Geometry

Area of Regular Polygons

Last updated: March 2026 · Intermediate
Before you start

You should be comfortable with:

Real-world applications
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Carpentry

Measurements, material estimation, cutting calculations

A regular polygon is a shape where every side has the same length and every interior angle is equal. Equilateral triangles, squares, regular pentagons, hexagons, and octagons are all regular polygons. These shapes appear everywhere β€” from honeycomb cells to stop signs to decorative floor tiles β€” and calculating their area is a practical skill in construction, design, and engineering.

While you can find the area of triangles and rectangles with simple base-times-height formulas, regular polygons with five or more sides need a different approach. The key concept is the apothem, and once you understand it, a single formula covers every regular polygon.

The Apothem

The apothem of a regular polygon is the perpendicular distance from the center of the polygon to the midpoint of any side. Think of it as the β€œinradius” β€” the radius of the largest circle that fits inside the polygon, touching every side.

For a regular polygon with nn sides of length ss, the apothem is:

a=s2tan⁑ ⁣(Ο€n)a = \frac{s}{2\tan\!\left(\dfrac{\pi}{n}\right)}

The apothem is always shorter than the circumradius (the distance from the center to a vertex). As the number of sides increases, the apothem gets closer to the circumradius, and the polygon looks more like a circle.

The General Area Formula

Every regular polygon can be divided into nn identical isosceles triangles, each with a base equal to the side length ss and a height equal to the apothem aa. Adding up those triangles gives the general formula:

A=12Γ—perimeterΓ—apothem=12Γ—nΓ—sΓ—aA = \frac{1}{2} \times \text{perimeter} \times \text{apothem} = \frac{1}{2} \times n \times s \times a

This formula works for any regular polygon β€” a triangle (n=3n = 3), a square (n=4n = 4), a pentagon (n=5n = 5), a hexagon (n=6n = 6), and beyond. The perimeter is simply nΓ—sn \times s.

Regular Hexagon with Apothem and Side Labeled

ascenterA = Β½ Γ— perimeter Γ— apothemEach triangle has base s and height a

The diagram above shows how the apothem connects the center to the midpoint of a side at a right angle. The dashed lines from the center to two adjacent vertices outline one of the six identical triangles that make up the hexagon.

Area of a Regular Hexagon

The regular hexagon is the most commonly searched polygon area formula. A hexagon has 6 sides, and its apothem is a=s32a = \frac{s\sqrt{3}}{2}. Substituting into the general formula:

A=12Γ—6sΓ—s32=332 s2A = \frac{1}{2} \times 6s \times \frac{s\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}\,s^2

Ahexagon=332 s2β‰ˆ2.598 s2\boxed{A_{\text{hexagon}} = \frac{3\sqrt{3}}{2}\,s^2 \approx 2.598\,s^2}

Hexagons are nature’s favorite shape β€” honeycombs, basalt columns, and turtle shells all use hexagonal geometry because hexagons tile a plane with zero wasted space while enclosing the maximum area for a given perimeter among all tileable shapes.

Example 1: Find the area of a regular hexagon with side length 8 cm.

A=332(8)2=332(64)=19232=963A = \frac{3\sqrt{3}}{2}(8)^2 = \frac{3\sqrt{3}}{2}(64) = \frac{192\sqrt{3}}{2} = 96\sqrt{3}

A=96Γ—1.7321β‰ˆ166.3Β cm2A = 96 \times 1.7321 \approx 166.3 \text{ cm}^2

Answer: The area is approximately 166.3 square centimeters.

Example 2: A hexagonal patio tile has an apothem of 6 inches. Find its area.

If the apothem is 6 in, we first find the side length. For a hexagon, a=s32a = \frac{s\sqrt{3}}{2}, so:

s=2a3=2(6)3=123=1233=43β‰ˆ6.928Β ins = \frac{2a}{\sqrt{3}} = \frac{2(6)}{\sqrt{3}} = \frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3} \approx 6.928 \text{ in}

The perimeter is P=6s=6(43)=243β‰ˆ41.57P = 6s = 6(4\sqrt{3}) = 24\sqrt{3} \approx 41.57 in. Now apply the general formula:

A=12Γ—PΓ—a=12(243)(6)=723β‰ˆ124.7Β in2A = \frac{1}{2} \times P \times a = \frac{1}{2}(24\sqrt{3})(6) = 72\sqrt{3} \approx 124.7 \text{ in}^2

Answer: The area is approximately 124.7 square inches.

Area of a Regular Pentagon

A regular pentagon (5 sides) has the area formula:

Apentagon=145(5+25)β€…β€Šs2β‰ˆ1.720 s2\boxed{A_{\text{pentagon}} = \frac{1}{4}\sqrt{5(5 + 2\sqrt{5})}\;s^2 \approx 1.720\,s^2}

This comes from the general formula with n=5n = 5. The apothem of a regular pentagon is a=s2tan⁑(36Β°)β‰ˆ0.6882 sa = \frac{s}{2\tan(36Β°)} \approx 0.6882\,s.

Example 3: Find the area of a regular pentagon with side length 10 m.

