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If H, S and V Denote Respectively the Height, Curved Surface Area and Volume of a Right Circular Cone, Then 3 π V H 3 − S 2 H 2 + 9 V 2 is Equal to - Mathematics

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Question

If h, S and V denote respectively  the height, curved surface area and volume of a right circular cone, then  `3 pi Vh^3 - S^2h^2 + 9V^2`   is equal to

Options

  • 8

  • 0

  • 4`pi`

  • 32`pi^2`

MCQ

Solution

Here we are asked to find the value for a given specific equation which is in terms of V, h and Srepresenting the volume, vertical height and the Curved Surface Area of a cone.

We know `V =1/3(pir^2h) and S=`pirl`.

Also, `l = sqrt(r^2 + h^2)` 
Now, the given equation is 

`3 piVh^3 - S^2h^2 + 9 V^2``

So,

`3 piVh^3 - S^2h^2 + 9 V^2``

`= 3pi (1/3 pi r^2h)h^3 -(pirl)^2 h^2 + 9(1/3 pi r^2 h)^2`

`=pi^2r^2h^4-pi^2r^2l^2h^2 + 9 (1/9pi^2r^4h^2)`

` = pi^2r^2h^4 - pi^2r^2h^2 (sqrt(r^2 + h^2))^2  + pi^2r^4h^2`

`=pi^2r^2h^4 - pi^2r^2h^2 (sqrt(r^2 +h^2))^2 + pi^2 r^4 h^2`

`=pi^2r^2h^4 - pi^2r^2h^2 (r^2 ++h^2) + pi^2r^4h^2`

`=pi^2 r^2h^4 - pi^2r^4h^2 - pi^2r^2h^4 + pi^2r^4h^2`

= 0 

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Chapter 20: Surface Areas and Volume of A Right Circular Cone - Exercise 20.4 [Page 25]

APPEARS IN

RD Sharma Mathematics [English] Class 9
Chapter 20 Surface Areas and Volume of A Right Circular Cone
Exercise 20.4 | Q 16 | Page 25

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