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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.

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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 follows:

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, the required area of solids is: 10.56 + 5.50 + 1.54 = 17.60 cm².

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

Posted by

Devendra Khairwa

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