1.The surface area increases approximately by 2k%
2.The volume increases approximately by 3k% for small k
Soln: Height of the new cone = H+K*H/100 where H is the height of the original cone
The semivertical angle 'A' is related as
cos A = H/L where L is the slant height
Also for the new cone ,
cos A = (H+kH/100)/L' where L' is the slant height of new cone
From above eqs.
L' = L+kL/100
We know
sin A = R/L
For the new cone,
sin A = R'/(L+kL/100)
From above eqs.
R' = R+kR/100
Total Surface Area = 2
RH+
R
2Total Surface area of new cone = 2
(R+kR/100)(H+kH/100) +
(R+kR/100)
2
Increase in area = ((Area of new cone)/Area of original cone)-1
On solving ,
Increase = 2k/100
Therefore , percentage increase = 2k%
Volume = 1/3
R
2H
Volume of new cone = 1/3
R'
2H' where R'=R+kR/100 , H'=H+kH/100
On solving
Percentage Increase =3k%
Moderation Team
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