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

Asked by Mdivya | 12th Mar, 2009, 11:21: PM

Expert Answer:

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

Answered by  | 13th Mar, 2009, 01:57: AM

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