Skip to main content

Grade 10 Mathematics Teachers’ Book

INNODEMS

Subsection 2.7.3 (*) Surface Area of Cylinders

A cylinder is a type of prism with circular faces at either ends. Jars, water tanks, and pipes are examples of cylinders. The surface area of a cylinder can be found by calculating the area of the two circular faces and the area of the curved face, as we will explore.
🔗

Learner Experience 2.7.4.

\(\textbf{Work in groups}\)
🔗
\(\textbf{What you require:}\) A pair of scissor and a piece of paper.
🔗
  1. Make a paper made of a cylinder.
    🔗
    🔗
  2. Measure and record the the length of the model.
    🔗
    🔗
  3. Open the model as illustrated below.
    🔗
    🔗
  4. Fold one circular end into two equal parts as shown;
    🔗
    🔗
  5. Measure and recod the diameter.
    🔗
    🔗
  6. Calculate the circumference of the circular end.
    🔗
    🔗
  7. Rotate the width of the rectangular part with the diameter of the circular part.
    🔗
    🔗
  8. Find the area of the rectangular part.
    🔗
    🔗
  9. Find the area of the circular ends.
    🔗
    🔗
  10. Find the surface area of the cylinder.
    🔗
    🔗
  11. Discuss and share your answer with other groups
    🔗
    🔗
🔗
🔗

Key Takeaway 2.7.20.

The total surface area of a cylinder of radius \(r\) and height \(h\text{,}\) is given by the sum of the areas of the two circular faces and the curved face. The circular faces each have area \(\pi r^2\text{.}\) The curved face has height \(h\) and width \(2\pi r\) (the circumference of the circular face). Thus,
🔗
\begin{align*} \textbf{Surface Area} \amp = 2\pi r^2+2\pi r h \\ \amp = 2\pi r(r+h) \end{align*}
🔗
Note: A cylinder is always considered closed, unless it is specified that it is open.
🔗

Example 2.7.21.

Calculate the surface area of the cylinder shown. Give your answer in terms of \(\pi\text{.}\)
🔗
Solution.
Given that, \(h=30\,\text{cm} \quad \text{and} \quad d=24\,\text{cm}\)
🔗
We first get the radius of the cylinder which is given by,
🔗
\begin{align*} r\amp =\frac{d}{2} \\ \amp = \frac{24}{2} \\ \amp = 12\,\text{cm} \end{align*}
🔗
Therefore,
🔗
\begin{align*} \text{Surface area}\amp = 2\pi r^2+2\pi r h\\ \amp = 2\pi r(r+h)\\ \amp = 2 \times \pi \times 12\,\text{cm}\,(12\,\text{cm}+30\,\text{cm} ) \\ \amp = 2 \times \pi \times 12\,\text{cm}\,\times 42\,\text{cm} \\ \amp = 1,008 \pi \,\text{cm}^2 \end{align*}
🔗
The surface area of the above cylinder is \(3\,168 \,\text{cm}^2\)
🔗
🔗
🔗

Example 2.7.22.

The figure below shows a cylinder with diameter 21 cm and height 91 cm. Calculate the surface area of the cylinder to two decimal places. (Use \(\pi=3.14\))
🔗
Cylindrical prism
Solution.
Given that, \(h=91\,\text{cm} \quad \text{and} \quad d=21\,\text{cm}\)
🔗
We first get the radius of the cylinder which is given by,
🔗
\begin{align*} r\amp =\frac{d}{2} \\ \amp = \frac{21}{2} \\ \amp = 10.5\,\text{cm} \end{align*}
🔗
Therefore,
🔗
\begin{align*} \text{Surface area}\amp = 2\pi r(r+h)\\ \amp = 2 \times \pi \times 10.5 \,\text{cm}(10.5\,\text{cm}+91\,\text{cm})\\ \amp = 2 \times 3.14 \times 10.5 \,\text{cm} \times 101.5 \,\text{cm}\\ \amp = 6692.91\,\text{cm}^2 \end{align*}
🔗
The surface area of the cylinder is \(6\,692.91\,\text{cm}^2\)
🔗
🔗
🔗

Open Cylinders.

An open cylinder, is a cylinder where the top is open. It means that you will only have one circle instead of two.
🔗
Therefore, to calculate the surface area of the cylinder you add the area of the curved surface and the circle.
🔗
An open cylinder has a curved surface and one circular face.
🔗
Therefore,
🔗
\begin{align*} \textbf{Surface area of an open cylinder } \amp = \pi r^2 +2 \pi rh \\ \amp = \pi r (r+2h) \end{align*}
🔗

Example 2.7.23.

