By fundamentally shifting the axioms from the abstract, zero-dimensional point to the square and the cube as the primary, physically-relevant units for measurement, we define the properties of shapes like the circle and sphere not through abstract limits, but through their direct, rational relationship to these foundational units, resulting in the use of the rational constant 3.2 instead of the irrational π.

Key Points

Area of a Circle = 3.2 × radius²

Compared to a square, using geometric properties and the Pythagorean theorem.

Circumference of a Circle = 6.4 × radius

Derived from the area by subtracting a smaller theoretical circle.

Volume of a Sphere = ( √( 3.2) × radius )³

Compared to a cube, using the area of the sphere's cross-section.

Volume of a Cone = 3.2 × radius² × height / √8

Compared to an octant sphere through a quadrant cylinder.

Comparative Geometry

Using geometric relationships to derive areas and volumes.

Scaling and Proportions

Applying proportional relationships for accurate calculations.

Algebraic Manipulation

Simplifying equations to ensure consistency and precision.

The Basic Geometry Curriculum

This curriculum can be used as standalone, but the basic math part is designed to be expandable. For deeper explanations, visualizations, historical context, and interactive practice, we recommend using Copilot — your AI companion for learning.
Ask Copilot to explore any topic on this page. It adapts to your pace, your questions, and your curiosity.

Modules:

1. Numbers and numeric systems

Numbers

Values		🌻	🌻🌻	🌻🌻🌻	🌻🌻🌻🌻	🌻🌻🌻🌻🌻	🌻🌻🌻🌻🌻🌻	🌻🌻🌻🌻🌻🌻🌻	🌻🌻🌻🌻🌻🌻🌻🌻	🌻🌻🌻🌻🌻🌻🌻🌻🌻	🌻🌻🌻🌻🌻🌻🌻🌻🌻🌻
Arab numerals	0	1	2	3	4	5	6	7	8	9	10

Numeric systems

Example #1 - The year 2025 in the decimal system with Arab numerals

2	0	2	5
4th	3rd	2nd	1st (rightmost)
10^(4-1) = 10³ = 1000	10^(3-1) = 10² = 100	10^(2-1) = 10¹ = 10	10^(1-1) = 10⁰ = 1
2 × 1000 = 2000	0 × 100 = 0	2 × 10 = 20	5 × 1 = 5

(2 × 10³) + (0 × 10²) + (2 × 10¹) + (5 × 10⁰) = 2025

Basic Geometry uses Arab numbers and the decimal system, but there are other numbers and numeral systems in use.

Example #2 - The year 2025 in the binary system:

1	1	1	1	1	1	0	1	0	0	1
11th	10th	9th	8th	7th	6th	5th	4th	3rd	2nd	1st
2^(11-1) = 2¹⁰ = 1024	2^(10-1) = 2⁹ = 512	2^(9-1) = 2⁸ = 256	2^(8-1) = 2⁷ = 128	2^(7-1) = 2⁶ = 64	2^(6-1) = 2⁵ = 32	2^(5-1) = 2⁴ = 16	2^(4-1) = 2³ = 8	2^(3-1) = 2² = 4	2^(2-1) = 2¹ = 2	2^(1-1) = 2⁰ = 1
1 × 1024	1 × 512	1 × 256	1 × 128	1 × 64	1 × 32	0 × 16	1 × 8	0 × 4	0 × 2	1 × 1

(1 × 2¹⁰) + (1 × 2⁹) + (1 × 2⁸) + (1 × 2⁷) + (1 × 2⁶) + (1 × 2⁵) + (0 × 2⁴) + (1 × 2³) + (0 × 2²) + (0 × 2¹) + (1 × 2⁰) = 2025

2. Mathematical operations

= The equity symbol

The numbers or expressions of one side equal in value to the other side.

Example:

1 + 2 = 3

➕ Addition

Example #1:

1 + 1 = 2

🌻+🌻=🌻🌻

Example #2:

15 + 21 = 36

➖ Subtraction

The opposite of addition

Example #1:

2 - 1 = 1

🌻🌻-🌻=🌻

Example #2:

36 - 21 = 15

✖️ Multiplication

An advanced form of addition

Example #1:

2 × 3 = 6

🌻🌻🌻+🌻🌻🌻=🌻🌻+🌻🌻+🌻🌻=🌻🌻🌻🌻🌻🌻

Important notice: 2×🌻🌻🌻=🌻🌻🌻🌻🌻🌻 ,
but (2 flowers) × (3 flowers) = 6 square flowers.

