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Grade 10 Mathematics Teachers’ Book

INNODEMS

Subsection 2.8.9 Volume in Real Life

Volume tells us how much space an object or container holds. In everyday life we use volume to know how much water fits in a jerrycan, how much grain a sack will carry, or how much fuel a drum can store. Knowing how to calculate volume helps families and communities measure and plan for storage, transport and household needs, and it is useful in farming, construction and cooking.
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In this section we practice simple formulas for common shapes (box, cylinder, cone) and learn to convert between litres and cubic metres so the results are useful in ordinary situations.
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Learner Experience 2.8.9.

  • Think about different containers you might find in school or at home, and that you might want to carry water or store flour in.
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  • For example, this could include jars, barrels, boxes, ...
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  • For each, determine what kind of shape it is and estimate its dimensions. Try to compute its volume.
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  • Which of these containers can store the largest amount of water or flour?
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Key Takeaway 2.8.55. Volume in Everyday Life.

  • Volume measures how much a container holds; use formulas for boxes and cylinders for common containers.
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  • Always use consistent units: convert centimetres to metres before computing volume, then convert cubic metres to litres by multiplying by 1000.
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  • Estimating volumes helps with storage, transport, planning for water and grain, and with simple cost calculations in rural and urban settings alike.
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Example 2.8.56.

A metal jerrycan is roughly a rectangular box of dimensions 35 cm by 20 cm by 30 cm. Estimate how many litres of water it can hold.
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Solution.

Compute the Volume.

Volume of a box is \(V = lwh\text{.}\) Convert centimetres to metres: \(l=0.35\,m,\; w=0.20\,m,\; h=0.30\,m.\) Thus \(V = 0.35\times0.20\times0.30 = 0.021\,\text{m}^3.\)
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Convert cubic metres to litres: \(1\,\text{m}^3 = 1000\,\text{L}\text{,}\) so \(V = 0.021\times1000 = 21\,\text{L}.\)
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Example 2.8.57.

A cylindrical water drum has radius 0.25 m and height 1.0 m. Find its volume in litres.
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Solution.
Volume of a cylinder: \(V=\pi r^2 h.\) So \(V=\pi\times(0.25)^2\times1.0\approx\pi\times0.0625\approx0.19635\,\text{m}^3.\) In litres: \(0.19635\times1000\approx196\,\text{L}.\)
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Example 2.8.58.

A small borehole is shaped like a cone with height 1.2 m and base radius 0.15 m. Estimate the volume of water it can hold in litres.
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Solution.
Volume of a cone: \(V=\tfrac{1}{3}\pi r^2 h\text{.}\) So \(V=\tfrac{1}{3}\pi(0.15)^2(1.2)\approx\tfrac{1}{3}\pi\times0.027\approx0.0283\,\text{m}^3\text{,}\) which is about \(28\,\text{L}.\)
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Example 2.8.59.

The storage space under a roof forms a triangular prism. The triangular cross-section has base 4 m and height 1.2 m, and the attic runs 6 m along the house. Estimate the storage volume in cubic metres and litres.
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Triangular prism roof
Solution.
Area of triangle = \(\tfrac{1}{2}bh = 0.5\times4\times1.2 = 2.4\,\text{m}^2\text{.}\) Volume = area × length = \(2.4\times6 = 14.4\,\text{m}^3\text{,}\) which is \(14{,}400\,\text{L}.\)
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Example 2.8.60. Composite Solid: Cylinder with Cone Top.

A container is made by placing a cone (height 0.5 m, base radius 0.3 m) on top of a cylinder (height 1.0 m, same radius 0.3 m). Find the total volume in litres.
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Solution.
Cylinder: \(V_c=\pi r^2 h=\pi\times0.3^2\times1.0=\pi\times0.09\approx0.283\,\text{m}^3.\)
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Cone: \(V_k=\tfrac{1}{3}\pi r^2 h=\tfrac{1}{3}\pi\times0.09\times0.5=0.015\pi\approx0.047\,\text{m}^3.\)
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Total: \(0.283+0.047\approx0.330\,\text{m}^3\approx330\,\text{L}.\)
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Example 2.8.61. Pond.

A shallow pond is shaped like a frustum of a square pyramid. The water surface is a square 12 m by 12 m, the bottom is a square 6 m by 6 m, and the depth is 1.5 m. Estimate the volume of water in the pond in cubic metres and litres.
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Pyramidal frustum pond
Solution.
For a pyramidal frustum the volume is \(V=\tfrac{h}{3}(A_1 + A_2 + \sqrt{A_1A_2})\) where \(A_1\) and \(A_2\) are the areas of the top and bottom squares. Here \(A_1=12^2=144\) and \(A_2=6^2=36\text{,}\) so \(\sqrt{A_1A_2}=\sqrt{144\times36}=72\text{.}\) Thus \(V=\tfrac{1.5}{3}(144+36+72)=0.5\times252=126\,\text{m}^3\text{,}\) which is about \(126{,}000\,\text{L}.\)
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Checkpoint 2.8.62.

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Checkpoint 2.8.63.

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Exercises Exercises

1.

A farmer fills a rectangular trough 2.0 m long, 0.5 m wide and 0.4 m deep with water. How many litres of water does the trough hold?
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Answer.
400 L
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Solution.
Volume: \(2.0\times0.5\times0.4=0.4\,\text{m}^3\text{.}\) In litres: \(0.4\times1000=400\,\text{L}.\)
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2.

A drum has radius 0.30 m and height 0.9 m. Estimate its capacity in litres. (Use \(\pi\approx3.14\text{.}\))
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Answer.
About 254 L
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Solution.
Volume: \(V=\pi r^2 h = 3.14\times0.3^2\times0.9 = 3.14\times0.09\times0.9 \approx 0.254\,\text{m}^3\text{.}\) In litres: \(0.254\times1000\approx254\,\text{L}.\)
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3.

A spherical water tank has diameter 1.2 m. How many litres can it hold? (Use \(\pi\approx3.14\text{.}\))
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Answer.
approximately \(904\,\text{L}.\)
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Solution.
Radius \(r=0.6\,\text{m}.\) Volume: \(V=\tfrac{4}{3}\pi r^3=\tfrac{4}{3}\times3.14\times0.6^3\approx\tfrac{4}{3}\times3.14\times0.216\approx0.904\,\text{m}^3\text{,}\) about \(904\,\text{L}.\)
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4.

A small store has a square pyramidal roof under which we want to store grain. The base of the pyramidal roof is 4 m by 4 m and its height 2.5 m. Estimate the volume of the the space under the roof in cubic metres.
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Answer.
Approximately 13.3 m^3 (≈ 13,333 L)
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Solution.
Volume of a pyramid: \(V=\tfrac{1}{3}Bh\) where \(B\) is base area. Here \(B=4\times4=16\,\text{m}^2\) and \(h=2.5\,\text{m}\text{,}\) so \(V=\tfrac{1}{3}\times16\times2.5=\tfrac{40}{3}\approx13.33\,\text{m}^3.\)
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5.

A conical sand pile has base radius 1.2 m and height 0.8 m. Estimate the volume of sand in cubic metres and litres. (Use \(\pi\approx3.14\text{.}\))
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Answer.
About 1.21 m^3 (≈ 1,210 L)
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Solution.
Volume of a cone: \(V=\tfrac{1}{3}\pi r^2 h\text{.}\) With \(r=1.2, h=0.8\) we get \(V=\tfrac{1}{3}\times3.14\times1.2^2\times0.8\approx1.206\,\text{m}^3\text{,}\) about \(1{,}206\,\text{L}.\)
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