CBSE Class 10 Answered
Please solve the NCERT question No.4 Exercise-13.5 (p-258)
Asked by Mdivya | 12 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 | 13 Mar, 2009, 01:57: AM
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