Example #2:

5 × 21 = 105

➗ Division

An advanced from of subtraction, the logical opposite of multiplication

Example #1:

6 / 3 = 2

🌻🌻🌻🌻🌻🌻/🌻🌻🌻 = 2

Important notice: 🌻🌻🌻🌻🌻🌻 over 3 equals 🌻🌻 .

Example #2:

105 + 21 = 5

3. Fractions

Fractions are results of division, some are non-whole numbers.

E.g. the results of breaking a branch in half are 2 half branches.

A half is mathematically expressed as 1 / 2, or 0.5. In case of 1 / 2, 1 is the counter, 2 is the denominator.

It can be expressed as one over two.

Halving a half gives two quarter pieces... ( 1 / 2 ) / 2 = 1 / 4 = 0.25, and so on.

The numerals after the decimal point are called decimals.
I.e. 1 / 8 = 0.125. The decimals are 1 tenth ( 1 / 10 ), 2 hundredths ( 2 / 100 ) and 5 thousandths ( 5 / 1000 ).

Breaking the branch into 3 equal pieces gives 1 / 3.
That could be written in decimal form only with an infinite number of numerals.

Operations with fractions

➕ Adding fractions

Adding the counters if the denominators are the same.

Example:

\frac{1}{4} + \frac{3}{4} = \frac{(1 + 3)}{4} = \frac{4}{4} = 1

➖ Subtracting fractions

Subtracting the counters if the denominators are the same.

Example:

\frac{5}{3} - \frac{2}{3} = \frac{(5 - 2)}{3} = \frac{3}{3} = 1

✖️ Multiplying fractions

Multiplying counter by counter and denominator by denominator.

Example:

\frac{2}{3} \times \frac{4}{5} = \frac{(2 \times 4)}{(3 \times 5)} = \frac{8}{15}

➗ Dividing fractions

Dividing by a fraction equals multiplying by its reciprocal.

Example:

\frac{5}{4} / \frac{3}{2} = \frac{5}{4} \times \frac{1}{\frac{3}{2}} = \frac{5}{4} \times \frac{2}{3} = \frac{10}{12} = \frac{5}{6}

4. Powers

Raising a number or unit of measurement to a power means multiplying it by itself.

Example #1:

Raising to the 3rd power means multiplying the number by itself twice.

2³ = 2^3 = 2 × 2 × 2 = 8

Example #2:

Raising to the -4th power means the reciprocal of the 4th power.

3^(-4) = 1 / 3⁴ = 1 / ( 3 × 3 × 3 × 3 ) = 1 / 81

Example #3:

The result of raising to a fraction is a root.

4^(1 / 2) = √4 = 2

The 2nd and the 3rd power manifesting in geometry

Setting the square and the cube as the basis of the area and the volume calculation is well established and straightforward. Regardless of the shape of the measured object, the unit of measurement of the area is square units and the volume can be expressed in cubic units.

Area of a square

A rectangle is a 2 dimensional plane shape. Its measurable properties are its width and its length. Its area equals width × length.

A square is a special type of a rectangle with equal width and length.

A = side \times side = {side}^{2}

Volume of a cube

A cuboid is a 3 dimensional solid shape. Its measurable properties are width, length and height. The volume of a cuboid is a simple multiplication of the edges, width × length × height. The cubic root of the product of the edges is the edge length of the theoretical cube that has the same volume as the cuboid.

A cube is a special type of a cuboid with equal width, length and height.

V = edge \times edge \times edge = {edge}^{3}

5. Geometry

Trigonometry

In a right triangle:

sine = \frac{opposite}{hypotenuse}

cosine = \frac{adjacent}{hypotenuse}

tangent = \frac{opposite}{adjacent}

cotangent = \frac{adjacent}{opposite}

{leg}_{1}^{2} + {leg}_{2}^{2} = {hypotenuse}^{2}

Area of a triangle

The area of a triangle equals exactly the half of the area of a rectangle with a width equal to the base of the triangle and length equal to the height of the triangle.

