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Regular pentagon calculator

- By Dr. Minas E. Lemonis, PhD - Updated: April 24, 2020

This tool calculates the basic geometric properties of a regular pentagon. Regular polygons are equilateral (all sides equal) and their angles are equal too. The tool can calculate the properties of the pentagon, given either the length of its side, or the inradius or the circumradius or the area or the height or the width. Enter below the shape dimensions. The calculated results will have the same units as your input. Please use consistent units for any input.

Known data:
Geometric properties:
Area =
Perimeter =
α =
Ri =
Rc =

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Bounding box:
Height h =
Width w =
Angles :
Interior φ =
Central θ =
shape details
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Table of Contents
See also
Properties of Hexagon
Properties of Heptagon
Properties of Octagon
Properties of Decagon
Properties of a N-gon
All Geometric Shapes

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Theoretical background

Table of contents

Definitions

Pentagon is a polygon with five sides and five vertices. A pentagon may be either convex or concave, as depicted in the next figure. When convex, the pentagon (or any closed polygon in that matter) does have all its interior angles lower than 180°. A concave polygon, to the contrary, does have one or more of its interior angles larger than 180°. A pentagon is regular when all its sides and interior angles are equal. Having only the sides equal is not adequate, because the pentagon can be concave with equal sides. In that case the pentagon is called equilateral. The next figure illustrates the classification of pentagons, also presenting equilateral ones that are concave. Any pentagon that is not regular is called irregular

Types of pentagons

The sum of the internal angles of a pentagon is constant and equal to 540°. This is true for either regular or irregular pentagons, convex or concave. It can be easily proved by decomposing the pentagon to individual, non overlapping triangles. If we try to draw straight lines between all vertices, avoiding any intersections, we divide the pentagon into three individual triangles. There are many different ways to draw lines between the vertices, resulting in different triangles, however their count is always three. In a single triangle the sum of internal angles is 180°, therefore, for 3 triangles, positioned side by side, the internal angles should measure up to 3x180°=540°.

A pentagon can be divided into three triangles

Properties of regular pentagons

Symmetry

A regular pentagon has five axes of symmetry. Each one of them passes through a vertex of the pentagon and the middle of the opposite edge, as shown in the following drawing. All axes of symmetry intersect at a common point, the center of the regular pentagon. This is in fact its center of gravity or centroid.

Axes of symmetry of regular pentagon

Interior angle and central angle

By definition the interior angles of a regular pentagon are equal. It is also a common property of all pentagons that the sum of their interior angles is always 540°, as explained previously. Therefore, the interior angle, \varphi , of a regular pentagon should be 108°:

\varphi={540^\circ\over5}=108^\circ

Five identical, isosceles, triangles are defined if we draw straight lines from the center of the regular pentagon towards each one of its vertices. The central angle, \theta , of each triangle is:

\theta={360^\circ\over5}=72^\circ

Focusing on one of the five triangles, its two remaining angles are identical and equal to 54°, so that the sum of all angles in the triangle is 180°, (72°+54°+54°). That is also the half of the interior angle \varphi , (108°/2=54°). It is not coincidence that the sum of interior and central angles is 180°:

\varphi+\theta=108^\circ+72^\circ=180^\circ

In other words \varphi and \theta are supplementary.

Interior and central angle of a regular pentagon
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The regular pentagon is divided into five identical isosceles triangles having a common vertex, the polygon center.

Circumcircle and incircle

It is possible to draw a circle that passes through all the five vertices of the regular pentagon . This is the so called cirmuscribed circle or circumcircle of the regular pentagon (indeed this is a common characteristic of all regular polygons). The center of this circle is also the center of the pentagon, where all the symmetry axes are intersecting also. The radius of circumcircle, R_c , is usually called circumradius.

Another circle can also be drawn, that touches tangentially all five edges of the regular pentagon at the midpoints (also a common characteristic of all regular polygons). This is the so called inscribed circle or incircle. Its center is the same with the center of the circumcircle and it is tangent to all five sides of the regular pentagon. The radius of incircle, R_i , is usually called inradius.

The following figure depicts both circumscribed circle of the regular pentagon and the inscribed one.

Circumcircle and incircle of a regular pentagon

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We will try to find the relationships between the side length a of the regular pentagon and its circumradius R_c and inradius R_i .To this end, we will examine the triangle with sides the circumradius, the inradius and half the pentagon edge, as highlighted in the figure below. This is a right triangle since by definition the incircle is tangential to all sides of the polygon.

Using basic trigonometry we find:

\begin{split} R_c & = \frac{a}{2 \sin{\frac{\theta}{2}}} \\ R_i & = \frac{a}{2 \tan{\frac{\theta}{2}}} \\ R_i & = R_c \cos{\frac{\theta}{2}} \end{split}

where \theta the central angle and a the side length. It turns out that these expressions are valid for any regular polygon, not just the pentagon. We can obtain specific expression for the regular pentagon by setting θ = 72°. These expressions are:

\begin{split} R_c & = \frac{a}{2 \sin{36^{\circ}}} \approx 0.851 a \\ R_i & = \frac{a}{2 \tan{36^{\circ}}} \approx 0.688 a \\ \\ R_i & = R_c \cos{36^{\circ}} \approx 0.809 R_c \end{split}

Area and perimeter

In order to find the area of a regular pentagon we have take into account that its total area is divided into five identical isosceles triangles. All. these triangle have one side a and two sides R_c , while their height, cast from the vertex lying at the pentagon center, is equal to R_i (remember that the incircle is tangential to all sides of the pentagon touching them at their midpoints). The area of each triangle is then: \frac{1}{2}a R_i . Therefore, the total area of the five triangles is found:

A = 5\frac{1}{2} a R_i = {5\over2}a\frac{a}{2 \tan{36^{\circ}}}\Rightarrow

A=\frac{5a^2}{4 \tan{36^{\circ}}}

An approximation of the last relationship is:

A \approx 1.720 a^2

The perimeter of any N-sided regular polygon is simply the sum of the lengths of all sides: P = N a . Therefore, for the regular pentagon :

P = 5a

Bounding box

The bounding box of a planar shape is the smallest rectangle that encloses the shape completely. For the regular pentagon the bounding box may be drawn intuitively, as shown in the next figure, but its exact dimensions need some calculations.

