Question 899858
sorry, in previous answer, I forgot to add inside area of each  of the four pieces

let
{{{SA}}}=surface area of a cone
{{{SA[I]}}}=Inside surface area of each of {{{4}}} pieces of cone

 total area:{{{SA+4*SA[I]}}}

first find {{{SA}}}

{{{SA = pi*r^2 + pi*r*l}}} ................surface area

{{{SA = 3.14*3^2 + 3.14*3*6}}}

{{{SA = 3.14*9 + 3.14*18}}}

{{{SA = 28.26 + 56.52}}}

{{{SA = 84.78}}}

now find {{{SA[I]}}}

if cut in {{{4}}} peaces,  you have {{{4}}} inside areas and each peace has {{{SA[I]=2*hr/2=hr}}}

where heit {{{h^2=l^2-r^2}}}

{{{h^2=6^2-3^2}}}

{{{h^2=36-9}}}

{{{h^2=27}}}

{{{h=sqrt(27)}}}

{{{h=5.2}}}.............plug it in {{{SA[I]=hr}}}

{{{SA[I]=5.2*3=15.6}}}....inside area of one peace, and

so, {{{4*SA[I]=4*15.6=62.4}}}

and {{{SA+4*SA[I]=84.78+62.4=147.18}}} which rounded gives you {{{SA+4*SA[I]=84.78+62.4=147cm^2}}}