Grade 8 Unit Mastery Quiz: Cylinder Volume Derivation

This quiz is a unit assessment for Grade 8 Geometry and Measurement on Cylinder Volume Derivation. It assumes the learner has already studied the derivation and examples in:

• rea.m08.geometry-measurement.volume.cylinder-derivation
• rea.m08.geometry-measurement.volume.cylinder-volume-derivation-lesson
• rea.m08.geometry-measurement.volume.cylinder-volume-derivation-misconceptions
• rea.m08.geometry-measurement.volume.cylinder-volume-derivation-practice-set

Use this record to measure mastery, not to replace those teaching records.

Quiz directions

• Show all reasoning.
• Use exact answers with pi unless the question asks for a decimal.
• Give units for every volume answer.
• Total: 40 marks

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Question 1: Why does the formula work? (4 marks)

A cylinder has radius r and height h. Explain why its volume is V = pi r^2 h using the idea of stacked circular layers.

Marking solution

• 1 mark: States that a cylinder can be thought of as many equal circular layers stacked from bottom to top.
• 1 mark: Identifies the area of each circular layer as pi r^2.
• 1 mark: Connects volume to base area x height because the same cross-section repeats through height h.
• 1 mark: Concludes correctly that V = pi r^2 h.

Full solution

A cylinder can be viewed as a stack of identical circular slices. Each slice has the same area as the base circle, which is pi r^2. When the same cross-sectional area extends through a height of h, the volume is the area of one layer multiplied by the total height. Therefore,

V = (area of base) x height = pi r^2 h.

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Question 2: Radius or diameter? (4 marks)

A cylinder has diameter 10 cm and height 8 cm.

1. Find its exact volume.
2. Find its volume to the nearest tenth.

Marking solution

• 1 mark: Converts diameter 10 cm to radius 5 cm.
• 1 mark: Substitutes into V = pi r^2 h correctly.
• 1 mark: Simplifies exact volume to 200pi cm^3.
• 1 mark: Gives decimal 628.3 cm^3 to the nearest tenth.

Full solution

Diameter = 10 cm, so radius r = 5 cm.

V = pi r^2 h

V = pi(5)^2(8)

V = pi(25)(8)

V = 200pi cm^3

Decimal form:

200pi ≈ 628.3

So the volume is:

• Exact: 200pi cm^3
• Approximate: 628.3 cm^3

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Question 3: Match the diagram description to the formula step (3 marks)

A student says, “I know the base circle has area 36pi cm^2, and the cylinder is 12 cm tall.” Find the volume.

Marking solution

• 1 mark: Recognizes that 36pi cm^2 is already the base area.
• 1 mark: Uses V = base area x height.
• 1 mark: Computes 432pi cm^3.

Full solution

The base area is already given: 36pi cm^2.

V = base area x height

V = 36pi x 12

V = 432pi cm^3

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Question 4: Find a missing height (4 marks)

A cylinder has volume 392pi m^3 and radius 7 m. Find its height.

Marking solution

• 1 mark: Writes 392pi = pi(7)^2h.
• 1 mark: Simplifies pi(49)h.
• 1 mark: Solves 392 = 49h.
• 1 mark: Gets h = 8 m.

Full solution

Start with the formula:

V = pi r^2 h

Substitute the known values:

392pi = pi(7)^2h

392pi = 49pi h

Divide both sides by 49pi:

h = 392pi / 49pi = 8

So the height is 8 m.

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Question 5: Compare two cylinders (4 marks)

Cylinder A has radius 3 cm and height 10 cm. Cylinder B has radius 6 cm and height 5 cm.

1. Find the exact volume of each cylinder.
2. Which cylinder has the greater volume?
3. By how much?

Marking solution

• 1 mark: Finds V_A = 90pi cm^3.
• 1 mark: Finds V_B = 180pi cm^3.
• 1 mark: Identifies Cylinder B as larger.
• 1 mark: Finds difference 90pi cm^3.

Full solution

Cylinder A:

V_A = pi(3)^2(10) = pi(9)(10) = 90pi cm^3

Cylinder B:

V_B = pi(6)^2(5) = pi(36)(5) = 180pi cm^3

Compare:

180pi > 90pi, so Cylinder B has the greater volume.

Difference:

180pi - 90pi = 90pi cm^3

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Question 6: Effect of changing dimensions (5 marks)

A cylinder has radius r and height h.

1. If the radius is doubled and the height stays the same, by what factor does the volume change?
2. If the height is doubled and the radius stays the same, by what factor does the volume change?
3. If both radius and height are doubled, by what factor does the volume change?

Marking solution

• 2 marks: Correctly shows that doubling radius gives pi(2r)^2h = 4pi r^2 h, so the factor is 4.
• 1 mark: Correctly shows doubling height gives factor 2.
• 2 marks: Correctly shows doubling both gives factor 8.

Full solution

Original volume:

V = pi r^2 h

1. Double the radius:

V_new = pi(2r)^2h = pi(4r^2)h = 4pi r^2 h = 4V

So the volume becomes 4 times as large.

2. Double the height:

V_new = pi r^2(2h) = 2pi r^2 h = 2V

So the volume becomes 2 times as large.