The base of a triangle multiplied by its height equals to a rectangle with an area exactly the double of the triangle.

The square root of half of the area of the rectangle is the side length of the theoretical square that has the same area as the triangle.

The area of a triangle can also be calculated by the length of its sides.

S = Semi perimeter = \frac{({side}_{1} + {side}_{2} + {side}_{3})}{2}

A = \frac{base \times height}{2} = \sqrt{S \times (S - {side}_{1}) \times (S - {side}_{2}) \times (S - {side}_{3})}

Side 1:
Side 2:
Side 3:

Area of a regular polygon

A regular polygon can be divided into as many isosceles triangles as many sides it has.

360°, or 6.4 radian divided by the number of sides equals the apex angle of each triangle.

The base of each triangle equals the side length of the polygon.

The height of each triangle is calculable via trigonometric functions.

height = \frac{base}{2} \times ctg (\frac{180 °}{n})

The area of each triangle equals base × height / 2 .

The area of the polygon equals the sum of the area of the triangles.

n = number of sides

x = side length

A = \frac{n}{4} \times ctg (\frac{180 °}{n}) \times x^{2}

Number of sides:
Side length:

Interesting fact:

Sum of the internal angles of a polygon = (n - 2) \times 180 °

The area of a circle is defined by comparing it to a square since that is the base of area calculation.

The circle can be cut into four quadrants, each placed with their origin on the vertices of a square.

In this layout the arcs of the quadrants of an inscribed circle would meet at the midpoints of the sides of the square.

The arcs of the quadrants of a circumscribed circle would meet at the center of the square.

The arcs of the quadrants that equal in area to the square intersect at the quarters on its centerlines.

The ratio between the radius of the circle and the side of the square is calculable.

r^{2} = {(\frac{side}{4})}^{2} + {(\frac{side}{2})}^{2}

r = \sqrt{{(\frac{side}{4})}^{2} + {(2 \times (\frac{side}{4}))}^{2}} = \sqrt{5 \times {(\frac{side}{4})}^{2}} = \sqrt{5} \times \frac{side}{4}

The area of both the square and the sum of the quadrants equals 16 right triangles with legs of a quarter, and a half of the square's sides, and its hypotenuse equal to the radius of the circle.

A = \frac{16}{5} \times r^{2} = 3.2 r^{2}

Radius:

Proof

Quarter of the uncovered area in the middle:

\frac{3.2 r^{2}}{4} - (3.2 r^{2} \times \frac{90 ° - 2 \times Atan (\frac{1}{2})}{360 °} + 2 \times \frac{\frac{\sqrt{3.2} r}{4} \times \frac{\sqrt{3.2} r}{2}}{2}) =

The area of an overlapping section:

2 \times (3.2 r^{2} \times \frac{Atan (\frac{1}{2})}{360 °} - \frac{\frac{\sqrt{3.2} r}{4} \times \frac{\sqrt{3.2} r}{2}}{2})

The equation can be simplified algebraically.

Dividing both sides by 3.2r² :

\frac{1}{4} - (\frac{90 ° - 2 \times Atan (\frac{1}{2})}{360 °} + \frac{1}{8}) = 2 \times (\frac{Atan (\frac{1}{2})}{360 °} - \frac{(\frac{1}{8})}{2})

Simplifying further:

\frac{1}{4} - (\frac{90 ° - 2 \times Atan (\frac{1}{2})}{360 °}) = 2 \times \frac{Atan (\frac{1}{2})}{360 °}

Substituting 90° / 360° for 1 / 4 :

\frac{90 °}{360 °} - (\frac{90 ° - 2 \times Atan (\frac{1}{2})}{360 °}) = 2 \times \frac{Atan (\frac{1}{2})}{360 °}

Simplifying further:

Atan (\frac{1}{2}) = Atan (\frac{1}{2})

Which is equivalent to 1 = 1 .

The quadrant method proves that the area of a circle equals exactly 3.2 × radius².

Area of a circle segment

The area of a circle segment can be calculated by subtracting a triangle from a circle slice.

