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From a solid cylinder whose height is $2.8$ cm and radius $2.1$ cm, a conical cavity of the same height and same radius is hollowed out. Find the volume and the total surface area of the remaining solid.
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Height of cylinder ($h$) = $2.8$ cm
Radius of cylinder ($r$) = $2.1$ cm
Volume of remaining solid = $\frac{22}{7} \times 2.1 \times 2.1 \times 2.8 - \frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.8$ (1 Mark)
$= 25.872$ cm$^3$ ($\frac{1}{2}$ Mark)
Slant height ($l$) = $\sqrt{(2.1)^2 + (2.8)^2} = 3.5$ cm ($\frac{1}{2}$ Mark)
Total surface area of the remaining solid
$= 2 \times \frac{22}{7} \times 2.1 \times 2.8 + \frac{22}{7} \times 2.1 \times 3.5 + \frac{22}{7} \times 2.1 \times 2.1$ (1 Mark)
$= 73.92$ cm$^2$ ($\frac{1}{2}$ Mark)
Radius of cylinder ($r$) = $2.1$ cm
Volume of remaining solid = $\frac{22}{7} \times 2.1 \times 2.1 \times 2.8 - \frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.8$ (1 Mark)
$= 25.872$ cm$^3$ ($\frac{1}{2}$ Mark)
Slant height ($l$) = $\sqrt{(2.1)^2 + (2.8)^2} = 3.5$ cm ($\frac{1}{2}$ Mark)
Total surface area of the remaining solid
$= 2 \times \frac{22}{7} \times 2.1 \times 2.8 + \frac{22}{7} \times 2.1 \times 3.5 + \frac{22}{7} \times 2.1 \times 2.1$ (1 Mark)
$= 73.92$ cm$^2$ ($\frac{1}{2}$ Mark)