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Grade 10 Mathematics Learners’ Book

INNODEMS

Subsection 2.7.13 Volume of a Cylinder

Activity 2.7.17.

\(\textbf{Work in groups}\)
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\(\textbf{What you require:}\) A pair of scissor and a piece of paper.
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  1. Make a paper made of a cylinder.
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  2. Measure and record the the length of the model.
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  3. Open the model as illustrated below.
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  4. Fold one circular end into two equal parts as shown;
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  5. Measure and recod the diameter.
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  6. Calculate the circumference of the circular end.
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  7. Rotate the width of the rectangular part with the diameter of the circular part.
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  8. Find the area of the rectangular part.
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  9. Find the area of the circular ends.
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  10. Find the surface area of the cylinder.
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  11. Discuss and share your answer with other groups
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\(\textbf{Key point}\)
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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 curves face. Thus,
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\begin{align*} \textbf{Total surface area of a cylinder}=\amp 2\pi r^2+2\pi r h \\ = \amp 2\pi r(r+h) \end{align*}
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\(\textbf{Note:}\) A cylinder is always considered closed, unless it is specified that it is open.
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Example 2.7.44.

Calculate the surface area of the cylinder shown.
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Solution.
Given that, \(h=30\,\text{cm} \quad \text{and} \quad d=24\,\text{cm}\)
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We first get the radius of the cylinder which is given by,
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\begin{align*} r=\amp\frac{d}{2} \\ =\amp \frac{24}{2} \\ =\amp 12\,\text{cm} \end{align*}
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Therefore,
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\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 2 \times 22 \times 12 \times 6\\ =\amp 3\,168 \,\text{cm}^2 \end{align*}
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The surface area of the above cylinder is \(3\,168 \,\text{cm}^2\)
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Example 2.7.45.

The figure below shows a cylinder. Calculate the surface area of the cylinder to two decimal places.(\(\textbf{Use}\, \pi=\frac{22}{7}\))
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Cylindrical prism
Solution.
Given that, \(h=91\,\text{cm} \quad \text{and} \quad d=21\,\text{cm}\)
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We first get the radius of the cylinder which is given by,
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\begin{align*} r=\amp\frac{d}{2} \\ =\amp \frac{21}{2} \\ =\amp 10.5\,\text{cm} \end{align*}
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Therefore,
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\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 \pi \times 10.5 \,\text{cm} \times 101.5 \,\text{cm}\\ =\amp 6\,699.00\,\text{cm}^2 \end{align*}
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The surface area of the cylinder\(=6\,699.00\,\text{cm}^2\)
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Example 2.7.46.

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 \(1.4\,\text{m}\text{.}\) (\(\textbf{Use}\, \pi=3.142\))
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Solution.
Given that, \(h=01.4\,\text{m} \quad \text{and} \quad r=0.2\,\text{m}\)
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Therefore,
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\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*}
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Therefore, surface area \(= 2.01088\,\text{m}^2\)
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Activity 2.7.18.

\(\textbf{Work in groups}\)
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\(\textbf{What you require:}\) A cylindrical container without a lid, a pair of scissores and a piece of paper.
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  1. Trace the bottom face of the cylindrical container on a piece of paper and cut out the shape.
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    1. Entirely cover the curved surfaceof the container with a piece of paper.
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    2. Cut off any parts of the piece of paper that extend beyond the curved surface. Also, ensure the piece of paper does not overlap.
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  3. Calculate the area of each of the two cutouts.
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  4. Work out the total area of all the cutouts. What does the area represent?
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  5. Discuss how to calculate the surface area of an open cylinder and share your findings with other groups in your class.
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\(\textbf{Key Takeaway}\)
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An open cylinder, is a cylinder where the top is open. It means that you will only have one circle instade of two.
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Therefore, to calculate the surface area of the cylinder you add the area of the curved surface and the circle.
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An open cylinder has a curved surface and one circular face.
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Therefore,
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\begin{align*} \textbf{Surface area of an open cylinder } = \amp \pi r^2 +2 \pi rh \\ = \amp \pi r (r+2h) \end{align*}
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Example 2.7.47.