A=145(5+25)Γ—(10)2A = \frac{1}{4}\sqrt{5(5 + 2\sqrt{5})} \times (10)^2

First, compute the constant:

25β‰ˆ4.47212\sqrt{5} \approx 4.4721

5+4.4721=9.47215 + 4.4721 = 9.4721

5Γ—9.4721=47.36055 \times 9.4721 = 47.3605

47.3605β‰ˆ6.8819\sqrt{47.3605} \approx 6.8819

6.88194β‰ˆ1.7205\frac{6.8819}{4} \approx 1.7205

Aβ‰ˆ1.7205Γ—100=172.05Β m2A \approx 1.7205 \times 100 = 172.05 \text{ m}^2

Answer: The area is approximately 172.1 square meters.

Area of a Regular Octagon

A regular octagon (8 sides) has the area formula:

Aoctagon=2(1+2) s2β‰ˆ4.828 s2\boxed{A_{\text{octagon}} = 2(1 + \sqrt{2})\,s^2 \approx 4.828\,s^2}

The apothem of a regular octagon is a=s2tan⁑(22.5Β°)β‰ˆ1.2071 sa = \frac{s}{2\tan(22.5Β°)} \approx 1.2071\,s.

Octagons are familiar as stop signs β€” every standard stop sign in the United States is a regular octagon.

Example 4: A stop sign has sides that are 12.4 inches long. Find its area.

A=2(1+2)(12.4)2A = 2(1 + \sqrt{2})(12.4)^2

A=2(1+1.4142)(153.76)A = 2(1 + 1.4142)(153.76)

A=2(2.4142)(153.76)A = 2(2.4142)(153.76)

A=4.8284Γ—153.76A = 4.8284 \times 153.76

Aβ‰ˆ742.4Β in2A \approx 742.4 \text{ in}^2

Answer: The area of the stop sign is approximately 742.4 square inches (about 5.16 square feet).

Example 5: An octagonal gazebo has an apothem of 6 feet. Find its area.

First, find the side length from the apothem. For an octagon, a=s2tan⁑(22.5°)a = \frac{s}{2\tan(22.5°)}, so:

s=2atan⁑(22.5°)=2(6)(0.41421)=4.971 fts = 2a\tan(22.5°) = 2(6)(0.41421) = 4.971 \text{ ft}

The perimeter is P=8s=8(4.971)=39.77P = 8s = 8(4.971) = 39.77 ft. Now apply the general formula:

A=12Γ—PΓ—a=12(39.77)(6)=119.3Β ft2A = \frac{1}{2} \times P \times a = \frac{1}{2}(39.77)(6) = 119.3 \text{ ft}^2

Answer: The area of the gazebo floor is approximately 119.3 square feet.

Formula Reference Table

PolygonSidesArea FormulaApproximate Multiplier
Equilateral Triangle3A=34 s2A = \frac{\sqrt{3}}{4}\,s^2β‰ˆ0.433 s2\approx 0.433\,s^2
Square4A=s2A = s^2=1.000 s2= 1.000\,s^2
Regular Pentagon5A=145(5+25) s2A = \frac{1}{4}\sqrt{5(5+2\sqrt{5})}\,s^2β‰ˆ1.720 s2\approx 1.720\,s^2
Regular Hexagon6A=332 s2A = \frac{3\sqrt{3}}{2}\,s^2β‰ˆ2.598 s2\approx 2.598\,s^2
Regular Heptagon7A=7s24cot⁑ ⁣(Ο€7)A = \frac{7s^2}{4}\cot\!\left(\frac{\pi}{7}\right)β‰ˆ3.634 s2\approx 3.634\,s^2
Regular Octagon8A=2(1+2) s2A = 2(1+\sqrt{2})\,s^2β‰ˆ4.828 s2\approx 4.828\,s^2
Regular nn-gonnnA=12nsa=ns24cot⁑ ⁣(Ο€n)A = \frac{1}{2}nsa = \frac{ns^2}{4}\cot\!\left(\frac{\pi}{n}\right)Depends on nn

As the number of sides increases, the multiplier grows and the polygon approaches a circle with area Ο€r2\pi r^2.

Real-World Applications

Hexagonal Tiles

Hexagonal tiles are popular in kitchens and bathrooms because they tessellate perfectly β€” they cover a surface with no gaps and no overlaps. To estimate how many tiles you need, calculate the area of one tile and divide it into the total floor area.

Example: A bathroom floor is 60 square feet. Each hexagonal tile has a side length of 4 inches. How many tiles are needed?

Area of one tile: A=332(4)2=332(16)=243β‰ˆ41.57Β in2A = \frac{3\sqrt{3}}{2}(4)^2 = \frac{3\sqrt{3}}{2}(16) = 24\sqrt{3} \approx 41.57 \text{ in}^2

Convert the floor area: 60Β ft2Γ—144Β in2/ft2=8640Β in260 \text{ ft}^2 \times 144 \text{ in}^2/\text{ft}^2 = 8640 \text{ in}^2

Number of tiles: 864041.57β‰ˆ208\frac{8640}{41.57} \approx 208 tiles

In practice, order 10-15% extra for cuts along walls and waste.

