Question 1120123
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1.  Base radius of the frustum cone is  r = 3.5 cm.

    The upper radius of the frustum cone is  R = 7 cm.

    The height of the frustum of the cone is  h = 21 - 7 = 14 cm.

    The slant height of the frustum of the cone is  s = {{{sqrt((R-r)^2 + h^2)}}} = {{{sqrt(3.5^2 + 14^2)}}} = 14.43 cm  (approximately).



2.  The latent surface of the frustum of the cone is  

    {{{A[1]}}} = {{{pi*(R + r)*s}}} = {{{(22/7)*(7 + 3.5)*14.43}}} = 476.19 square centimeters.

       (regarding this general formula see the link <A HREF=http://mathworld.wolfram.com/ConicalFrustum.html>http://mathworld.wolfram.com/ConicalFrustum.html</A> )



3.  The surface area of the semi-sphere is  {{{A[2]}}} = {{{2*pi*R^2}}} = {{{2*(22/7)*7^2}}} = 308 square centimeters.


4.  The area of the base is  {{{A[3]}}} = {{{pi*r^2}}} = {{{(22/7)*3.5^2}}} = 38.5 square centimeters.



5.  The answer to the problem question is the sum  


    {{{A[1]}}} + {{{A[2]}}} + {{{A[3]}}} = 476.19 + 308 + 38.5 = 822.69 square centimeters.
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Solved.