Given, a cone inscribed in a sphere of radius a.
Let the radius, height and slant height of cone be r , h , l r,h,l r , h , l respectively.
O B = O A = a , B C = r , A C = h , O C = h − a , A B = l OB=OA=a,BC=r,AC=h,OC=h-a,AB=l OB = O A = a , BC = r , A C = h , OC = h − a , A B = l
Using pythagoras theorem in triangle BOC,
B C 2 + C O 2 = O B 2 ⇒ r 2 + ( h − a ) 2 = a 2 ⇒ r 2 + h 2 + a 2 − 2 h a = a 2 ⇒ r 2 + h 2 = 2 h a . . . ( i ) BC^2+CO^2=OB^2
\\\Rightarrow r^2+(h-a)^2=a^2
\\\Rightarrow r^2+h^2+a^2-2ha=a^2
\\\Rightarrow r^2+h^2=2ha \ ...(i) B C 2 + C O 2 = O B 2 ⇒ r 2 + ( h − a ) 2 = a 2 ⇒ r 2 + h 2 + a 2 − 2 ha = a 2 ⇒ r 2 + h 2 = 2 ha ... ( i )
Using pythagoras theorem in triangle BAC,
B C 2 + C A 2 = A B 2 ⇒ r 2 + h 2 = l 2 . . . ( i i ) BC^2+CA^2=AB^2
\\\Rightarrow r^2+h^2=l^2\ ...(ii) B C 2 + C A 2 = A B 2 ⇒ r 2 + h 2 = l 2 ... ( ii )
From (i) and (ii)
2 h a = l 2 2ha=l^2 2 ha = l 2
Now, lateral area of cone, L = π r l L=\pi rl L = π r l
L 2 = π 2 r 2 l 2 = π 2 ( l 2 − h 2 ) l 2 [ F r o m ( i i ) ] = π 2 ( 2 h a − h 2 ) ( 2 h a ) = π 2 ( 4 h 2 a 2 − 2 h 3 a ) L^2=\pi^2 r^2 l^2
\\=\pi^2 (l^2-h^2)l^2\ [From\ (ii)]
\\=\pi^2(2ha-h^2)(2ha)
\\=\pi^2(4h^2a^2-2h^3a) L 2 = π 2 r 2 l 2 = π 2 ( l 2 − h 2 ) l 2 [ F ro m ( ii )] = π 2 ( 2 ha − h 2 ) ( 2 ha ) = π 2 ( 4 h 2 a 2 − 2 h 3 a )
On differentiating w.r.t. h
2 L d L d h = π 2 ( 8 h a 2 − 6 h 2 a ) . . . ( i i i ) ⇒ d L d h = π 2 ( 2 h a π 2 ( 4 a − 3 h ) ) 2 L 2L\dfrac{dL}{dh}=\pi^2(8ha^2-6h^2a) ...(iii)
\\ \Rightarrow\dfrac{dL}{dh}=\dfrac{\pi^2 (2ha\pi^2 (4a-3h))}{2L} 2 L d h d L = π 2 ( 8 h a 2 − 6 h 2 a ) ... ( iii ) ⇒ d h d L = 2 L π 2 ( 2 ha π 2 ( 4 a − 3 h ))
Now, put d L d h = 0 \dfrac{dL}{dh}=0 d h d L = 0
⇒ π 2 ( 2 h a π 2 ( 4 a − 3 h ) ) 2 L = 0 ⇒ 4 a − 3 h = 0 ⇒ h = 4 a 3 . . . ( i v ) \\ \Rightarrow\dfrac{\pi^2 (2ha\pi^2 (4a-3h))}{2L}=0
\\ \Rightarrow 4a-3h=0
\\ \Rightarrow h=\dfrac{4a}3\ ...(iv) ⇒ 2 L π 2 ( 2 ha π 2 ( 4 a − 3 h )) = 0 ⇒ 4 a − 3 h = 0 ⇒ h = 3 4 a ... ( i v )
Again, differentiating (iii) w.r.t h,
2 L d 2 L d h 2 + 2 ( d L d h ) 2 = π 2 ( 8 a 2 − 12 h a ) = π 2 ( 8 a 2 − 12 a ( 4 a 3 ) ) [ U s i n g ( i v ) ] = − 8 π 2 a 2 < 0 ⇒ m a x i m a 2L\dfrac{d^2L}{dh^2}+2(\dfrac{dL}{dh})^2=\pi^2(8a^2-12ha)
\\=\pi^2(8a^2-12a(\dfrac{4a}3))\ [Using\ (iv)]
\\=-8\pi^2a^2<0
\\\Rightarrow\ maxima 2 L d h 2 d 2 L + 2 ( d h d L ) 2 = π 2 ( 8 a 2 − 12 ha ) = π 2 ( 8 a 2 − 12 a ( 3 4 a )) [ U s in g ( i v )] = − 8 π 2 a 2 < 0 ⇒ ma x ima
Put (iv) in (i)
r 2 + ( 4 a 3 ) 2 = 2 ( 4 a 3 ) a ⇒ r 2 = 8 a 2 3 − 16 a 2 9 ⇒ r 2 = 8 a 2 9 ⇒ r = 2 2 a 3 \\r^2+(\dfrac{4a}3)^2=2(\dfrac{4a}3)a
\\\Rightarrow r^2=\dfrac{8a^2}3-\dfrac{16a^2}9
\\\Rightarrow r^2=\dfrac{8a^2}9
\\\Rightarrow r=\dfrac{2\sqrt2a}3 r 2 + ( 3 4 a ) 2 = 2 ( 3 4 a ) a ⇒ r 2 = 3 8 a 2 − 9 16 a 2 ⇒ r 2 = 9 8 a 2 ⇒ r = 3 2 2 a
Thus, the dimensions of cone are r = 2 2 a 3 ; h = 4 a 3 r=\dfrac{2\sqrt2a}3;h=\dfrac{4a}3 r = 3 2 2 a ; h = 3 4 a
Comments