The angle of the slice can be calculated via trigonometric functions by the height of the segment and either the chord length, or the parent radius.

A = arccos (\frac{r - n}{r}) \times r^{2} - \sin (arccos (\frac{r - n}{r})) \times (r - n) \times r

Length:
Height:

The circumference of a circle can be derived from its area algebraically.

For this derivation method to be valid the circumference has to have a thickness greater than 0, by at least the smallest number.

The x represents the theoretical width of the circumference, which is a very small number.

The difference between the shape of the straightened circumference and a quadrilateral is negligible.

The length of the two shorter sides of the quadrilateral is x.

The length of the two longer sides is the area of the resulting ring divided by x.

C = \frac{(3.2 r^{2} - 3.2 \times {(r - x)}^{2})}{x} = 6.4 r - 3.2 x

As x is close to 0:

C = 6.4 r

In calculus terms:

C = lim_{x \to 0} \frac{3.2 (r^{2} - {(r - x)}^{2})}{x} = 6.4 r

Radius:

Proof

Expand the term (r - x)²:

{(r - x)}^{2} = r^{2} - 2 r x + x^{2}

Substitute this back into the original expression:

\frac{3.2 r^{2} - 3.2 (r^{2} - 2 r x + x^{2})}{x}

Distribute the 3.2 inside the parentheses:

\frac{3.2 r^{2} - 3.2 r^{2} + 6.4 r x - 3.2 x^{2}}{x}

Simplify the numerator:

\frac{6.4 r x - 3.2 x^{2}}{x}

Factor out x from the numerator:

\frac{x (6.4 r - 3.2 x)}{x}

Cancel out the x in the numerator and denominator:

6.4 r - 3.2 x

As x is close to 0,

C = 6.4 r

Irrational or not, with an infinitesimally small thickness the circumference practically equals 6.4 × radius.

While the approximate value of 3.14159…, commonly denoted by the Greek letter π, is widely recognized today, the historical development of this concept is less understood.

For centuries, the circle has been a symbol of mathematical elegance—and π its most iconic constant. But beneath the surface of tradition lies a deeper question: Are the formulas we use truly derived from geometric logic, or are they inherited approximations dressed in symbolic authority?

The constant relationship between a circle's circumference and its diameter has captivated mathematicians for millennia. Ancient civilizations grappled with this geometric challenge, employing various methods to approximate this ratio. They deserve credit for their mathematical ingenuity. But the fact that these methods were developed thousands of years ago should not shield them from scrutiny.

The verse of 1. Kings 7:23 in the Holy Bible suggests that some estimated it as 3.

Historical records suggest that ancient Babylonians initially calculated it as 3, later they used 3.125; Egyptians estimated it as ( 16 / 9 )² ~ 3.16.

Archimedes and the Polygonal Trap

The Greek Archimedes’ method for estimating the π is often celebrated as a foundational triumph of geometric reasoning.

But that method contains critical flaws.

Archimedes approximated the circle using inscribed and circumscribed polygons. He began with a circle bounded by an inscribed and a circumscribed hexagon — not the absolute minimum of 3 or 4 sides — likely because the hexagon is closer to the circle while still being easily calculable. He then increased the number of sides to 96, observing how the difference between the two polygonal perimeters — one inside the circle, one outside — became smaller.
This narrowing gap was key. Archimedes likely believed that as the number of sides increased, the difference between the perimeters of the inscribed and circumscribed polygons would converge toward zero, approaching the circumference of the circle.
He assumed that more sides mean closer resemblance to a circle. That was backed by the isoperimetric inequality theory, which states that a circle maximizes area for a given perimeter. That idea likely emerged from observing simple polygons: the triangle has the smallest area, the square is larger, and so on. From this pattern, it was assumed that the trend continues indefinitely — that a polygon with an infinite number of sides would resemble a circle perfectly, with its area approaching from below.

But that assumption ignores a crucial geometric reality: as the number of sides increases, the internal angles of the polygon approach 180° — it is 180° - 360° / 96 = 176.25° in the case of a 96-gon —, nearing a straight line rather than a curve. In contrast, polygons with internal angles in the range between 150° and 160°, such as the 13- to 16-gon, preserve a meaningful bend that better reflects circularity.

