# 8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

Firstly we need to calculate the slant height of cone :

$l^2\ =\ r^2\ +\ h^2$

or                                                         $=\ (0.7)^2\ +\ (2.4)^2$

or                                                    $l^2\ =\ 6.25$

or                                                   $l\ =\ 2.5\ cm$

Now, the total surface area of solid can be calculated as :

The surface area of solid   =   Surface area of cylinder  +   Surface area of cone   +    Area of base of the cylinder

The surface area of the cylinder is  $=\ 2 \pi rh$

or                                          $=\ 2 \pi \times 0.7\times 2.4$

or                                         $=\ 10.56\ cm^2$

Now, the surface area of a cone  $=\ \pi rl$

or                                         $=\ \pi \times 0.7\times 2.5$

or                                         $=\ 5.50\ cm^2$

And the area of the base of the cylinder is   $=\ \pi r^2$

or                                                      $=\ \pi \times 0.7\times 0.7$

or                                                      $=\ 1.54\ cm^2$

Thus required area of solid  =  10. 56  +  5.50  +  1.54   =  17.60  cm2.

Thus total surface area of remaining solid to nearest cm2 is 18 cm2.

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