficult to differentiate Let Z = A2 Z = r2 Γ h2 Since A is positive, A is maximum if A2 is maximum So, we maximize Z = A2 Now, Z = r2 Γ h2 Z = (2βπ ββ^2) Γ β^2 Z = 2β^3 π ββ^4 Maximizing Z Differentiating Z w.r.t h ππ/πβ=2π Γ3β^2β4β^3 ππ/πβ=6π Γβ^2β4β^3 Putting ππ/πβ = 0 6π Γβ^2β4β^3 = 0 6π Γβ^2β4β^3 = 0 β^2 (6π β4β) = 0 So, h = 0 and h = 3π /2 Since h cannot be 0 h = 3π /2 Now, checking sign of (π^2 π)/(πβ^2 ) " " ππ/πβ=6π Γβ^2β4β^3 Differentiating again w.r.t h (π^2 π)/(πβ^2 ) = 6π Γ2ββ12β^2 Putting h = 3π /2 (π^2 π)/(πβ^2 ) = 6π Γ2Γ3π /2β12(3π /2)^2 (π^2 π)/(πβ^2 ) = 6π Γ3π β12Γ(9π ^2)/4 (π^2 π)/(πβ^2 ) = 18π ^2β36π ^2 (π^2 π)/(πβ^2 ) = β18π ^2 β΄ (π^2 π)/(πβ^2 ) < 0 for h = 3π /2 β΄ Z is maximum when h = 3π /2 Now, Solving for h and r π^2 = 2βπ ββ^2 Putting h = 3π /2 π^2 = 2Γ3π /2Γπ β(3π /2)^2 π^2 = 3π ^2β(9π ^2)/4 π^2 = (3π ^2)/4 π = (β3 π )/2 So, Side BC = 2r = β3R Now, In right triangle ABD AB2 = AD2 + BD2 AB2 = h2 + r2 AB2 = (3π /2)^2 + ((β3 π )/2)^2 AB2 = (9π ^2)/4 + (3π ^2)/4 AB2 = (12π ^2)/4 AB2 = 3R^2 AB = β3R Similarly, AC = β3R Since BC = AB = AC = β3R Hence, Ξ ABC is an equilateral triangle