On the day of her examination, Riya sharpened her pencil from both ends as shown below : The diameter of the cylindrical and conical part of the pencil is $\displaystyle 4.2$ mm. If the height of each conical part is $\displaystyle 2.8$ mm and length of entire pencil is $\displaystyle 105.6$ mm , find the total surface area of the pencil.

\(\displaystyle \mathrm{d}=4.2 \mathrm{~mm}\) \(\displaystyle \mathrm{r}=2.1 \mathrm{~mm}\) Height of cylindrical part \(\displaystyle (\mathrm{h})=100 \mathrm{~mm}\) Height of conical part = 2.8mm Slant height \(\displaystyle (l)=\sqrt{(2.8)^{2}+(2.1)^{2}}=3.5 \mathrm{~mm}\) Curved Surface Area of Cylinder \(\displaystyle =\mathbf{2} \times \frac{\mathbf{2 2}}{\mathbf{7}} \times 2.1 \times 100=1320 \mathrm{~mm}^{2}\) Curved Surface Area of Cone \(\displaystyle =\frac{22}{7} \times 2.1 \times 3.5=23.1 \mathrm{~mm}^{2}\) Total Surface Area of pencil = Curved Surface Area of cylinder + Curved Surface Area of two cones \[=1366.2 \mathrm{~mm}^{2} \]