Rasheed got a playing top as …

Surface area to colour = surface area of hemisphere + curved surface area of cone

Diameter of hemisphere = 3.5 cm

So radius of hemispherical portion of the lattu = r =  {tex}\frac { 3.5 } { 2 } \mathrm { cm }{/tex} = 1.75 
r = Radius of the concial portion = {tex}\frac{3.5}2{/tex} = 1.75 
 Height of the conical  portion = height of top - radius of hemisphere = {tex}{/tex} 5 - 1.75  = 3.25 cm
Let I be the slant height of the conical part. Then,
{tex}l^2=h^2+r^2{/tex}

{tex}\begin{array}{l}l^2=(3.25)^2+(1.75)^2\\\Rightarrow l^2\;=10.5625+3.0625\\\Rightarrow l^2=13.625\\\Rightarrow l=\sqrt{13.625}\\\Rightarrow l=3.69\end{array}{/tex}
Let S be the total surface area of the top. Then,
{tex}S = 2 \pi r ^ { 2 } + \pi r l{/tex}
{tex}\Rightarrow \quad S = \pi r ( 2 r + l ){/tex}
{tex}\begin{array}{l}\Rightarrow S=\frac{22}7\times1.75(2\times1.75+3.7)\\\;\;\;\;=\;5.5(3.5+3.7)\\=5.5(7.2)\\=39.6\;cm^2\end{array}{/tex}