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From a solid wooden cylinder of height $10\text{ cm}$ and radius $14\text{ cm}$, a cylinder of radius $7\text{ cm}$ and height $5\text{ cm}$ is scooped out to form a cavity inside the solid cylinder. Find the total surface area of the remaining solid.
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Total surface area of the remaining solid
$= \text{CSA of outer cylinder} + \text{CSA of inner cylinder} + \text{area of bases} + \text{Area of ring}$
$= 2\pi RH + 2\pi rh + (\pi R^2 + \pi r^2) + (\pi R^2 - \pi r^2)$
$= 2\pi RH + 2\pi rh + 2\pi R^2$
$= 2 \times \frac{22}{7} \times 14 \times 10 + 2 \times \frac{22}{7} \times 7 \times 5 + 2 \times \frac{22}{7} \times 14 \times 14$ [$1\frac{1}{2} + 1\frac{1}{2} + 1\frac{1}{2}$ marks]
$= 880 + 220 + 1232$
$= 2332\text{ sq. cm}$ [$\frac{1}{2}$ mark]
$= \text{CSA of outer cylinder} + \text{CSA of inner cylinder} + \text{area of bases} + \text{Area of ring}$
$= 2\pi RH + 2\pi rh + (\pi R^2 + \pi r^2) + (\pi R^2 - \pi r^2)$
$= 2\pi RH + 2\pi rh + 2\pi R^2$
$= 2 \times \frac{22}{7} \times 14 \times 10 + 2 \times \frac{22}{7} \times 7 \times 5 + 2 \times \frac{22}{7} \times 14 \times 14$ [$1\frac{1}{2} + 1\frac{1}{2} + 1\frac{1}{2}$ marks]
$= 880 + 220 + 1232$
$= 2332\text{ sq. cm}$ [$\frac{1}{2}$ mark]