Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12 Important
Ex 6.3,13
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19 Important You are here
Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Last updated at June 12, 2023 by Teachoo
Ex 6.3, 19 Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.Let radius be r of the circle & let 𝑥 be the length & 𝑦 be the breadth of the rectangle Now, Δ ABC is right angle triangle (AB)2 + (BC)2 = (AC)2 𝑥^2+𝑦^2 = (2𝑟)^2 𝑥^2+𝑦^2= 4𝑟2 𝑦2 = 4𝑟2 – 𝑥2 𝑦= √(4𝑟"2 – " 𝑥"2" ) We need to maximize Area of rectangle Let A be the area rectangle Area of rectangle = Length × Breadth A = xy A = 𝑥 √(4𝑟^2−𝑥^2 ) Since A has square root It will be difficult to differentiate So, we take Z = A2 Let Z = A2 Z = 𝑥^2× (√(4𝑟^2−𝑥^2 ))^2 Z = 𝑥^2×(4𝑟^2−𝑥^2 ) Z = 4𝑟^2 𝑥^2−𝑥^4 Since A is positive, A is maximum if A2 is maximum So, we maximize Z = A2 Diff. Z w.r.t 𝑥 𝑑Z/𝑑𝑥=𝑑(4𝑟^2 𝑥^2−𝑥^4 )/𝑑𝑥 𝑑Z/𝑑𝑥=4𝑟^2×2𝑥−4𝑥^3 𝑑Z/𝑑𝑥=8𝑟^2 𝑥−4𝑥^3 Putting 𝑑Z/𝑑𝑥 = 0 8𝑟^2 𝑥−4𝑥^3 = 0 4𝑥^3−8𝑟^2 𝑥 = 0 𝑥^3−2𝑟^2 𝑥 = 0 𝑥 (𝑥^2−2𝑟^2 ) = 0 Therefore, Finding (𝒅^𝟐 𝐙)/(𝐝𝒙^𝟐 ) 𝑑Z/𝑑𝑥=8𝑟^2 𝑥−4𝑥^3 Diff w.r.t 𝑥 (𝑑^2 Z)/(𝑑𝑥^2 ) = 𝑑/𝑑𝑥 [8𝑟^2 𝑥−4𝑥^3 ] (𝑑^2 Z)/(𝑑𝑥^2 ) = 8𝑟^2−4×3𝑥^2 (𝑑^2 Z)/(𝑑𝑥^2 ) = 8𝑟^2−12𝑥^2 Putting 𝒙^𝟐=𝟐𝒓^𝟐 (𝑑^2 Z)/(𝑑𝑥^2 ) = 8𝑟^2−12×2𝑟^2 (𝑑^2 Z)/(𝑑𝑥^2 ) = 8𝑟^2−24𝑟^2 (𝑑^2 Z)/(𝑑𝑥^2 ) = −16𝑟^2 < 0 Hence, (𝑑^2 Z)/(𝑑𝑥^2 ) < 0 at 𝑥^2=2𝑟^2 Thus area is maximum when 𝑥^2=2𝑟^2 Now, finding y 𝑦 = √(4𝑟^2 −𝑥^2 ) Putting 𝑥^2=2𝑟^2 𝑦 = √(4𝑟^2 −〖2𝑟〗^2 ) 𝑦 = √(〖2𝑟〗^2 ) 𝑦 = √2 𝑟 Therefore 𝑥 = 𝑦 = √2 𝑟 Hence area is maximum when 𝒙 = 𝒚 ∴ The rectangle is a square.