Show that the triangle of maximum area that can be inscribed in a given circle is an equilateral triangle.
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
CBSE Class 12 Sample Paper for 2020 Boards
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
CBSE Class 12 Sample Paper for 2020 Boards
Last updated at May 29, 2023 by Teachoo
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
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 ∴ OD be perpendicular to chord BC Since perpendicular from chord bisects the chord BD = 𝐵𝐶/2 = r Now, Δ OBD is right angle triangle OB2 = OD2 + BD2 𝑅^2=𝑟^2+(ℎ−𝑅)^2 𝑅^2=𝑟^2+ℎ^2+𝑅^2−2ℎ𝑅 0=𝑟^2+ℎ^2−2ℎ𝑅 2ℎ𝑅−ℎ^2=𝑟^2 𝑟^2 = 2ℎ𝑅−ℎ^2 Now, we need to maximize Area of Triangle Area of Triangle = 1/2 × Base × Height A = 1/2 × 2r × h A = r × 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 × h2 Since A is positive, A is maximum if A2 is maximum So, we maximize Z = A2 Now, Z = r2 × h2 Z = (2ℎ𝑅−ℎ^2) × ℎ^2 Z = 2ℎ^3 𝑅−ℎ^4 Maximizing Z Differentiating Z w.r.t h 𝑑𝑍/𝑑ℎ=2𝑅×3ℎ^2−4ℎ^3 𝑑𝑍/𝑑ℎ=6𝑅×ℎ^2−4ℎ^3 Putting 𝑑𝑍/𝑑ℎ = 0 6𝑅×ℎ^2−4ℎ^3 = 0 6𝑅×ℎ^2−4ℎ^3 = 0 ℎ^2 (6𝑅−4ℎ) = 0 So, h = 0 and h = 3𝑅/2 Since h cannot be 0 h = 3𝑅/2 Now, checking sign of (𝑑^2 𝑍)/(𝑑ℎ^2 ) " " 𝑑𝑍/𝑑ℎ=6𝑅×ℎ^2−4ℎ^3 Differentiating again w.r.t h (𝑑^2 𝑍)/(𝑑ℎ^2 ) = 6𝑅×2ℎ−12ℎ^2 Putting h = 3𝑅/2 (𝑑^2 𝑍)/(𝑑ℎ^2 ) = 6𝑅×2×3𝑅/2−12(3𝑅/2)^2 (𝑑^2 𝑍)/(𝑑ℎ^2 ) = 6𝑅×3𝑅−12×(9𝑅^2)/4 (𝑑^2 𝑍)/(𝑑ℎ^2 ) = 18𝑅^2−36𝑅^2 (𝑑^2 𝑍)/(𝑑ℎ^2 ) = −18𝑅^2 ∴ (𝑑^2 𝑍)/(𝑑ℎ^2 ) < 0 for h = 3𝑅/2 ∴ Z is maximum when h = 3𝑅/2 Now, Solving for h and r 𝑟^2 = 2ℎ𝑅−ℎ^2 Putting h = 3𝑅/2 𝑟^2 = 2×3𝑅/2×𝑅−(3𝑅/2)^2 𝑟^2 = 3𝑅^2−(9𝑅^2)/4 𝑟^2 = (3𝑅^2)/4 𝑟 = (√3 𝑅)/2 So, Side BC = 2r = √3R Now, In right triangle ABD AB2 = AD2 + BD2 AB2 = h2 + r2 AB2 = (3𝑅/2)^2 + ((√3 𝑅)/2)^2 AB2 = (9𝑅^2)/4 + (3𝑅^2)/4 AB2 = (12𝑅^2)/4 AB2 = 3R^2 AB = √3R Similarly, AC = √3R Since BC = AB = AC = √3R Hence, Δ ABC is an equilateral triangle