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 = 2RH+R²

Total 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 = 1/3R²H

Volume of new cone = 1/3R'²H'    where R'=R+kR/100 , H'=H+kH/100

On solving

Percentage Increase =3k%