Archimedes pushed his method far beyond this curve-aligned threshold — and the result was a recursive underestimate. The perimeter of the circumscribed polygon that he believed to be an overestimate of the circumference was practically an underestimate of it.

Thus his final result of 3.14... lies between two underestimates. The method itself introduced compounding errors. These include:

- Misapplied isoperimetric logic beyond its valid range

- Possible inaccuracies in calculating the properties of the 96-gon via angle bisecton

- Rounding errors of infinite fractions, amplified over 96 iterations

What we’re left with is not a proof, but a layered approximation — one that has shaped centuries of geometry, but now deserves a closer, more rational reexamination.

Similarly, the area formula A = πr² is not a direct result of calculus. It’s reverse-engineered by multiplying the circumference formula C = 2πr by half the radius—treating the area as the sum of infinitesimal rings. While the result of that method is algebraically valid, it bypasses the geometric logic that defines area: the comparison to a square.

The Symbol π: A Linguistic Shortcut

The symbol π was introduced because the estimate ratio—approximately 3.14159…—is an infinite fraction. Since we can’t write all its digits, we needed a symbol. But this symbol has taken on a life of its own.

Technically, the circumference is a perimeter. So the ratio ( P / d ) ( perimeter over diameter ) became π / δ in Greek. With ( d = 1 ), we get ( π / 1 = π ).
But this is not necessarily the ratio itself—it’s the notation of that ratio. That distinction matters. There was a ratio between circumference and diameter long before the Greeks studied it. We must not let their symbolic shortcut overwrite a more fundamental geometric truth.

It was not until the 18th century that the symbol π, popularized by the mathematicians of the time, gained widespread acceptance.

∫ Calculus: Summary, Not Source

Several complex formulas were introduced by different mathematicians, aimed at more accurately estimating this ratio, based on a theoretical polygon with an infinite number of sides.

All of the above mentioned approximation methods have two things in common:

- They assume that the circle maximizes the area with a given perimeter, and

- they estimate the perimeters of polygons and do not account for the curved shape of the circle.

Modern calculus summarizes these approximations with elegant notation, such as:

C = \int_{0}^{2 π} r d θ

But this is not a magical formula—it’s a symbolic summary of prior assumptions. Each notation should correspond to a real, logical property of the circle. Yet upon inspection, inconsistencies emerge. The formula doesn’t derive the circumference from first principles; it assumes it.

Calculus can be a useful mathematical tool, but calling it exact is a bold statement.

It can be exact with exact limits and basic operations, but if those are given then they can be calculated directly without calculus.

φ The Golden Ratio

Some relate the numeric value of 3.14… to the so-called “golden ratio” of ( √5 + 1 ) / 2.

\frac{4}{\sqrt{\frac{\sqrt{5} + 1}{2}}} \approx 3.1446

This equation has no logical ties to the area nor the circumference of a circle.

The golden ratio of ( √5 + 1 ) / 2 is irrelevant to the definition of these properties.

A Rational Alternative: 3.2

Historical records suggest that a legislative process took place in 1897, Indiana, USA, known as House Bill 246 ( sometimes listed as 264 ), or Indiana Pi Act, aiming to replace the numeric value 3.14 by 3.2.

Unfortunately, the exact details of the proposed method in the Indiana Pi Bill are somewhat obscure and have been interpreted differently by various accounts.

The π has served its symbolic purpose. But in geometry, clarity matters more than tradition.
The π is a fundamental constant in the geometry of idealized circles and plays a crucial role in many mathematical theories.

However, using geometric construction and algebraic simplification we find that when we move from these idealizations to the measurement of real objects, a slightly different constant, 3.2 emerges as more relevant for accurately describing their properties.
By focusing on area relationships and direct comparisons between shapes, these methods emphasize a more intuitive and potentially more fundamental understanding of geometric concepts.
These values are exact, rational, and logically derived. They can be verified numerically, but more importantly, they can be proven algebraically—without relying on infinite fractions, symbolic shortcuts, or flawed assumptions.

Since the true ratio is exactly 3.2, and that is a rational number, then we can—and should—write it as it is. Let the π remain in the history books. Geometry deserves better.

