Link to Prism Volume: Extended Worked Examples

Grade 8 focus

This record is for Grade 8 Geometry and Measurement. It does not re-derive the cylinder formula from scratch; for that conceptual development, see [rea.m08.geometry-measurement.volume.cylinder-derivation.prism-link-lesson].

It also assumes some fluency with volume structure from:

• [rea.m08.geometry-measurement.volume.basic-formulas.worked-examples]
• [rea.m08.geometry-measurement.volume.basic-formulas.prism-bh.worked-examples]

The point here is narrower: treat a cylinder as a base-area-times-height solid, just like a prism.

Core idea used in every example

For a prism, volume is:

• V = B x h

For a cylinder, the same structure works:

• V = B x h, where the base is a circle
• Since B = pi r^2, this becomes V = pi r^2 h

The key choice is always:

1. Identify the base area.
2. Identify the perpendicular height.
3. Multiply B x h.

Worked Example 1: Direct cylinder volume from radius and height

Problem. A cylinder has radius 4 cm and height 9 cm. Find its volume.

Solution. Use the prism-style structure V = B x h.

The base is a circle, so:

• B = pi r^2 = pi(4)^2 = 16pi cm^2

Now multiply by the height:

• V = B x h = 16pi x 9 = 144pi cm^3

Approximate value:

• 144pi ≈ 452.4

Answer: 144pi cm^3 or about 452.4 cm^3

Commentary. The important move is not memorizing a separate rule first. It is noticing that the cylinder has the same cross-sectional area all the way up, so the prism rule applies unchanged.

Worked Example 2: Start from base area, not radius

Problem. The circular base of a cylinder has area 50pi cm^2. The height is 12 cm. Find the volume.

Solution. Because volume is B x h, use the base area directly.

• B = 50pi cm^2
• h = 12 cm

So:

• V = B x h = 50pi x 12 = 600pi cm^3

Approximate value:

• 600pi ≈ 1885.0

Answer: 600pi cm^3

Commentary. This is one of the clearest ways to see the prism link. Once the base area is known, it does not matter whether the base is rectangular, triangular, or circular. The structure is still V = B x h.

Worked Example 3: Cylinder and prism with the same base area and height

Problem. A cylinder has base area 36pi cm^2 and height 7 cm. A prism also has base area 36pi cm^2 and height 7 cm. Compare their volumes.

Solution. For the cylinder:

• V = B x h = 36pi x 7 = 252pi cm^3

For the prism:

• V = B x h = 36pi x 7 = 252pi cm^3

So both solids have the same volume.

Answer: The volumes are equal; each is 252pi cm^3.

Commentary. This example is the central idea of the topic. The shape of the base can differ, but if the base area and the perpendicular height are the same, the volume is the same.

Worked Example 4: Diameter given instead of radius

Problem. A cylindrical can has diameter 10 cm and height 15 cm. Find its volume.

Solution. The formula uses radius, not diameter.

• Diameter = 10 cm
• Radius = 10 / 2 = 5 cm

Find the base area:

• B = pi r^2 = pi(5)^2 = 25pi cm^2

Multiply by height:

• V = 25pi x 15 = 375pi cm^3

Approximate value:

• 375pi ≈ 1178.1

Answer: 375pi cm^3 or about 1178.1 cm^3

Commentary. A common error is to square the diameter by accident. The repair is simple: pause before substituting and check whether the formula needs r or d.

Worked Example 5: Find a missing height from the volume

Problem. A cylinder has volume 320pi cm^3 and radius 8 cm. Find its height.

Solution. Use V = B x h.

First find the base area:

• B = pi r^2 = pi(8)^2 = 64pi cm^2

Now solve for height:

• 320pi = 64pi x h
• h = 320pi / 64pi
• h = 5

Answer: 5 cm

Commentary. Writing the formula as V = B x h helps with reverse problems. Once the circular base area is found, the problem becomes a one-step prism equation: h = V / B.

Worked Example 6: Find a missing radius from the volume

Problem. A cylinder has volume 196pi m^3 and height 4 m. Find its radius.

Solution. Start with the prism structure:

• V = B x h

So the base area is:

• B = V / h = 196pi / 4 = 49pi m^2

But for a circular base:

• B = pi r^2

So:

• pi r^2 = 49pi
• r^2 = 49
• r = 7

Answer: 7 m

Commentary. The reasoning is in two layers:

1. Use the prism relationship to get the base area.
2. Use circle area to recover the radius.

That two-step structure is often cleaner than substituting everything at once.

Worked Example 7: Same height, different radii

Problem. Cylinder A and Cylinder B both have height 10 cm. Cylinder A has radius 3 cm. Cylinder B has radius 6 cm. How many times as large is the volume of Cylinder B compared with Cylinder A?

Solution. Cylinder A:

• V_A = pi(3)^2(10) = 90pi cm^3

Cylinder B:

• V_B = pi(6)^2(10) = 360pi cm^3

Compare:

• V_B / V_A = 360pi / 90pi = 4

Answer: Cylinder B has 4 times the volume of Cylinder A.

