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Informal Arguments for
Circle & Solid Formulas

Discover why the formulas for circumference, area, and volume actually work: using dissection, Cavalieri's principle, and limit arguments.

Geometry

What Are Informal Arguments?

An informal argument is a convincing explanation for why a formula is true, without writing a full rigorous proof. We use visual reasoning, physical analogies, and logical steps to build intuition.

Dissection
Cut a shape into pieces and rearrange them into a simpler shape whose area or volume you already know.
Cavalieri's Principle
Two solids with equal cross-sectional areas at every height have the same volume.
Limit Arguments
Approximate a curved shape with many straight pieces: as the count grows to infinity, the approximation becomes exact.

Why Do Formulas Need Justification?

Memorizing \( C = 2\pi r \) or \( V = \pi r^2 h \) is useful, but understanding why those formulas work lets you derive them from scratch, apply them to unfamiliar problems, and catch mistakes.

Circumference: \( C = 2\pi r \)

\(r\) \( C = 2\pi r \) (unrolled circumference) Inscribed polygon → circle

Area of a Circle: Dissection Argument

Slice into wedges → \( \text{base} \approx \pi r \) \( \text{height} = r \) Rearrange into "rectangle"

Area: The Limit Argument

Think of the circle as built from concentric rings, each of width \( \Delta r \). "Unroll" a ring at radius \( t \): it becomes a thin rectangle of length \( 2\pi t \) and height \( \Delta r \).

Concentric rings → \( 2\pi r \) \(r\) \( A = \tfrac{1}{2} \cdot 2\pi r \cdot r = \pi r^2 \)

Volume of a Cylinder: \( V = \pi r^2 h \)

\(r\) \(h\) Each cross-section = circle of area \( \pi r^2 \)

Understanding Cavalieri's Principle

Solid A = same volume Solid B

Volume of a Pyramid: \( V = \tfrac{1}{3}Bh \)

B (base) \(h\) Cube = 3 congruent pyramids (one shown highlighted)

Volume of a Cone: \( V = \tfrac{1}{3}\pi r^2 h \)

\(h\) \(r\) cross-section \( V = \tfrac{1}{3}\pi r^2 h \)

Prisms vs. Pyramids: The \( \tfrac{1}{3} \) Factor

Prism / Cylinder Pyramid / Cone
Cross-section Constant at every height Shrinks toward apex
Volume formula \( V = Bh \) \( V = \frac{1}{3}Bh \)
Why the factor? Full base at every level Tapers: 3 fit inside the prism
Circular base \( V = \pi r^2 h \) \( V = \frac{1}{3}\pi r^2 h \)

How to Build an Informal Argument

1
Identify the shape. Is it a circle, cylinder, pyramid, or cone?
2
Choose your strategy: dissection (cut & rearrange), Cavalieri (match cross-sections), or limit (approximate with simple shapes).
3
Set up the simpler shape. Show that your approximation or rearrangement has a known formula.
4
Take the limit or sum. Argue that as pieces become infinitely small, the approximation becomes exact.
5
State the formula. Write the final result clearly and check units.

Example 1: Circumference via Polygons

Problem: Use a regular 12-gon inscribed in a circle of radius \( r = 5 \) to approximate the circumference.

1
Each side of the 12-gon subtends a central angle of \( \frac{360^\circ}{12} = 30^\circ \).
2
Side length: \( s = 2r\sin(15^\circ) = 2(5)\sin(15^\circ) \approx 2(5)(0.2588) \approx 2.588 \).
3
Perimeter: \( P = 12 \times 2.588 \approx \) \( 31.06 \).
4
Exact circumference: \( C = 2\pi(5) \approx 31.42 \). The 12-gon is already within 1.2% of the true value.

Example 2: Area by Dissection

Problem: A circle has radius \( r = 7 \). Use the "wedge rearrangement" to find its area.

1
Slice the circle into many thin wedges (imagine 100+ slices).
2
Rearrange them alternating up/down. The shape approximates a rectangle.
3
Base = half the circumference = \( \pi r = 7\pi \). Height = \( r = 7 \).
4
\( A = \text{base} \times \text{height} = 7\pi \times 7 = 49\pi \approx \) \( 153.94 \) square units.

Example 3: Cone Inside a Cylinder

Problem: A cylinder and a cone share the same base radius \( r = 4 \) and height \( h = 9 \). How many cones of sand fill the cylinder?

Cylinder: \( \pi r^2 h \) = 3 × Cone: \( \tfrac{1}{3}\pi r^2 h \)

Common Mistakes

✕ Forgetting \( \frac{1}{3} \)
Writing \( V = \pi r^2 h \) for a cone or \( V = Bh \) for a pyramid. The taper means only one-third of the prism's volume.
✕ Confusing \( C \) and \( A \)
Circumference is \( 2\pi r \) (linear, in \( \text{cm} \)); area is \( \pi r^2 \) (squared, in \( \text{cm}^2 \)). Mixing them leads to wrong units.
✕ Using diameter as radius
If a problem gives diameter \( d = 10 \), remember \( r = 5 \). Plugging 10 into \( \pi r^2 \) gives \( 4 \times \) the correct area.

Formula Summary & Arguments

Shape Formula Argument Type
Circumference \( C = 2\pi r \) Limit (polygon perimeter)
Circle Area \( A = \pi r^2 \) Dissection (wedge → rectangle)
Cylinder \( V = \pi r^2 h \) Cavalieri (stacked circles)
Pyramid \( V = \frac{1}{3}Bh \) Dissection (cube → 3 pyramids)
Cone \( V = \frac{1}{3}\pi r^2 h \) Cavalieri + pyramid rule

Question 1 of 5

In the dissection argument for the area of a circle, the circle is cut into thin wedges and rearranged into a shape that approximates a:

Question 2 of 5

Cavalieri's Principle states that two solids have the same volume if:

Question 3 of 5

A cone and a cylinder share the same base radius and height. The volume of the cone is what fraction of the cylinder's volume?

Question 4 of 5

In the limit argument for circumference, what happens to the perimeter of a regular \( n \)-gon inscribed in a circle as \( n \to \infty \)?

Question 5 of 5

A pyramid has a square base with side length 6 and height 10. What is its volume?