Volume Formula Fundamentals: Extended Worked Examples

Position in the unit

This record is for Grade 8 Geometry and Measurement and is meant to be used alongside, not instead of, the main concept records:

• rea.m08.geometry-measurement.volume.basic-formulas.lesson for the core procedures and meaning of the formulas.
• rea.m08.geometry-measurement.volume.basic-formulas.overview for the unit-level structure.
• rea.m08.geometry-measurement.volume.cylinder-derivation for why the cylinder formula works.
• rea.m08.geometry-measurement.surface-area.overview if you are mixing up volume with surface area.

Here, the goal is to see many complete examples and the choices behind each step.

Working rules to keep in mind

For every problem in this record:

1. Identify the solid.
2. Choose the correct formula.
3. Check that all measurements are in compatible units.
4. Substitute carefully.
5. Write the final answer in cubic units.
6. Ask whether the answer is reasonable.

Useful formulas used here:

• Right rectangular prism: V = lwh
• Any right prism: V = Bh where B is the area of the base
• Cylinder: V = pi r^2 h

Worked Example 1: Right rectangular prism with whole numbers

Problem. A storage box is 8 cm long, 5 cm wide, and 3 cm high. Find its volume.

Reasoning. This is a right rectangular prism, so all three dimensions multiply directly. This is the most basic use of volume: count layers of equal-sized rectangular arrays of cubes.

Solution. V = lwh

V = 8 x 5 x 3

V = 40 x 3

V = 120

So the volume is 120 cm^3.

Commentary. Writing cm^3 matters because volume measures space in three dimensions, not just length or area.

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Worked Example 2: Rectangular prism with decimals

Problem. A fish tank measures 12.5 cm by 4 cm by 7.2 cm. Find the volume.

Reasoning. The shape is still a right rectangular prism. Decimals do not change the formula; they only change the arithmetic.

Solution. V = lwh

V = 12.5 x 4 x 7.2

First multiply the easier pair:

12.5 x 4 = 50

Then:

50 x 7.2 = 360

So the volume is 360 cm^3.

Commentary. Choosing 12.5 x 4 first is strategic. It creates a clean number and lowers the chance of arithmetic error.

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Worked Example 3: Missing dimension in a rectangular prism

Problem. A rectangular prism has volume 180 cm^3, length 9 cm, and width 4 cm. Find the height.

Reasoning. This is a reverse problem. Instead of finding volume from dimensions, we find a missing dimension from the volume. Start with the same formula and solve for the unknown.

Solution. V = lwh

Substitute what is known:

180 = 9 x 4 x h

180 = 36h

Divide both sides by 36:

h = 180 / 36

h = 5

So the height is 5 cm.

Commentary. The unit of the missing dimension is linear, so the answer is cm, not cm^3.

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Worked Example 4: Finding a missing side from base area times height

Problem. A rectangular prism has volume 294 m^3, width 6 m, and height 7 m. Find its length.

Reasoning. This is still a rectangular prism, but it helps to think of the formula as V = Bh, where the base is a rectangle. Then B = length x width.

Solution. V = lwh

294 = l x 6 x 7

294 = 42l

l = 294 / 42

l = 7

So the length is 7 m.

Commentary. Either V = lwh or V = Bh works. Switching forms can make the structure clearer without changing the mathematics.

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Worked Example 5: Cylinder using radius directly

Problem. A can is shaped like a cylinder with radius 3 cm and height 10 cm. Find its volume in exact form and as a decimal approximation.

Reasoning. Cylinders use base area times height, but the base is a circle. The circular base area is pi r^2, so the volume formula is pi r^2 h.

Solution. V = pi r^2 h

V = pi x 3^2 x 10

V = pi x 9 x 10

V = 90pi

Exact volume: 90pi cm^3

Approximate value:

90pi ≈ 90 x 3.14 = 282.6

So the approximate volume is 282.6 cm^3.

Commentary. Keeping 90pi until the end preserves accuracy. Rounding too early can create avoidable error.

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Worked Example 6: Cylinder when the diameter is given

Problem. A cylindrical candle has diameter 14 cm and height 8 cm. Find its volume.

Reasoning. The formula needs the radius, not the diameter. So the first choice is to convert correctly before substituting.

Solution. Diameter = 14 cm, so radius r = 7 cm.

Now use the formula:

V = pi r^2 h

V = pi x 7^2 x 8

V = pi x 49 x 8

V = 392pi

Approximate value:

392pi ≈ 392 x 3.14 = 1230.88

So the volume is 392pi cm^3, or about 1230.88 cm^3.