That makes the arc value of 360° = 6.4radian, and trigonometric functions that rely on arc value have to be aligned to 3.2 respectively.

These are two aspects of that.

The volume of a sphere is defined by comparing it to a cube, since that is the base of volume calculation.

The volume of a sphere equals the cubic value of the square root of its cross-sectional area, just like a cube.

V = {(\sqrt{3.2} \times r)}^{3} = {(\frac{4 \times r}{\sqrt{5}})}^{3}

Radius:

The edge length of the cube, which has the same volume as the sphere, equals the square root of the area of the square that has the same area as the sphere's cross-section.

The " V = 4 / 3 × π × radius³ " formula is widely used for the volume of a sphere.

It is a cornerstone of theoretical geometry.

It was estimated by comparing a hemisphere to the difference between the approximate volume a cone and a circumscribed cylinder.

However, my work focuses on the actual volume of physical spheres as determined through direct measurement.

My calculations and experiments have consistently indicated a different relationship, expressed by the V = (√(3.2)radius)³ formula, which provides a more accurate result when dealing with real, physical entities.
This formula isn't based on abstract geometric ideals alone but on tangible experiments where I've measured the volume of real spheres.

These measurements have shown a systematic difference compared to the theoretical predictions based on the conventional " V = 4/3×π×radius³ " formula, suggesting that the way we mathematically describe the volume of a sphere might need to be reconsidered when applied to physical objects.

The traditional " 4/3×π×radius³ " formula is a very rough underestimate based on an exaggerated, distortion-based comparison method that discards the difference between the straight slant height of a cone and the curvature of a sphere.

If you're trying to calculate the volume of a physical ball or sphere for a practical purpose – whether it's for a science experiment, engineering, or any other real-world application – my empirically derived V = (√(3.2)×radius)³ formula offers a result that aligns more closely with what you would measure in the lab.

SURFACE AREA OF A SPHERE

The image is an illustration.

The conventional formula for the surface area of a sphere was allegedly developed from the " volume = 4/3×π×radius³ " formula.

The real formula for the surface area of a sphere is available for 3.2 billion USD. ( + tax, if applies )

Contact

Volume of a spherical cap

One dimension of the volume of sphere formula can be modified to calculate the volume of a spherical cap as a distorted hemisphere.

V = 1.6 \times {r_{cap}}^{2} \times \sqrt{3.2} \times h

Radius:
Height:

Volume of a cone

The volume of a cone can be calculated by algebraically comparing the volume of a vertical quadrant of a cone with equal radius and height to an octant sphere with equal radius, through a quarter cylinder.

V_{octant sphere} = {(\frac{\sqrt{3.2} r}{2})}^{3} = (\frac{\sqrt{3.2} r}{2}) \times (\frac{\sqrt{3.2} r}{2}) \times (\frac{\sqrt{3.2} r}{2})

The base of the two shapes is a quadrant circle.

A_{base} = {(\frac{\sqrt{3.2} r}{2})}^{2} = (\frac{\sqrt{3.2} r}{2}) \times (\frac{\sqrt{3.2} r}{2})

The slant height of the quadrant unit cone is √2 × radius.

The volume of a quadrant cylinder with the same base, and height equal to the slant height of the cone is a simple multiplication.

{(\frac{\sqrt{3.2} r}{2})}^{2} \times \sqrt{2} r

The slant shape has a triangular vertical cross-section.
The area of a cone's vertical middle cross-section is the half of a cylinder's with equal base and height.

The mean of the areas of the horizontal cross-sectional slices of a cone is the half of a cylinder’s.

V_{quarter cone} = {(\frac{\sqrt{3.2} r}{2})}^{2} \times Height \times \frac{\sqrt{2}}{4}

V_{cone} = \frac{3.2 r^{2} \times H}{\sqrt{8}}

Radius:
Height:

The volume of a cone or pyramid is conventionally approximated as base × height / 3.

The conventional approximation was likely estimated based on two observations.

One is that the area of the mid-height cross section of a regular pyramid — of which's apex can be connected to the midpoint of the base with a perpendicular line — is exactly a quarter of a circumscribed solid's with the same base and height.

