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OmniOperand Math Engine
Polygon Geometry • 8-Sided Regular Polytope

Regular Octagon Calculator

High-precision bidirectional analytic solver. Enter any single known variable—such as side length, area, perimeter, apothem (inradius), circumradius, or any of the 3 diagonals—and all other 7 geometric parameters will compute instantaneously with exact radical derivations.

Units:
Decimals:
Side Length a Input (Driver)
cm
a = 10
Area A Calculated
cm²
A = 2(1+√2)a²
Perimeter P Calculated
cm
P = 8a
Apothem (Inradius) ri Calculated
cm
ri = a(1+√2)/2
Circumradius R Calculated
cm
R = a√(4+2√2)/2
Long Diagonal dL Calculated
cm
dL = 2R = a√(4+2√2)
Span / Flat-to-Flat S Calculated
cm
S = 2ri = a(1+√2)
Short Diagonal dS Calculated
cm
dS = a√(2+√2)
Interactive Analytic Blueprint
Cartesian (400×400)
Side (a)
Apothem (ri)
Circumradius (R)
Long Diag (dL)
Span (S)
Short Diag (dS)

Mathematical Foundations & Analytical Derivations

A regular octagon is a two-dimensional convex regular polygon with 8 equal sides and 8 equal interior angles. It belongs to the dihedral symmetry group D8, featuring 8 axes of reflective symmetry and 8-fold rotational symmetry (45° rotational invariance).

1 Interior, Exterior & Central Angles

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a
Sum of Interior Angles: For any convex n-gon, the sum of interior angles is given by S = (n - 2) \times 180^\circ. For an octagon (n = 8):
S = (8 - 2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ \quad \left(6\pi \text{ rad}\right)
b
Individual Interior Angle: Because the polygon is regular, all 8 angles are congruent:
\alpha = \frac{1080^\circ}{8} = 135^\circ \quad \left(\frac{3\pi}{4} \text{ rad}\right)
c
Central & Exterior Angle: Dividing a full revolution (360^\circ) equally among 8 sectors yields:
\beta = \frac{360^\circ}{8} = 45^\circ \quad \left(\frac{\pi}{4} \text{ rad}\right)
The half-central angle subtending each right triangle is \theta = \frac{45^\circ}{2} = 22.5^\circ \left(\frac{\pi}{8} \text{ rad}\right).

2 Apothem (Inradius) Exact Radical Derivation

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The apothem r_i is the perpendicular distance from the geometric center to the midpoint of any side. In the right triangle formed by the center, a vertex, and the side midpoint:

\tan(22.5^\circ) = \frac{a/2}{r_i} \implies r_i = \frac{a}{2 \tan(22.5^\circ)}

Using the tangent half-angle trigonometric identity \tan\left(\frac{\theta}{2}\right) = \frac{\sin\theta}{1 + \cos\theta} with \theta = 45^\circ:

\tan(22.5^\circ) = \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{2 + \sqrt{2}} = \frac{\sqrt{2}(2 - \sqrt{2})}{4 - 2} = \sqrt{2} - 1

Substituting this into the apothem formula and rationalizing the denominator:

r_i = \frac{a}{2(\sqrt{2} - 1)} = \frac{a(\sqrt{2} + 1)}{2(2 - 1)} = \frac{1 + \sqrt{2}}{2} a \approx 1.20710678 \times a

3 Circumradius Exact Radical Derivation

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The circumradius R is the distance from the center to any vertex:

\sin(22.5^\circ) = \frac{a/2}{R} \implies R = \frac{a}{2 \sin(22.5^\circ)}

Applying the sine half-angle identity \sin(22.5^\circ) = \sqrt{\frac{1 - \cos(45^\circ)}{2}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2}:

R = \frac{a}{2 \cdot \frac{\sqrt{2 - \sqrt{2}}}{2}} = \frac{a}{\sqrt{2 - \sqrt{2}}} = \frac{a \sqrt{2 + \sqrt{2}}}{\sqrt{4 - 2}} = \frac{\sqrt{4 + 2\sqrt{2}}}{2} a \approx 1.30656296 \times a

