## Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle.

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CBSE Class 12 Sample Paper for 2020 Boards

Paper Summary

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10 Important

Question 11 Important

Question 12

Question 13

Question 14 (OR 1st Question)

Question 14 (OR 2nd Question) Important

Question 15 (OR 1st Question)

Question 15 (OR 2nd Question) Important

Question 16

Question 17 Important

Question 18 (OR 1st Question)

Question 18 (OR 2nd Question)

Question 19

Question 20

Question 21 (OR 1st Question) Important

Question 21 (OR 2nd Question) Important

Question 22

Question 23

Question 24 (OR 1st Question)

Question 24 (OR 2nd Question)

Question 25

Question 26 Important

Question 27

Question 28 (OR 1st Question) Important

Question 28 (OR 2nd Question)

Question 29

Question 30 Important

Question 31 (OR 1st Question) Important

Question 31 (OR 2nd Question)

Question 32 Important

Question 33 (OR 1st Question) Important

Question 33 (OR 2nd Question) Important

Question 34

Question 35 (OR 1st Question) Important

Question 35 (OR 2nd Question) Important You are here

Question 36 Important

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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