Commentary. A common error is to substitute 14 for r. That would square the diameter by mistake and make the answer far too large.

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Worked Example 7: Unit conversion before using the formula

Problem. A rectangular container is 40 cm long, 25 cm wide, and 0.5 m high. Find its volume in cubic centimetres.

Reasoning. The dimensions are not all in the same unit, so substitution must wait until the units match.

Solution. Convert the height:

0.5 m = 50 cm

Now all dimensions are in centimetres.

V = lwh

V = 40 x 25 x 50

40 x 25 = 1000

1000 x 50 = 50000

So the volume is 50000 cm^3.

Commentary. Do not multiply mixed units such as cm x cm x m and expect a meaningful final unit. Convert first, then calculate.

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Worked Example 8: Comparing two solids by volume

Problem. Which holds more: a rectangular prism measuring 6 cm by 5 cm by 8 cm, or a cylinder with radius 3 cm and height 7 cm?

Reasoning. This is a comparison problem. Find each volume separately, then compare. The important choice is to avoid comparing formulas directly without numbers.

Solution. Rectangular prism:

V = lwh

V = 6 x 5 x 8 = 240

So its volume is 240 cm^3.

Cylinder:

V = pi r^2 h

V = pi x 3^2 x 7

V = pi x 9 x 7

V = 63pi

63pi ≈ 63 x 3.14 = 197.82

So its volume is about 197.82 cm^3.

Comparison:

240 > 197.82

So the rectangular prism holds more.

Difference:

240 - 197.82 = 42.18

It holds about 42.18 cm^3 more.

Commentary. When the question asks “which holds more,” a numerical comparison is required. Computing both volumes completely avoids guesswork.

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Worked Example 9: Multi-step cylinder problem from volume and height

Problem. A cylinder has volume 154pi cm^3 and height 14 cm. Find its radius.

Reasoning. This is another reverse problem. Because the volume is written with pi, the algebra stays neat.

Solution. Start with:

V = pi r^2 h

Substitute:

154pi = pi r^2 x 14

154pi = 14pi r^2

Divide both sides by 14pi:

r^2 = 154pi / 14pi

r^2 = 11

r = sqrt(11)

So the radius is sqrt(11) cm, which is about 3.32 cm.

Commentary. This example shows that not every missing-dimension problem gives an integer answer. The exact form sqrt(11) is mathematically correct, and a decimal approximation can be added if useful.

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Worked Example 10: Challenge problem with a hollow space removed

Problem. A solid rectangular block measures 12 cm by 10 cm by 6 cm. A square hole measuring 2 cm by 2 cm passes straight through the full height of the block. Find the remaining volume.

Reasoning. This is a composite-volume problem. Find the full volume first, then subtract the volume of the missing prism. The “through the full height” phrase tells us the hole has the same height as the block.

Solution. Volume of whole block:

V = 12 x 10 x 6 = 720 cm^3

Volume of removed hole:

The hole is a rectangular prism with dimensions 2 cm x 2 cm x 6 cm.

V_hole = 2 x 2 x 6 = 24 cm^3

Remaining volume:

720 - 24 = 696

So the remaining volume is 696 cm^3.

Commentary. In multi-step problems, drawing or imagining the solid helps decide whether to add or subtract volumes.

Common misconceptions and repairs

• Using square units for volume. Repair: volume always uses cubic units such as cm^3, m^3, or in^3.
• Using diameter as radius in cylinder problems. Repair: if diameter is given, divide by 2 before using r^2.
• Forgetting to match units. Repair: convert first, calculate second.
• Mixing surface area and volume formulas. Repair: ask whether the problem is about covering the outside or filling the inside; see rea.m08.geometry-measurement.surface-area.overview if needed.
• Rounding too early with pi. Repair: keep answers in terms of pi until the last step unless the question requires a decimal.

Quick practice

Try these after studying the worked examples.

1. Find the volume of a rectangular prism with dimensions 11 cm, 4 cm, and 9 cm.
2. Find the volume of a cylinder with radius 5 cm and height 12 cm.
3. A rectangular prism has volume 256 cm^3, length 8 cm, and width 4 cm. Find the height.
4. A cylinder has diameter 10 m and height 3 m. Find its volume in exact form.

Answers

1. 396 cm^3
2. 300pi cm^3 or about 942 cm^3
3. 8 cm
4. 75pi m^3

What this record develops

This example set builds from direct substitution to reverse problems, unit conversion, comparison, and simple composite volume. After mastering these, a learner should be ready to move between formula use and explanation, not just perform isolated calculations.