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CBSE Class 12 Math 2022 Term I Solved Paper

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Section C
Attempt any 8 questions out of the Questions 41-50. Each question is of one mark.
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Question : 41 of 50
Marks: +1, -0
In a sphere of radius rr, a right circular cone of height hh having maximum curved surface area is inscribed. The expression for the square of curved surface of cone is
Solution:     👈: Video Solution
Here,     \;\; CSA of cone =πRl=\pi R l
Radius of sphere =r=r
height of cone =h=h
In â–³AOC\triangle AOC ,
AO2=AC2+OC2A O^2=A C^2+O C^2
⇒    r2=R2+(h−r)2\Rightarrow \;\; r^2=R^2+(h-r)^2
⇒    R2=2hr−h2\Rightarrow \;\; R^2=2 h r-h^2
∴    \therefore \;\; Radius of cone, R=2hr−h2      ⋅⋅⋅⋅⋅⋅⋅(i)R=\sqrt{2 h r - h^2}\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (i)
In â–³ABC\triangle ABC ,
AB2  =AC2+BC2A B^2\;=A C^2+B C^2
⇒l2  =R2+h2\Rightarrow l^2\;=R^2+h^2
⇒l2  =2hr−h2+h2\Rightarrow l^2\;=2 h r-h^2+h^2
∴ slant height     =2hr      ⋅⋅⋅⋅⋅⋅⋅(ii)\;\;=\sqrt{2 h r}\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot (ii)
CSA of cone =Ï€Rl=\pi R l
  =π2hr−h22hr\;=\pi \sqrt{2 h r - h^2} \sqrt{2 h r}
(  CSA of cone  )2  =π2(2hr−h2)(2hr)(\;\text{CSA of cone}\;)^2\;=\pi^2 (2 h r - h^2) (2 h r)
  =2π2hr(2hr−h2)\;=2 \pi^2 h r (2 h r - h^2)
  =2π2r(2rh2−h3)\;=2 \pi^2 r (2 r h^2 - h^3)
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