Commentary. Doubling the radius does not double the volume when height stays fixed. Because area depends on r^2, doubling the radius multiplies the base area, and therefore the volume, by 4.

Worked Example 8: Same radius, doubled height

Problem. A cylinder has radius 5 cm and height 8 cm. A second cylinder has the same radius but height 16 cm. Compare their volumes.

Solution. First cylinder:

• V_1 = pi(5)^2(8) = 25pi x 8 = 200pi cm^3

Second cylinder:

• V_2 = pi(5)^2(16) = 25pi x 16 = 400pi cm^3

Compare:

• V_2 / V_1 = 400pi / 200pi = 2

Answer: The second cylinder has twice the volume.

Commentary. With the base unchanged, volume changes directly with height. This mirrors prism reasoning exactly: if B is constant, doubling h doubles V.

Worked Example 9: Multi-step filling problem

Problem. A cylindrical water container has radius 7 cm and height 20 cm. It is filled to a height of 12 cm. How much more water, in cubic centimetres, is needed to fill it completely?

Solution. The base area stays constant throughout the container.

Base area:

• B = pi r^2 = pi(7)^2 = 49pi cm^2

Full volume:

• V_full = 49pi x 20 = 980pi cm^3

Current water volume:

• V_now = 49pi x 12 = 588pi cm^3

More water needed:

• 980pi - 588pi = 392pi cm^3

Approximate value:

• 392pi ≈ 1231.5

Answer: 392pi cm^3 or about 1231.5 cm^3

Commentary. This works because each horizontal slice has the same area. The water already present forms a shorter cylinder with the same base, so the prism-style structure still applies.

Worked Example 10: Reverse comparison through equal volume

Problem. A rectangular prism has base area 40 cm^2 and height 9 cm. A cylinder has the same height and the same volume as the prism. Find the radius of the cylinder in exact form.

Solution. First find the prism volume:

• V = B x h = 40 x 9 = 360 cm^3

The cylinder has the same volume and the same height 9 cm, so:

• pi r^2 x 9 = 360

Solve for r^2:

• pi r^2 = 40
• r^2 = 40 / pi

So:

• r = sqrt(40 / pi) cm

Approximate value:

• r ≈ sqrt(12.73) ≈ 3.57 cm

Answer: sqrt(40 / pi) cm approximately 3.57 cm

Commentary. This is a strong prism-link problem because the comparison starts with a prism and transfers directly to a cylinder by matching V and h. The missing quantity is hidden inside the circular base area.

Worked Example 11: Challenge problem with unit conversion

Problem. A cylinder has radius 0.3 m and height 120 cm. Find its volume in cubic metres.

Solution. The units must match before using the formula.

Convert height:

• 120 cm = 1.2 m

Now use V = pi r^2 h:

• V = pi(0.3)^2(1.2)
• V = pi(0.09)(1.2)
• V = 0.108pi m^3

Approximate value:

• 0.108pi ≈ 0.339 m^3

Answer: 0.108pi m^3 or about 0.339 m^3

Commentary. The geometry is easy here; the real difficulty is unit discipline. Since volume is three-dimensional, mixing metres and centimetres without converting first will produce a wrong answer even if the formula is correct.

Worked Example 12: Challenge reasoning from a changed base area

Problem. A cylinder and a prism have the same height. The cylinder’s base area is 1.5 times the prism’s base area. What can you say about their volumes?

Solution. For both solids, volume is B x h.

Let the prism’s base area be B. Then the cylinder’s base area is 1.5B. Since both heights are the same, say h:

• Prism volume: V_p = Bh
• Cylinder volume: V_c = 1.5Bh

So:

• V_c = 1.5V_p

Answer: The cylinder’s volume is 1.5 times the prism’s volume.

Commentary. This is the most abstract example in the set. No radius is needed. The key is recognizing that once height is fixed, volume changes in the same ratio as base area.

Common misconceptions and repairs

1. Using diameter as if it were radius

Error: Substituting d directly into pi r^2.

Repair: Write a separate line: r = d / 2 before finding base area.

2. Forgetting that height must be perpendicular to the base

Error: Using a slanted segment or some other interior measure as the height.

Repair: Ask, “What is the distance straight from one base to the other?” That is the h in B x h.

3. Treating the cylinder formula as unrelated to prism volume

Error: Seeing pi r^2 h as a completely separate fact.

Repair: Rewrite it as V = B x h first, then replace B with pi r^2 only after the structure is clear.

4. Mixing units

Error: Combining cm and m in one substitution.

Repair: Convert all linear measures to one unit before calculating.

What to practice next

After working this set, a learner should be ready to:

• move between V = B x h and V = pi r^2 h fluently
• solve missing-dimension problems for cylinders
• compare prisms and cylinders through shared base area and height

For more independent practice, continue with [rea.m08.geometry-measurement.volume.cylinder-volume-derivation-practice-set].