3. Double both:

V_new = pi(2r)^2(2h) = pi(4r^2)(2h) = 8pi r^2 h = 8V

So the volume becomes 8 times as large.

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Question 7: Error analysis (4 marks)

A student solves a problem like this:

Radius = 4 cm, height = 9 cm

V = 2pi rh

V = 2pi(4)(9) = 72pi cm^3

Identify the mistake and find the correct volume.

Marking solution

• 2 marks: Explains that the student used a circumference formula idea instead of the cylinder volume formula; volume needs pi r^2 h.
• 1 mark: Substitutes correctly.
• 1 mark: Gets 144pi cm^3.

Full solution

The mistake is using 2pi rh, which is not a cylinder volume formula. It comes from mixing up volume with a circle or cylinder side measurement idea. For volume, we must use:

V = pi r^2 h

Now substitute:

V = pi(4)^2(9)

V = pi(16)(9)

V = 144pi cm^3

The correct volume is 144pi cm^3.

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Question 8: Applied problem with unit interpretation (4 marks)

A soup can is shaped like a cylinder with radius 3.5 cm and height 11 cm.

1. Find its volume to the nearest tenth of a cubic centimetre.
2. Explain what that volume means in context.

Marking solution

• 1 mark: Substitutes correctly into V = pi r^2 h.
• 1 mark: Computes 134.75pi cm^3 or equivalent intermediate work.
• 1 mark: Gives decimal 423.3 cm^3.
• 1 mark: Correctly interprets as the space inside the can.

Full solution

V = pi r^2 h

V = pi(3.5)^2(11)

V = pi(12.25)(11)

V = 134.75pi cm^3

V ≈ 423.3 cm^3

So the can's volume is about 423.3 cm^3.

In context, this means the can can hold about 423.3 cubic centimetres of soup if it is filled to capacity.

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Question 9: Build from volume and radius (4 marks)

A cylinder must hold exactly 250pi cm^3. Its radius is 5 cm. Find the required height.

Marking solution

• 1 mark: Writes 250pi = pi(5)^2h.
• 1 mark: Simplifies to 250pi = 25pi h.
• 1 mark: Solves for h.
• 1 mark: Gets 10 cm.

Full solution

Use V = pi r^2 h.

250pi = pi(5)^2h

250pi = 25pi h

Divide both sides by 25pi:

h = 250pi / 25pi = 10

So the required height is 10 cm.

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Question 10: Multi-step mastery problem (4 marks)

A solid is made from two identical cylinders. Each cylinder has radius 2 cm and height 15 cm.

1. Find the total exact volume.
2. A student claims the total volume is 2pi(2)^2(15) = 60pi cm^3. Explain whether the claim is correct.

Marking solution

• 2 marks: Finds one cylinder's volume as 60pi cm^3 and total volume as 120pi cm^3.
• 2 marks: Explains that 60pi cm^3 is only the volume of one cylinder, so the claim is incorrect for the total solid.

Full solution

Volume of one cylinder:

V = pi r^2 h = pi(2)^2(15) = pi(4)(15) = 60pi cm^3

There are two identical cylinders, so total volume:

2 x 60pi = 120pi cm^3

The student's claim is incorrect for the total solid. 60pi cm^3 is the volume of one cylinder, not both together.

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Answer key summary

1. V = pi r^2 h because a cylinder is stacked equal circular layers of area pi r^2 through height h.
2. 200pi cm^3, about 628.3 cm^3
3. 432pi cm^3
4. 8 m
5. V_A = 90pi cm^3, V_B = 180pi cm^3, B is larger by 90pi cm^3
6. Factors: 4, 2, 8
7. Correct volume: 144pi cm^3
8. About 423.3 cm^3; it is the capacity of the can
9. 10 cm
10. 120pi cm^3; the claim is wrong because it counts only one cylinder

Recommended mastery threshold

Recommended threshold: 32/40 (80%), with these additional conditions:

• At least 3/4 on Question 1, because the learner should explain why the formula works, not only use it.
• At least 3/4 total across Questions 7 and 10, because error analysis shows whether the learner can detect common misconceptions independently.

Interpretation guide

• 36-40 marks: Strong mastery. The learner can explain the derivation idea, use the formula accurately, and detect common errors.
• 32-35 marks: Secure mastery. The learner is ready to move on, though one weak spot may still need review.
• 24-31 marks: Partial mastery. Revisit the lesson and misconceptions records, then complete more problems from the practice set.
• Below 24 marks: Not yet secure. Return first to rea.m08.geometry-measurement.volume.cylinder-volume-derivation-lesson, then use rea.m08.geometry-measurement.volume.cylinder-volume-derivation-misconceptions and rea.m08.geometry-measurement.volume.cylinder-volume-derivation-practice-set before re-attempting a quiz.

Most likely remediation links

• If the learner confuses radius and diameter, review rea.m08.geometry-measurement.volume.cylinder-volume-derivation-misconceptions.
• If the learner can calculate but cannot explain why V = pi r^2 h, review rea.m08.geometry-measurement.volume.cylinder-derivation and rea.m08.geometry-measurement.volume.cylinder-volume-derivation-lesson.
• If the learner needs more independent practice before reassessment, use rea.m08.geometry-measurement.volume.cylinder-volume-derivation-practice-set.