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? (\(\, \pi = 3.142\))
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Solution.
You are given the height and diameter of the container to be \(h=12\,\text{cm} \, \text{and} \, d=2\,\text{cm}\)
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You first identify the radius of the container that is,
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\begin{align*} r=\amp \frac{2}{2} \\ =\amp 1 \,\text{cm} \end{align*}
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Therefore,
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\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*}
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Therefore, surface area of the outer container is \(= 78.55\,\text{m^}2\text{.}\)
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Example 2.7.48.

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 \(\textbf{1 decimal place}\text{.}\) (\(\textbf{Use} \, \pi =\frac{22}{7}\))
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Solution.
The open cylindrical bucket has a diameter and height of \(d=32.4\,\text{cm} \quad \text{and} \quad h=35\,\text{cm}\text{.}\)
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You first calculate the radius of the cylindrical bucket which is,
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\begin{align*} r=\amp \frac{d}{2} \\ = \amp \frac{32.4}{2}\\ =\amp 16.2 \,\text{cm} \end{align*}
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Therefore,
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\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*}
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The surface area \(= 4388.8\,\text{cm}^2\)
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Example 2.7.49.

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. (\(\textbf{Use} \, \pi =3.142\)).
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Solution.
You are given,
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\(\textbf{Surface are}=594\,\text{cm}^2\)
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\(r=7\,cm\)
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\(h=\text{?}\)
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To find the height of the jar we substitute the above values in the formula below.
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\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*}
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The height of the jar \(= 10 \,\text{cm }\)
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\(\textbf{Extended Exercise}\)
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The figure alongside 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.
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Example 2.7.50.

The cylinder below has a radius of 4 cm and a height of 10 cm. Use \(\pi = 3.142\)
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Cylindrical prism
\begin{align*} V =\amp \text{Base Area }\times h \\ V = \amp \pi r^2 \times h\\ = \amp 3.142 \times (4)^2 \, \text{cm} \times 10\, \text{cm}\\ = \amp 502.65\, \text{cm}^3 \end{align*}
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Example 2.7.51.

A pipe has a radius of 3 cm and a height of 15 cm. Find the volume of the pipe.
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Solution.
\begin{align*} V =\amp \text{Base Area} \times h \\ V = \amp \pi r^2 \times h\\ = \amp 3.142 \times (3)^2 \, \text{cm} \times 15\, \text{cm}\\ = \amp 424.12\, \text{cm}^3 \end{align*}
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Example 2.7.52.

Mueni’s water tank has a radius of 7 m and a height of 20 m. Find the capacity of the water tank in litres.
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Solution.
\begin{align*} V =\amp \text{Base Area} \times h \\ V = \amp \pi r^2 \times h\\ = \amp \frac{22}{7} \times (7)^2\, \text{m} \times 20 \, \text{m}\\ = \amp 3080\, \text{m}^3 \end{align*}
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To convert from \(m^3 \) to litres we multiply by \(1000 \, \text{m}^3\)
\begin{align*} 1 \text{litre} = \amp 1000\, \text{m}^3 \\ = \amp 3080\, \text{m}^3 \times 1000\, \text{m}^3\\ = \amp 3,080,000 \, \text{ litres} \end{align*}
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\(\textbf{Exercise}\)
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1. A cylinder has a volume of 1,570 cm³ and a diameter of 30 cm. Find its height.Use (\(\pi = 3.14\))
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2. Alongside is a cross-sectional view of an open cylinder that has a radius of 7 cm and a height of 10 cm. Find its volume.
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3. A hollow cylindrical pipe has an outer radius of 8 cm, an inner radius of 6 cm, and a length of 100 cm. Find the volume of material used to make the pipe.
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4. A cylindrical water tank has a radius of 2 m and a height of 5 m. How many liters of water can it hold when full?
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5. If the radius of a cylinder is doubled while keeping the height the same, by what factor does the volume increase?
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Checkpoint 2.7.53.

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Checkpoint 2.7.54.

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