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A Formula for Transforming Square Formulas Into Circle Formulas

·6 mins

Let us assume that we do not know anything about circles besides the following equation:

\[ \begin{align} S = C \cdot\dfrac{\pi}{2D} \mid s=2r \end{align} \tag{1} \]

Where \(\textit{S}\) stands for the formula of a sphere or circle, \(\textit{C}\) stands for the formula of a cube or square, \(\textit{D}\) stands for the number of dimensions in the geometric shape, \(\textit{s}\) stands for the side length of the square, and \(\textit{r}\) stands for the radius of the circle or sphere.

If the circle is inscribed in the square, then the length of the square’s side will be \(s = 2r\) as shown in Figure 1.

s = 2rs
Figure 1: Illustration of a circle inscribed in a square.

The logic for designing the equation 1 is simple: as long as the dimensions of a square or a cube are given, the dimensions of its inscribed circle or sphere can be calculated. Analyzing 2-dimensionally, squares have a larger area than their inscribed circles because circles are missing four corners; however, the missing area of those four corners is accounted for in equation 1.

Applying the Formula #

Equation 1 can be used to find the volume, surface area, area, circumference, side length, arc length, and sector area, of spheres and circles.

Volume #

To see how the equation works, consider the formula for the volume of a cube and the number of dimensions a cube has:

\[ C_{volume} = s^{3} \tag{2} \]

\[ D = 3 \tag{3} \]

In order to find the corresponding volume of a sphere inscribed in such a cube, we must plug \(D\), the number of dimensions, into equation 1 as shown below:

\[ \begin{aligned} S_{volume} &= C_{volume} \cdot\dfrac{\pi}{2D} \mid s=2r \newline & = s^{3} \cdot\dfrac{\pi}{2\cdot 3} \mid s=2r \newline & = (2r)^{3} \cdot\dfrac{\pi}{6} \newline & = \dfrac{8\pi r^{3}}{6} \newline & = \dfrac{4\pi r^{3}}{3} \newline \end{aligned} \tag{4} \]

Surface Area #

Consider the formula for the surface area of a cube and the number of dimensions a cube has:

\[ C_{surfaceArea} = 6s^{2} \tag{5} \] \[ D = 3 \tag{6} \]

Then the corresponding surface area of a sphere inscribed in such cube can be found using equation 1, as shown below:

\[ \begin{align*} S_{surfaceArea} &= C_{surfaceArea} \cdot\dfrac{\pi}{2D} \mid s=2r \newline &= 6s^{2} \cdot\dfrac{\pi}{2\cdot 3} \mid s=2r \newline &= 6(2r)^{2} \cdot\dfrac{\pi}{6} \newline &= 4\pi r^{2} \end{align*} \tag{7} \]

Area #

Consider the formula for the area of a square (the cross-section of a cube) and the number of dimensions a square has:

\[ C_{area} = s^{2} \tag{8} \]

\[ D = 2 \tag{9} \]

Then the corresponding area of a circle (the cross-section of a sphere) inscribed in such a square can be found using equation 1, as shown below:

\[ \begin{align*} S_{area} &= C_{area} \cdot\dfrac{\pi}{2D} \mid s=2r \newline &= s^{2} \cdot\dfrac{\pi}{2\cdot 2} \mid s=2r \newline &= (2r)^{2} \cdot\dfrac{\pi}{4} \newline &= \dfrac{4\pi r^{2}}{4} \newline &= \pi r^{2} \end{align*} \tag{10} \]

Circumference #

Consider the formula for the perimeter of a square and the number of dimensions a square has:

\[ C_{perimeter} = 4s \tag{11} \]

\[ D = 2 \tag{12} \]

Then the corresponding perimeter (circumference) of a circle inscribed in such a square can be found using equation equation 1, as shown below:

\[ \begin{align*} S_{circumference} &= C_{perimeter} \cdot\dfrac{\pi}{2D} \mid s=2r \newline &= 4s \cdot\dfrac{\pi}{2\cdot 2} \mid s=2r \newline &= 4(2r) \cdot\dfrac{\pi}{4} \newline &= 2 \pi r \end{align*} \tag{13} \]

Figure 2 clarifies the formula’s meaning:

4s2πr
Figure 2: Illustration showing that a square's perimeter corresponds to its inscribed circle's circumference.

Side Length #

Consider the formula for the side length of a square and the number of dimensions a square has:

\[ C_{sideLength} = s \tag{14} \]

\[ D = 2 \tag{15} \]

Then the corresponding side length (arc length) of a circle inscribed in such square can be found using equation equation 1, as shown below:

\[ \begin{align*} S_{arcLength} &= C_{sideLength} \cdot\dfrac{\pi}{2D} \mid s=2r \newline &= s \cdot\dfrac{\pi}{2\cdot2} \mid s=2r \newline &= (2r) \cdot\dfrac{\pi}{4} \newline &= \dfrac{1}{2} \pi r \end{align*} \tag{16} \]

This formula’s meaning is clarified in figure 3:

s 1 2 π r
Figure 3: Illustration showing the relation between the side length of a square and the arc length of a circle inscribed in a square.