An open cylindrical cotainer has a height of \(12\,\text{cm}\) and a diameter of \(2\,\text{cm}\text{.}\) What is the surface area of the outer surfaces of the container? (\(\, \pi = 3.142\))
🔗
Solution.
You are given the height and diameter of the container to be \(h=12\,\text{cm} \, \text{and} \, d=2\,\text{cm}\)
🔗
You first identify the radius of the container that is,
🔗
\begin{align*} r\amp = \frac{2}{2} \\ \amp = 1 \,\text{cm} \end{align*}
🔗
Therefore,
🔗
\begin{align*} \text{surface area}\amp = \pi r (r+2h)\\ \amp = 3.142 1(1+2 \times 12) \\ \amp = 3.142 1(1+24) \,\text{cm}\\ \amp = 3.142 \times 25 \,\text{cm}\\ \amp = 78.55\, \text{cm}^2 \end{align*}
🔗
Therefore, surface area of the outer container is \(= 78.55\,m^2\text{.}\)
🔗
🔗
🔗

Example 2.7.24.

Ekadeli filled an open cylindrical bucket with water. The internal diameter of the bucket was \(32.4 \,\text{cm}\) and the internal height was \(35 \,\text{cm}\text{.}\) Calculate the area of the bucket that was in contact with the water. Write the answer correct to \(\textbf{1 decimal place}\text{.}\) (\(\textbf{Use} \, \pi =\frac{22}{7}\))
🔗
Solution.
The open cylindrical bucket has a diameter and height of \(d=32.4\,\text{cm} \quad \text{and} \quad h=35\,\text{cm}\text{.}\)
🔗
You first calculate the radius of the cylindrical bucket which is,
🔗
\begin{align*} r\amp = \frac{d}{2} \\ \amp = \frac{32.4}{2}\\ \amp = 16.2 \,\text{cm} \end{align*}
🔗
Therefore,
🔗
\begin{align*} \text{Surface area} \amp = \pi r (r+2h) \\ \amp = \frac{22}{7} \times 16.2 \,\text{cm} \times (16.2\,\text{cm}+2\times 35\,\text{cm}) \\ \amp = \frac{22}{7} \times 16.2 \,\text{cm}\times (16.2\,\text{cm}+70\,\text{cm})\\ \amp = \frac{22}{7} \times 16.2 \,\text{cm}\times 86.2 \,\text{cm} \\ \amp = 4388.81142857\,\text{cm}^2 \end{align*}
🔗
The surface area \(= 4388.8\,\text{cm}^2\)
🔗
🔗
🔗
🔗

Exercises Exercises

1.

Calculate the surface area of unsharpened circular pencil in the shape of a cylinder whose radius is \(0.2\,\text{m}\) and height is height is \(1.4\,\text{m}\text{.}\) (\(\textbf{Use}\, \pi=3.142\))
🔗
Solution.
Given that, \(h=1.4\,\text{m}\) and \(r=0.2\,\text{m}\)
🔗
Therefore,
🔗
\begin{align*} \text{Surface area}\amp = 2\pi r(r+h)\\ \amp = 2 \times 3.142 \times 0.2 \,\text{m}(0.2\,\text{m}+1.4\,\text{m})\\ \amp = 2 \times 3.142 \times 0.2 \,\text{m} \times 1.6 \,\text{m}\\ \amp = 2.01088\,\text{m}^2 \end{align*}
🔗
Therefore, surface area \(= 2.01088\,\text{m}^2\)
🔗
🔗
🔗

2.

The surface area of an open jar is \(594\,\text{cm}^2\text{.}\) The radius of the jar is \(7\,\text{cm}\text{.}\) calculate the height of the jar. (\(\textbf{Use} \, \pi =3.142\)).
🔗
Solution.
You are given,
🔗
\(\textbf{Surface are}=594\,\text{cm}^2\)
🔗
\(r=7\,cm\)
🔗
\(h=\text{?}\)
🔗
To find the height of the jar we substitute the above values in the formula below.
🔗
\begin{align*} \text{Surface area}\amp = \pi r (r+2h) \\ 594\,\text{cm}^2\amp = \frac{22}{7} \times 7\,\text{cm} (7\,\text{cm} +2\times h)\\ \frac{594}{22}\amp = 7\,\text{cm} +2\times h\\ (27 - 7)\, \text{cm} \amp = 2 \times h \\ 20\, \text{cm} \amp = 2 \times h\\ \frac{20}{2}\, \text{cm}\amp = h\\ h\amp = 10 \,\text{cm} \end{align*}
🔗
The height of the jar \(= 10 \,\text{cm }\)
🔗
🔗
🔗

3.

