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<h3style="margin:7px">The 2nd and the 3rd power manifesting in geometry</h3>
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<h3style="margin:7px">The 2nd and the 3rd Powers manifesting in Geometry</h3>
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<pstyle="margin:12px">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.
<h3style="margin:7px">Area and volume of basic shapes</h3>
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<h3style="margin:7px">Area and Volume of Basic Shapes</h3>
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<sectionid="triangle">
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<h3style="margin:7px">Area of a triangle</h3>
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<h3style="margin:7px">Area of a Triangle</h3>
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<pstyle="margin:12px">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.
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<sectionid="polygon">
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<h3style="margin:7px">Area of a regular polygon</h3>
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<h3style="margin:7px">Area of a regular Polygon</h3>
In this layout the arcs of the quadrants of an inscribed circle would meet at the midpoints of the sides of the square, leaving some of the square uncovered.
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The arcs of the quadrants of a circumscribed circle would overlap, and meet at the center of the square, covering it all.
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The arcs of the quadrants of a circumscribed circle would overlap, and intersect at the center of the square, covering it all.
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<strong>The arcs of the quadrants of the circle that equals in area to the square intersect right in between those limits, at the quarters on its centerlines.</strong>
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<sectionid="circle-segment">
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<h3style="margin:7px">Area of a circle segment</h3>
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<h3style="margin:7px">Area of a Circle Segment</h3>
<pstyle="margin:12px">For this derivation method to be valid the circumference has to have a thickness greater than 0, by at least the smallest number.
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<pstyle="margin:12px">The circumference of a circle can be derived from its area algebraically with an undefined thickness greater than 0.
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The x represents the theoretical width of the circumference, which is a very small number.
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The traditional method of polygon approximation fails not due to rounding errors, but due to a fundamental divergence of shape that invalidates its own geometric ordering.
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The polygon method attempts to define the perfect circle using imperfect, flawed limits. This destroys the basic geometric ordering that the method is based on, proving it is unsuitable for determining the true circumference to diameter ratio of a circle.
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<details>
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<summary><h4style="margin:7px">
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<summary><h4style="margin:12px">
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What we’re left with is not a proof, but a flawed approximation — one that has shaped centuries of geometry, but now deserves a closer, more rational reexamination.
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</h4></summary>
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While the number of sides is only 3, the perimeter is equal to the circumference, yet the ratio flipped.
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Rather than treating inscribed and circumscribed polygons separately and relying on assumptions about how their perimeter gaps behave as the number of sides increases, we introduce a creative and grounded condition: equal distance between the polygon’s sides, vertices, and the circle’s arc.
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We begin with a strong geometric foundation: the area of a circle is exactly 3.2r². This gives us reason to suspect that the true circumference is 6.4r, not 2πr. To test this, we reframe the polygon approximation method.
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<pstyle="margin:12px">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.</p>
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<summary><h4style="margin:7px">The " V = 4 / 3 × π × radius³ " formula is widely used for the volume of a sphere.</h4></summary>
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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.
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The conventional formula for the surface area of a sphere was allegedly developed from the " volume = 4/3×π×radius³ " formula.
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<pstyle="margin:6px">Advertisement</p>
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<h3style="margin:7px">SURFACE AREA OF A SPHERE</h3>
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<h3style="margin:7px">Surface Area of a Sphere</h3>
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</div>
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<pstyle="margin:6px"><strong>The conventional formula for the surface area of a sphere was allegedly developed from the " volume = 4/3×π×radius³ " formula.
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The real formula for the surface area of a sphere is available for 3.2 billion USD. ( + tax, if applies )</strong>
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<pstyle="margin:6px"><strong>
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The formula for the real surface area of a sphere is available for 3.2 billion USD.
<pstyle="margin:12px">The surface area of a cone is calculated as a circle slice with a radius equal to the slant height and the angle determined by the ratio of the height and the slant height.
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