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a right cone is inscribed in a sphere of radius R. Let S=f(x) be the functional relationship between the lateral surface area S of the cone and its gereratrix x then the find the value of f(R) and if x is such that x to the power n, n>= 6 is negligible then find S.

a right cone is inscribed in a sphere of radius R. Let S=f(x) be the functional relationship between the lateral surface area S of the cone and its gereratrix x then the find the value of f(R) and if x is such that x to the power n, n>= 6 is negligible then find S.

Grade:12

1 Answers

Askiitians_Expert Yagyadutt
askIITians Faculty 74 Points
10 years ago

Hii vashuda

First of all before reading this...draw the geometry of the condition given....i am assuming that u have drawn a sphere a cone ...with

 

Slant height = x  radius of cone = r  and heigh of cone = h

Radiius of sphere = R after this read my solution

 

The height of the cone h = R + (something)

something = distance of centre from the base of the cone....you need to find this something okk

Let it be z

then by applying a few strength on mind you will get ..

z = root( R^2 - r^2)   [ because there is a small right angle triangle formed by side R , r and z with R as largest side )

 

Now surface area  S = π.r.x= f(x)

 

again x , h and r form a rt.angled triangle..

 

x^2 = h^2 + r^2  => x^2 = (R+z)^2 + r^2    =  R^2 + z^2 + 2.z.R + r^2

 

x^2 = R^2 + R^2 -r^2 +  2.root(R^2-r^2).R + r^2 

x^2 = 2.R^2  + 2.root(R^2-r^2).R 

now put x=R

R^2 = 2.R^2 + 2.root(R^2 -r^2).R

 

-R^2 = 2.root(R^2 -r^2).R

-R = 2.root( R^2 -r^2)

 

R^2 = 4.(R^2 - r^2)

4r^2  = 3R^2

 

or  r = R.root(3)/2 --------(1)

 

f(x) = pi.x.r

 

So f(R) = pi.R.( R.root(3)/2 ) =  root(3).pi.R^2 / 2    ans ....

 

If u r unable to understand my steps...then please ask furthure...

 

Regard

Yagya

aksiitians_expert

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