Please solve the NCERT question No.4 Exercise-13.5 (p-258)

Calculations for lateral surface area:

--------------------- +P --------------------
       |                  /|\                  |
       |                 / | \                 |
       |                /  |  \ S-s            |H-h
       |               /   |   \               |
       |              /    |    \              |
       |             /*****|**r**\             |
       |            **    C+-----*+D--------------
      H|         S /  ***********  \           |
       |          /        |        \          |
       |         /         |         \         |
       |        /          |          \ s      |
       |       /           |           \       |h
       |      /            |            \      |
       |     /             |             \     |
       |    /    **********|**********    \    |
       |   /*****          |       R  *****\   |
     -----*               A+----------------+B----
           ******                     ******
                 *********************
The formula for a complete 
cone is:

     A = pi R S

where R is the radius and S is the slant height of the whole cone. For 
the frustum, we will subtract the area of the cut-off cone (whose 
slant height is S-s) from the whole:

     A = pi R S - pi r (S-s)

       = pi (RS - rS + rs)

       = pi ((R-r)S + rs)

By the same similar triangles as before, we can write

     AB   CD        R    r
     -- = --   or   - = ---
     PB   PD        S   S-s

Again solving for S,

      R(S-s) = rS

     RS - Rs = rS

     RS - rS = Rs

      (R-r)S = Rs

               Rs
           S = ---
               R-r

Now the area is

                  Rs
     A = pi ((R-r)--- + rs) = pi (Rs + rs) = pi(R+r)s
                  R-r