Admittedly, this function is not very useful without taking into account how much of the circle is inscribed into the square. Luckily, the side length can be expressed as a function of of \(\theta\) to find any arc length.

Functions of \(\theta\) #

It is possible to express equation 1 as functions of \(\theta\), which is required for finding the arc length or the sector area of a circle:

\[ \begin{align*} S &= C \cdot\dfrac{\pi}{2D} \mid s=2r \newline S(\theta) &= C \cdot\dfrac{\pi}{2D} \cdot \dfrac{\theta}{2\pi} \mid s=2r \newline S(\theta) &= C \cdot\dfrac{\theta}{4D} \mid s=2r \end{align*} \tag{17} \]

Arc Length #

The formula for the arc length of a circle can be obtained from the perimeter formula of a square:

\[ C_{perimeter} = 4s \tag{18} \]

\[ D = 2 \tag{19} \]

After the conversion formula is applied as normal, the final result needs to be multiplied by \(\dfrac{\theta}{2\pi}\) to take into account how much of the circle is inscribed into the square, as shown below:

\[ \begin{align*} S_{arcLength} &= C_{perimeter} \cdot\dfrac{\pi}{2D} \cdot \dfrac{\theta}{2\pi} \mid s=2r \newline &= 4s \cdot\dfrac{\pi}{2\cdot 2} \cdot \dfrac{\theta}{2\pi} \mid s=2r \newline &= (2r) \cdot\dfrac{\theta}{2} \newline &= \theta r \end{align*} \tag{20} \]

Sector Area #

The formula for the sector area of a circle can be obtained from the area formula of a square:

\[ C_{area} = s^{2} \tag{21} \]

\[ D = 2 \tag{22} \]

The conversion formula is applied as normal, but the final result is multiplied by \(\dfrac{\theta}{2\pi}\) to take into account how much of the circle is inscribed into the square, as shown below:

\[ \begin{align*} S_{sectorArea} &= C_{area} \cdot\dfrac{\pi}{2D} \cdot \dfrac{\theta}{2\pi} \mid s=2r \newline &= s^{2} \cdot\dfrac{\pi}{2\cdot 2} \cdot \dfrac{\theta}{2\pi} \mid s=2r \newline &= (2r)^2 \cdot\dfrac{\theta}{8} \newline &= 4r^2 \cdot\dfrac{\theta}{8} \newline &= \dfrac{1}{2} \theta r^2 \end{align*} \tag{23} \]

Formulas #

3D Shapes #

\[ \begin{align*} S_{volume} &= C_{volume} \cdot\dfrac{\pi}{2D} \mid s=2r \newline &= s^{3} \cdot\dfrac{\pi}{2\cdot 3} \mid s=2r \newline &= (2r)^{3} \cdot\dfrac{\pi}{6} \newline &= \dfrac{8\pi r^{3}}{6} \newline &= \dfrac{4\pi r^{3}}{3} \newline \end{align*} \]

\[ \begin{align*} S_{surfaceArea} &= C_{surfaceArea} \cdot\dfrac{\pi}{2D} \mid s=2r \newline &= 6s^{2} \cdot\dfrac{\pi}{2\cdot 3} \mid s=2r \newline &= 6(2r)^{2} \cdot\dfrac{\pi}{6} \newline &= 4\pi r^{2} \end{align*} \]

2D Shapes #

\[ \begin{align*} S_{area} &= C_{area} \cdot\dfrac{\pi}{2D} \mid s=2r \newline &= s^{2} \cdot\dfrac{\pi}{2\cdot 2} \mid s=2r \newline &= (2r)^{2} \cdot\dfrac{\pi}{4} \newline &= \dfrac{4\pi r^{2}}{4} \newline &= \pi r^{2} \newline S(\theta)_{sectorArea} &= \pi r^{2} \cdot \dfrac{\theta}{2\pi} \newline &= \dfrac{1}{2} \theta r^{2} \end{align*} \]

\[ \begin{align*} S_{circumference} &= C_{perimeter} \cdot\dfrac{\pi}{2D} \mid s=2r \newline &= 4s \cdot\dfrac{\pi}{2\cdot 2} \mid s=2r \newline &= 4(2r) \cdot\dfrac{\pi}{4} \newline &= 2 \pi r \newline S(\theta)_{arcLength} &= 2 \pi r \cdot \dfrac{\theta}{2\pi} \newline &= \theta r \end{align*} \]