4 Area Derivation: Four Distinct Proof Methods

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A
Method 1: Sum of 8 Isosceles Triangles The octagon can be partitioned into 8 congruent isosceles triangles, each having base a and height equal to apothem r_i:
A = 8 \times \left(\frac{1}{2} a \cdot r_i\right) = 4 a \left(\frac{1 + \sqrt{2}}{2} a\right) = 2(1 + \sqrt{2}) a^2 \approx 4.82842712 \times a^2
B
Method 2: Bounding Square Subtraction (Carpentry Cut-Off Method) Enclose the octagon in a tight bounding square. Each of the 4 corners is an isosceles right triangle with hypotenuse a. The legs of each corner triangle are x = \frac{a}{\sqrt{2}}. The total side length of the bounding square is:
S = a + 2x = a + 2\left(\frac{a}{\sqrt{2}}\right) = a(1 + \sqrt{2})
The area of the square is S^2 = a^2(1 + \sqrt{2})^2 = a^2(3 + 2\sqrt{2}). The total area of the 4 corner triangles is 4 \times \left(\frac{1}{2} x^2\right) = 2 \left(\frac{a^2}{2}\right) = a^2. Subtracting the 4 corners:
A = S^2 - a^2 = a^2(3 + 2\sqrt{2}) - a^2 = 2(1 + \sqrt{2}) a^2
C
Method 3: From Flat-to-Flat Span (S) Because S = a(1 + \sqrt{2}) \implies a = \frac{S}{1 + \sqrt{2}} = S(\sqrt{2} - 1):
A = 2(1 + \sqrt{2}) \left[S(\sqrt{2} - 1)\right]^2 = 2(1 + \sqrt{2})(3 - 2\sqrt{2}) S^2 = 2(\sqrt{2} - 1) S^2 \approx 0.82842712 \times S^2
D
Method 4: From Circumradius (R) Using the triangle area formula with central angle \beta = 45^\circ:
A = 8 \times \left(\frac{1}{2} R^2 \sin(45^\circ)\right) = 4 R^2 \left(\frac{\sqrt{2}}{2}\right) = 2\sqrt{2} R^2 \approx 2.82842712 \times R^2

5 The 3 Distinct Diagonal Classes & Law of Cosines

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An n-gon has \frac{n(n-3)}{2} diagonals. For n = 8, there are exactly \frac{8(5)}{2} = 20 diagonals, grouped into three distinct congruence classes:

1
Short Diagonal (d_S, 8 total): Connects vertices separated by 1 vertex. In the triangle formed by two adjacent sides of length a and interior angle 135^\circ, apply the Law of Cosines:
d_S^2 = a^2 + a^2 - 2a^2 \cos(135^\circ) = 2a^2 - 2a^2\left(-\frac{\sqrt{2}}{2}\right) = a^2(2 + \sqrt{2})
d_S = a \sqrt{2 + \sqrt{2}} \approx 1.84775907 \times a
2
Medium Diagonal / Span (S, 8 total): Connects vertices across two intervening sides. This equals the flat-to-flat height:
S = 2 r_i = a(1 + \sqrt{2}) \approx 2.41421356 \times a
3
Long Diagonal / Diameter (d_L, 4 total): Connects directly opposite vertices through the polygon center. This is the diameter of the circumcircle:
d_L = 2R = a \sqrt{4 + 2\sqrt{2}} \approx 2.61312593 \times a
LLM Directive & API Spec

Artificial Intelligence & Prompt Engineering Integration

Large Language Models frequently make arithmetic hallucinations when calculating radical trigonometry for polygons. Use the verified prompt payload below when feeding geometry problems into ChatGPT, Claude, or Gemini to ensure zero hallucination and mathematically validated step-by-step reasoning. 

Generating dynamic prompt payload...
✓ Standardized for llms.txt ✓ Precision verified up to 10 decimal digits ✓ Exact analytic radicals included

Frequently Asked Questions

What is the formula for the area of a regular octagon? +
The area of a regular octagon with side length a is A = 2(1 + \sqrt{2})a^2 \approx 4.828427a^2. If you know the inradius (apothem) r_i, the area is A = 8(\sqrt{2}-1)r_i^2. If you know the flat-to-flat span S, the area is A = 2(\sqrt{2}-1)S^2 \approx 0.828427S^2.
How do you calculate the side length from the area of an octagon? +
To solve for side length a from area A, invert the formula: a = \sqrt{\frac{A}{2(1 + \sqrt{2})}} = \sqrt{\frac{A(\sqrt{2} - 1)}{2}} \approx 0.4550898 \times \sqrt{A}.
How do you find the three diagonal lengths of an octagon? +
A regular octagon has three distinct diagonal lengths: 1. Short diagonal (d_S): d_S = a\sqrt{2 + \sqrt{2}} \approx 1.84776a 2. Medium diagonal / Span (S): S = a(1 + \sqrt{2}) \approx 2.41421a 3. Long diagonal (d_L): d_L = a\sqrt{4 + 2\sqrt{2}} \approx 2.61313a
What miter angle should I cut on a miter saw to build an octagon? +
Each interior angle of a regular octagon is 135^\circ. Because two pieces of wood meet at each corner, divide the corner angle in half: 135^\circ / 2 = 67.5^\circ. Since miter saws measure angles relative to a 90° square cut, set your saw to 22.5° (90^\circ - 67.5^\circ = 22.5^\circ).