Octagonal Stop Signs

The standard US stop sign is a regular octagon with a flat-to-flat width of 30 inches. Knowing the area helps when calculating material costs for sign manufacturing or reflective sheeting.

Pentagonal Structures

Regular pentagons appear in gazebo designs, decorative windows, and architectural details. The Pentagon building in Arlington, Virginia, is the most famous pentagonal structure, though its shape is a concentric set of pentagons rather than a single regular pentagon.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Problem 1: Find the area of a regular hexagon with side length 5 cm.

A=332(5)2=332(25)=7532=37.53A = \frac{3\sqrt{3}}{2}(5)^2 = \frac{3\sqrt{3}}{2}(25) = \frac{75\sqrt{3}}{2} = 37.5\sqrt{3}

Aβ‰ˆ37.5Γ—1.7321β‰ˆ64.95Β cm2A \approx 37.5 \times 1.7321 \approx 64.95 \text{ cm}^2

Answer: β‰ˆ64.95Β cm2\approx 64.95 \text{ cm}^2

Problem 2: A regular pentagon has side length 7 m. Find its area.

Aβ‰ˆ1.720Γ—(7)2=1.720Γ—49β‰ˆ84.3Β m2A \approx 1.720 \times (7)^2 = 1.720 \times 49 \approx 84.3 \text{ m}^2

Using the exact formula:

A=145(5+25)Γ—49=6.88194Γ—49β‰ˆ1.7205Γ—49β‰ˆ84.3Β m2A = \frac{1}{4}\sqrt{5(5 + 2\sqrt{5})} \times 49 = \frac{6.8819}{4} \times 49 \approx 1.7205 \times 49 \approx 84.3 \text{ m}^2

Answer: β‰ˆ84.3Β m2\approx 84.3 \text{ m}^2

Problem 3: A regular octagonal table top has sides of 1.5 feet. Find its area.

A=2(1+2)(1.5)2=2(2.4142)(2.25)A = 2(1 + \sqrt{2})(1.5)^2 = 2(2.4142)(2.25)

A=4.8284Γ—2.25β‰ˆ10.86Β ft2A = 4.8284 \times 2.25 \approx 10.86 \text{ ft}^2

Answer: β‰ˆ10.86Β ft2\approx 10.86 \text{ ft}^2

Problem 4: A regular hexagon has an apothem of 10 inches. Find its area using the general formula.

First, find the side length: s=2a3=2(10)3=2033β‰ˆ11.547s = \frac{2a}{\sqrt{3}} = \frac{2(10)}{\sqrt{3}} = \frac{20\sqrt{3}}{3} \approx 11.547 in

Perimeter: P=6s=6Γ—11.547β‰ˆ69.28P = 6s = 6 \times 11.547 \approx 69.28 in

Area: A=12Γ—PΓ—a=12(69.28)(10)=346.4Β in2A = \frac{1}{2} \times P \times a = \frac{1}{2}(69.28)(10) = 346.4 \text{ in}^2

Answer: β‰ˆ346.4Β in2\approx 346.4 \text{ in}^2

Problem 5: A regular heptagon (7 sides) has side length 6 cm. Use the general formula to find its area. (Hint: the apothem is a=s2tan⁑(Ο€/7)a = \frac{s}{2\tan(\pi/7)}. Use tan⁑(180Β°/7)=tan⁑(25.714Β°)β‰ˆ0.4816\tan(180Β°/7) = \tan(25.714Β°) \approx 0.4816.)

Apothem: a=62(0.4816)=60.9632β‰ˆ6.229a = \frac{6}{2(0.4816)} = \frac{6}{0.9632} \approx 6.229 cm

Perimeter: P=7Γ—6=42P = 7 \times 6 = 42 cm

A=12(42)(6.229)=12(261.62)β‰ˆ130.8Β cm2A = \frac{1}{2}(42)(6.229) = \frac{1}{2}(261.62) \approx 130.8 \text{ cm}^2

Answer: β‰ˆ130.8Β cm2\approx 130.8 \text{ cm}^2

Key Takeaways

  • A regular polygon has all sides equal and all angles equal
  • The apothem is the perpendicular distance from the center to the midpoint of a side
  • The universal formula A=12Γ—perimeterΓ—apothemA = \frac{1}{2} \times \text{perimeter} \times \text{apothem} works for every regular polygon
  • Hexagon: A=332 s2β‰ˆ2.598 s2A = \frac{3\sqrt{3}}{2}\,s^2 \approx 2.598\,s^2
  • Pentagon: A=145(5+25) s2β‰ˆ1.720 s2A = \frac{1}{4}\sqrt{5(5+2\sqrt{5})}\,s^2 \approx 1.720\,s^2
  • Octagon: A=2(1+2) s2β‰ˆ4.828 s2A = 2(1+\sqrt{2})\,s^2 \approx 4.828\,s^2
  • As the number of sides increases, a regular polygon approaches a circle β€” and its area approaches Ο€r2\pi r^2
  • These formulas apply to real-world objects: hexagonal tiles, octagonal stop signs, and pentagonal architectural features

Return to Geometry for more topics in this section.

Last updated: March 28, 2026