Question

Find the total surface area of a hollow cylinder of internal radius $3cm$, thickness $1cm$ and height $14cm$.
A) $330c{m^2}$
B) $660c{m^2}$
C) $990c{m^2}$
D) $1320c{m^2}$

Answer 1

Hint: Here it is said that the given cylinder is a hollow one. Since the thickness and internal radius are given we can
calculate the external radius by adding them. Then we can calculate the total surface area using internal radius, external
radius and thickness. By assuming the fractional approximation of $\pi $ we get the final answer.

Formula used:
For a cylinder with internal radius $r$, external radius $R$ and height $h$,
Total surface area, $TSA = 2\pi (R + h)(h + R - r)$

Complete step-by-step solution:
Given a hollow cylinder of internal radius $3cm$, thickness $1cm$ and height $14cm$.
Therefore we can see that the external radius will be the sum of internal radius and thickness.
So external radius of the cylinder, $R = 3 + 1 = 4cm$
And also $r = 3cm,h = 14cm$
For a cylinder with internal radius $r$, external radius $R$ and height $h$,
Total surface area, $TSA = 2\pi (R + r)(h + R - r)$
Substituting we get,
$TSA = 2\pi (4 + 3)(14 + 4 - 3)$
Simplifying we get,
$TSA = 2\pi \times 7 \times 15$
We can assume the value of $\pi $ as $\dfrac{{22}}{7}$.
So we have,
$TSA = 2 \times \dfrac{{22}}{7} \times 7 \times 15$
Simplifying we get,
$TSA = 2 \times 22 \times 15 = 660$
So, the total surface area of the hollow cylinder is $660c{m^2}$

$\therefore $ The answer is option B.

Additional information:
For a cylinder with radius $r$ and height $h$,
Curved surface area, $CSA = 2\pi rh$
Total surface area, $TSA = {\text{CSA + 2}} \times {\text{base area}}$
Area of the circle, $A = \pi {r^2}$
$ \Rightarrow TSA = {\text{CSA + 2}}\pi {r^2}$

Note:
Here the cylinder is hollow is important. If it is not, we can use the general formula for total surface area. For hollow
cylinders, there is a surface inside the cylinder too. So we have to consider that in calculating the total surface area.
That’s why the general formula does not work here.