The figure below shows the roof of a motorbike shade. The roof is painted on the outer curved surface and the two semi-circle faces. Calculate the surface area of the part of roof that is painted, to two decimal places. (Use \(\pi=3.14\)).
🔗
Solution.
Given that the diameter of the semicircle (top) is \(3.5\,\text{m}\) and the height (length) is \(7\,\text{m}\text{.}\)
🔗
The radius of the semicircular ends is
🔗
\begin{align*} r\amp=\frac{d}{2} \\ \amp=\frac{3.5}{2} \\ \amp=1.75\,\text{m} \end{align*}
🔗
The painted part is a half-cylinder: its curved surface is half the curved surface of a full cylinder and the two ends are two semicircles (which together make one full circle).
🔗
\begin{align*} \text{Painted area}\amp=\pi r h + \; 2\times\left(\tfrac{1}{2}\pi r^2\right) \\ \amp=\pi r h + \pi r^2\\ \amp=\pi r(r+h) \end{align*}
🔗
Substituting \(\pi=3.14\text{,}\) \(r=1.75\,\text{m}\) and \(h=7\,\text{m}\) gives
🔗
\begin{align*} \text{Painted area}\amp=3.14\times1.75\times(1.75+7)\\ \amp=3.14\times1.75\times8.75\\ \amp=48.08125\,\text{m}^2 \end{align*}
🔗
To two decimal places, the painted area is \(48.08\,\text{m}^2\text{.}\)
🔗
🔗
🔗

4.

A cylinder has radius \(r\) and its height is twice the radius. Express the surface area of the cylinder in terms of \(r\text{.}\)
🔗
Solution.
Using the formula for the surface area of a closed cylinder,
🔗
\begin{align*} \text{Surface area} \amp= 2\pi r(r+h) \end{align*}
🔗
Substitute \(h=2r\text{:}\)
🔗
\begin{align*} \text{Surface area} \amp= 2\pi r(r+2r)\\ \amp= 2\pi r(3r)\\ \amp= 6\pi r^2 \end{align*}
🔗
🔗
🔗

5.

The surface area of a closed cylinder is given as \(18\pi\,\text{cm}^2\text{.}\) If the radius of the cylinder is \(2\,\text{cm}\text{,}\) find the height \(h\) of the cylinder.
🔗
Solution.
Use the surface area formula for a closed cylinder: \(\text{SA}=2\pi r(r+h)\text{.}\)
🔗
Substitute \(\text{SA}=18\pi\) and \(r=2\text{:}\)
🔗
\begin{align*} 18\pi\amp=2\pi\times 2\,(2+h)\\ 18\pi\amp=4\pi(2+h) \end{align*}
🔗
Divide both sides by \(\pi\) and simplify:
🔗
\begin{align*} 18\amp=4(2+h)\\ 18\amp=8+4h\\ 18-8\amp=4h\\ 10\amp=4h\\ h\amp=\tfrac{10}{4}=2.5 \end{align*}
🔗
Therefore, the height of the cylinder is \(h=2.5\,\text{cm}\text{.}\)
🔗
🔗
🔗

6.

Calculate the surface area of a closed cylinder with radius \(5\,\text{cm}\) and height \(10\,\text{cm}\text{.}\) (Use \(\pi=3.14\text{.}\))
🔗
Solution.
\begin{align*} \text{SA}=2\pi r(r+h)\\ \amp=2\times 3.14 \times 5\,(5+10)\\ \amp=2\times 3.14 \times 5 \times 15\\ \amp=471\,\text{cm}^2 \end{align*}
🔗
🔗
🔗

7.

An open cylinder (one base) has radius \(4\,\text{cm}\) and height \(8\,\text{cm}\text{.}\) Find the surface area of the outer surfaces. (Use \(\pi=3.14\text{.}\))
🔗
Solution.
For an open cylinder, \(\text{SA}=\pi r^2 + 2\pi r h\text{.}\)
🔗
\begin{align*} \text{SA}=3.14\times 4^2 + 2\times 3.14\times 4\times 8\\ \amp=3.14\times 16 + 2\times 3.14\times 32\\ \amp=50.24 + 200.96\\ \amp=251.2\,\text{cm}^2 \end{align*}
🔗
🔗
🔗

8.

The surface area of a closed cylinder is \(100\pi\) and its height is \(5\text{.}\) Find the radius \(r\text{.}\)
🔗
Solution.
Use \(\text{SA}=2\pi r(r+h)\text{.}\) Divide both sides by \(\pi\text{:}\)
🔗
\begin{gather*} 100 = 2r(r+5)\\ 100 = 2r^2 + 10r\\ 2r^2 +10r -100 =0\\ r^2 +5r -50 =0 \end{gather*}
🔗
The quadratic factors as \((r+10)(r-5)=0\text{.}\) Hence \(r=5\) (rejecting the negative root).
🔗
🔗
🔗

9.

A cylinder has diameter \(14\,\text{cm}\) and height \(7\,\text{cm}\text{.}\) Find its surface area in terms of \(\pi\text{.}\)
🔗
Solution.
The radius is \(r=\tfrac{14}{2}=7\,\text{cm}\text{.}\) Using \(\text{SA}=2\pi r(r+h)\text{:}\)
🔗
\begin{gather*} \text{SA}=2\pi \times 7(7+7)=2\pi\times 7\times 14=196\pi\,\text{cm}^2 \end{gather*}
🔗
🔗
🔗
🔗
🔗
PrevTopNext