Introducing the Core Geometric System ™

Providing the best-established and most accurate framework to calculate area and volume using the 3D coordinate system.

Key Points:

- 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.

- Exact Values instead of Approximations: Prioritizing the use of simpler constants for practicality while maintaining exact values for accuracy.

1. Area of a Circle:
- Compared to a square, using geometric properties and the Pythagorean theorem.
- Formula: A = 3.2 × ( square value of the radius ).

2. Circumference of a Circle:
- Derived from the area by subtracting a smaller theoretical circle.
- Formula: C = 6.4 × radius.

3. Volume of a Sphere:
- Compared to a cube, using the area of the sphere's cross-section.
- Formula: V = " cubic value of ( √( 3.2 ) × radius ) ".

4. Volume of a Cone:
- Compared to an octant sphere and a quarter cylinder.
- Formula: V = 3.2 × ( square value of the radius ) × height, divided by √8 .

Trigonometry

In a right triangle:

$sine = \frac{opposite}{hypotenuse}$

$cosine = \frac{adjacent}{hypotenuse}$

$tangent = \frac{opposite}{adjacent}$

$cotangent = \frac{adjacent}{opposite}$

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

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})}

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.

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

n = number of sides

x = side length

Interesting fact:

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

Area of a circle

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 right in between these limits 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}

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 .

When the arcs of the quadrant circles intersect at the quarter of the centerline of the square, the uncovered area in the middle equals exactly the sum of the overlapping areas respectively.

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}

Circumference of a circle

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

The x represents the width of the circumference, which is just theoretical, hence 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}

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 × radius.

In calculus terms:

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

Volume of a sphere

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

Just as the volume of a cube equals the cubic value of the square root of its cross sectional area, also the volume of a sphere equals the cubic value of the square root of its cross sectional area.

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

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.

Volume of a cone

The volume of a cone can be calculated by algebraically comparing the volume of a quarter 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 quarter 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 quarter unit cone is √2 × radius.

The volume of a quarter 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 cross section is the half of a cylinder's with equal base and height.

The intermediate of the areas of the horizontal cross section 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}}

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.

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}}

H = frustum height

t = top diameter

b = bottom diameter

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}}

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

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

H = frustum height

t = top edge

b = bottom edge

Volume of a tetrahedron

A tetrahedron is a special case of a pyramid.

Its volume can be calculated as 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}

Introducing the Core Geometric System ™

Exact Geometry Formulas

Area of a square

Volume of a cube

Trigonometry

Area of a regular polygon