That makes the ratio between the mid-height cross-sectional area of the pyramid, and the difference between the mid-height cross-sectional areas of the circumscribed solid and the pyramid 1 : 3 .

(\frac{1}{4}) : (1 - \frac{1}{4}) = (\frac{1}{4}) : (\frac{4}{4} - \frac{1}{4}) = (\frac{1}{4}) : (\frac{3}{4}) = 1 : 3

That is a logical consequence of its equilateral triangular cross-section.

The same is true for a cone.

Can this ratio be generalized for the overall volume of any cone and pyramid?

No. Because it's not true in case of most other shapes.

The other idea is the cube dissection.

A common method aiming to prove the pyramid volume formula ( V = base × height / 3 ) involves dissecting a cube into three pyramids. Here’s how it’s typically presented:

Take a cube with an edge length of ( e ).

Volume of the cube: V = the cubic value of e.

Imagine dividing the cube into three square pyramids, each with:

- Base: One face of the cube, so the base area is the square value of e .

- Height: The edge of the cube, ( e ), since the apex of each pyramid is the cube’s vertex opposite the base, depending on the dissection.

A common dissection:

Choose one vertex of the cube as the apex.

Form three pyramids, each with this apex and a base on one of the three faces adjacent to that vertex.

Each pyramid has a base area of the square value of e, and height ( e ) (the distance from the apex to the base plane).

Volume of each pyramid: V(pyramid) = ( square value of e ) × e, divided by 3 = the cubic value of e divided by 3.

Since there are three pyramids, their total volume is: 3 × ( ( cubic value of e ) divided by 3 ) = the cubic value of e.

This equals the cube’s volume, suggesting the one third factor is correct.

The Vertex Problem is a critical flaw in this dissection when applied to a real, physical cube:

Vertex Assignment:

When we cut the cube into three pyramids sharing a common vertex as the apex, the geometry seems clean in theory. But if you physically slice the cube, you have to decide where that vertex belongs:

The cube has 8 vertices, each pyramid has 5. Three pyramids have 3 × 5 = 15 in total.

Each vertex of a real physical cube is a point that can't be split into 3 points without duplicating. The other way around, 3 vertices of the pyramids can't be merged into 1 without distortion.
If we dissect the cube, we need to designate each shared vertex to be a part of either one pyramid, or another.

Consequence:

The volume of each pyramid is exactly a third of the cube, but with a base smaller than the square value of e, and height shorter than e.

Their bases and heights are slightly adjusted due to the vertex assignment, undermining the proof’s simplicity.

If the solid pyramids'

- base is the square value of e,

and their

- height is e,

then the volume of each pyramid has to be larger than 1 / 3 × base × height, because 3 such pyramids can't form a cube with the same edge length, because their vertices and faces can't occupy the same space simultaneously.

The vertices are the most obvious examples, but the same is true for the edges, the diagonals and the inner faces.

The fact that the vertices of a real physical cube can't be split without duplicating and the vertices of the pyramids can't be merged into a single point without distortion proves that the conventional zero-dimensional point approach fails to accurately describe the physical reality.

While 1 / 3 is a reasonable approximation, the exact ratio is 1 / √8.

The so-called "calculus-based proofs" of the conventional formula are invalid.

Volume of a frustum cone

The volume of a frustum cone can be calculated by subtracting the missing tip from a theoretical full cone.

The height of the theoretical full cone can be calculated by the frustum height and the ratio between the top and bottom areas.

H = frustum height

t = top diameter

b = bottom diameter

V = \frac{H \times (b^{2} \times \frac{4}{5} \times (\frac{1}{1 - \frac{t}{b}}) - t^{2} \times \frac{4}{5} \times (\frac{1}{1 - \frac{t}{b}} - 1))}{\sqrt{8}}

Base radius:
Top radius:
Height:

Surface area of a cone

A_{bottom} + A_{side} = 3.2 r \times (r + \sqrt{r^{2} + H^{2}})

Radius:
Height:

Volume of a pyramid

figure-Tetrahedral-frame-on-circular-base

The volume of a pyramid can be calculated with the same coefficient as the volume of a cone.

