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

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

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Show that triangle of maximum area that can be inscribed in a circle

Question 35 (OR 2nd Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 2
Question 35 (OR 2nd Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 3
Question 35 (OR 2nd Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 4
Question 35 (OR 2nd Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 5
Question 35 (OR 2nd Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 6
Question 35 (OR 2nd Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 7
Question 35 (OR 2nd Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 8
Question 35 (OR 2nd Question) - CBSE Class 12 Sample Paper for 2020 Boards - Part 9


Transcript

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 = 𝐡𝐢/2 = r Now, Ξ” OBD is right angle triangle OB2 = OD2 + BD2 𝑅^2=π‘Ÿ^2+(β„Žβˆ’π‘…)^2 𝑅^2=π‘Ÿ^2+β„Ž^2+𝑅^2βˆ’2β„Žπ‘… 0=π‘Ÿ^2+β„Ž^2βˆ’2β„Žπ‘… 2β„Žπ‘…βˆ’β„Ž^2=π‘Ÿ^2 π‘Ÿ^2 = 2β„Žπ‘…βˆ’β„Ž^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 = (2β„Žπ‘…βˆ’β„Ž^2) Γ— β„Ž^2 Z = 2β„Ž^3 π‘…βˆ’β„Ž^4 Maximizing Z Differentiating Z w.r.t h 𝑑𝑍/π‘‘β„Ž=2𝑅×3β„Ž^2βˆ’4β„Ž^3 𝑑𝑍/π‘‘β„Ž=6π‘…Γ—β„Ž^2βˆ’4β„Ž^3 Putting 𝑑𝑍/π‘‘β„Ž = 0 6π‘…Γ—β„Ž^2βˆ’4β„Ž^3 = 0 6π‘…Γ—β„Ž^2βˆ’4β„Ž^3 = 0 β„Ž^2 (6π‘…βˆ’4β„Ž) = 0 So, h = 0 and h = 3𝑅/2 Since h cannot be 0 h = 3𝑅/2 Now, checking sign of (𝑑^2 𝑍)/(π‘‘β„Ž^2 ) " " 𝑑𝑍/π‘‘β„Ž=6π‘…Γ—β„Ž^2βˆ’4β„Ž^3 Differentiating again w.r.t h (𝑑^2 𝑍)/(π‘‘β„Ž^2 ) = 6𝑅×2β„Žβˆ’12β„Ž^2 Putting h = 3𝑅/2 (𝑑^2 𝑍)/(π‘‘β„Ž^2 ) = 6𝑅×2Γ—3𝑅/2βˆ’12(3𝑅/2)^2 (𝑑^2 𝑍)/(π‘‘β„Ž^2 ) = 6𝑅×3π‘…βˆ’12Γ—(9𝑅^2)/4 (𝑑^2 𝑍)/(π‘‘β„Ž^2 ) = 18𝑅^2βˆ’36𝑅^2 (𝑑^2 𝑍)/(π‘‘β„Ž^2 ) = βˆ’18𝑅^2 ∴ (𝑑^2 𝑍)/(π‘‘β„Ž^2 ) < 0 for h = 3𝑅/2 ∴ Z is maximum when h = 3𝑅/2 Now, Solving for h and r π‘Ÿ^2 = 2β„Žπ‘…βˆ’β„Ž^2 Putting h = 3𝑅/2 π‘Ÿ^2 = 2Γ—3𝑅/2Γ—π‘…βˆ’(3𝑅/2)^2 π‘Ÿ^2 = 3𝑅^2βˆ’(9𝑅^2)/4 π‘Ÿ^2 = (3𝑅^2)/4 π‘Ÿ = (√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, Ξ” ABC is an equilateral triangle

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.