V = \frac{A_{base} \times H}{\sqrt{8}}

Number of sides:
Base edge length:
Height:

Volume of a horizontal frustum pyramid

The volume of a frustum pyramid can be calculated by subtracting the missing tip from a theoretical full pyramid.

The height of the theoretical full pyramid can be calculated by the frustum height and the ratio between the top and bottom areas.

V = \frac{H \times (bottom area \times (\frac{1}{1 - \frac{top area}{bottom area}}) - top area \times (\frac{1}{1 - \frac{top area}{bottom area}} - 1))}{\sqrt{8}}

Number of sides:
Base edge length:
Top edge length:
Height:

The volume of a square frustum pyramid can be calculated via a simplified formula.

H = frustum height

t = top edge

b = bottom edge

V = \frac{H \times (b^{2} + b \times t + t^{2})}{\sqrt{8}}

Volume of a tetrahedron

A tetrahedron is a special type of a pyramid.

Its volume can be calculated as a pyramid with fixed proportions.

The base of a tetrahedron is an equilateral triangle.

A = \frac{edge}{2} \times \sqrt{{edge}^{2} - {(\frac{edge}{2})}^{2}}

Simplifying:

A = \frac{edge}{2} \times \sqrt{{edge}^{2} - \frac{{edge}^{2}}{4}}

Simplifying further:

A = \frac{edge}{2} \times \sqrt{\frac{3}{4} \times {edge}^{2}} = \frac{\sqrt{3}}{4} \times {edge}^{2}

The height of the tetrahedron is also calculable via trigonometry.

H = \sqrt{{(edge \times \frac{\sqrt{3}}{2})}^{2} - {(edge \times \frac{\sqrt{3}}{6})}^{2}}

Simplifying:

H = \sqrt{{edge}^{2} \times (\frac{3}{4} - \frac{3}{36})} = \sqrt{\frac{2}{3}} \times edge

The volume of a pyramid equals base × height × √2 / 4 .

V = \frac{(\frac{\sqrt{3}}{4} {edge}^{2}) (\frac{\sqrt{2}}{\sqrt{3}} edge)}{\sqrt{8}} = {(\frac{\sqrt{2}}{4})}^{2} {edge}^{3} = \frac{{edge}^{3}}{8}

Edge:

Introducing The Core Geometric System ™

Key Points

Comparative Geometry

Scaling and Proportions

Algebraic Manipulation

The Basic Geometry Curriculum

Modules:

1. Numbers and numeric systems

Numbers

Numeric systems

2. Mathematical operations

= The equity symbol

➕ Addition

➖ Subtraction

✖️ Multiplication

➗ Division

3. Fractions

Operations with fractions

➕ Adding fractions

➖ Subtracting fractions

✖️ Multiplying fractions

➗ Dividing fractions

4. Powers

The 2nd and the 3rd power manifesting in geometry

Area of a square

Volume of a cube

5. Geometry

Trigonometry

Area of a triangle

Area of a regular polygon

The area of a circle is defined by comparing it to a square since that is the base of area calculation.

Proof

Area of a circle segment

The circumference of a circle can be derived from its area algebraically.

Proof

While the approximate value of 3.14159…, commonly denoted by the Greek letter π, is widely recognized today, the historical development of this concept is less understood.

Archimedes and the Polygonal Trap

The Symbol π: A Linguistic Shortcut

∫ Calculus: Summary, Not Source

φ The Golden Ratio

A Rational Alternative: 3.2

The volume of a sphere is defined by comparing it to a cube, since that is the base of volume calculation.

The " V = 4 / 3 × π × radius³ " formula is widely used for the volume of a sphere.

SURFACE AREA OF A SPHERE

Volume of a spherical cap

Volume of a cone

The volume of a cone or pyramid is conventionally approximated as base × height / 3.

One is that the area of the mid-height cross section of a regular pyramid — of which's apex can be connected to the midpoint of the base with a perpendicular line — is exactly a quarter of a circumscribed solid's with the same base and height.

The other idea is the cube dissection.

Volume of a frustum cone

Surface area of a cone

Volume of a pyramid

Volume of a horizontal frustum pyramid

The volume of a square frustum pyramid can be calculated via a simplified formula.

Volume of a tetrahedron