Height

The height h of the regular pentagon is the distance from one of its vertices to the opposite edge. It is indeed perpendicular to the opposite edge and passes through the center of the pentagon. By definition though the distance from the center to a vertex is the circumradius R_c of the pentagon while the distance from the center to an edge is the inradius R_i . Therefore the following expression is derived:

h=R_c+R_i

It is possible to express the height h in terms of the circumradius R_c , or the inradius R_i or the side length a , using the respective analytical expressions for these quantities. The following formulas are derived:

h=R_c\left(1+\cos(\theta/2)\right)

h=R_i\left(1+{1\over \cos(\theta/2)}\right)

h={a\over2}{1+\cos(\theta/2)\over\sin(\theta/2)}

where \theta=72^\circ .

Substituting the value of \theta to the last expressions we get the following approximations:

h\approx 1.809 R_c

h\approx 2.236 R_i

h\approx 1.539 a

Width

The width w is the distance between two opposite vertices of the regular pentagon (the length of its diagonal). In order to find this distance we will employ the right triangle highlighted with dashed line in the figure above. The hypotenuse of the triangle is actually the side length of the pentagon, which is a . Also, one of the triangle angles is supplementary to the adjacent interior angle \varphi of the pentagon. It has been explained before, though, that the supplementary of \varphi is indeed the central angle \theta . Therefore, we may find the length w_1 of the triangle side:

w_1=a \cos\theta

Finally, we can determine the total width w by adding twice the length w_1 to the side length a (due to symmetry the triangle to the right of the pentagon is identical to the one examined).

w=a+2a \cos\theta

Substituting, \theta=72^\circ we get an approximation of the last formula:

w=1.618a

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The diagonal of a regular pentagon is related through the golden ratio with its side

How to draw a regular pentagon

You can draw a regular pentagon given the side length a , using simple drawing tools. Follow the steps described below:

  1. First draw a linear segment with length a , equal to the desired pentagon side length.
  2. Extend the linear segment to the left.
  3. Construct a circular arc, with center point at the right end of the linear segment and radius equal to the segment length.
  4. Repeat the last step, changing the center-point at the left end of the linear segment. Radius is the same.
  5. Draw a line, perpendicular to the linear segment a , passing through the intersection of the two arcs. It crosses the linear segment at its middle.
  6. Also draw a line, perpendicular to the linear segment, passing through left end of the linear segment a . Mark the intersection point with the circular arc (the one drawn at step 4)
  7. Draw another circular arc, by placing one needle of the compass at the middle of the linear segment a , (that was found in step 5) and the drawing tip at the intersection marked in step 6. Rotate the compass, until it crosses the extension of the linear segment, drawn in step 2. Mark this new intersection too.
  8. Draw another circular arc, by placing one needle of the compass at the right end of the linear segment a and the drawing tip at the intersection marked in step 7. Rotate the compass clockwise. Mark two intersections, one with the arc, drawn in step 4, and the other with the line, drawn in step 5. These are two vertices of the pentagon.
  9. Placing the compass needle at the 2nd intersection, and the drawing tip at the 1st one (both intersections marked in the last step) draw a circular arc until it crosses the arc, drawn in step 3. Mark this new intersection, which is a vertex of the pentagon.
  10. The two ends of the linear segment a , as well as the three intersections marked at steps 8 and 9 are the five vertices of the regular pentagon. Draw linear segments between them to construct the final shape.

The following figure illustrates the drawing procedure step by step.

Drawing a regular pentagon given its side length a

Note, that the described procedure is not strictly a by “ruler and compass” construction. In steps 5 and 6, a triangle was used in order to draw perpendicular lines from points of another line. This was selected for simplicity and in order to shorten the number of required steps. Drawing a perpendicular line, is a straightforward geometric construction, using ruler and compass alone, and one could replace the use of triangle in steps 5 and 6, if a strict geometric drawing by “ruler and compass” is required.

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Examples

Example 1

Determine the circumradius, the inradius and the area of a regular pentagon, with side length a=5''

We will use the exact analytical expressions for the circumradius and the inradius, in terms of the side length a , that have been described in the previous sections. These are:

R_c = \frac{a}{2 \sin{36^{\circ}}}

R_i = \frac{a}{2 \tan{36^{\circ}}}

Since the side length a is given, all we have to do is substitute its value 5'' to these expressions. The circumradius is:

R_c= \frac{5''}{2 \sin{36^{\circ}}}\approx 4.253'' ,

and the inradius:

R_i= \frac{5''}{2 \tan{36^{\circ}}}\approx 3.441'' .

The area of a regular pentagon is also given in terms of the side length a , by this formula:

A = \frac{5a^2}{4 \tan{36^{\circ}}}

Substituting a=5'' we find:

A = \frac{5\ (5'')^2}{4 \tan{36^{\circ}}} \approx 43.0\ \text{in}^2

Example 2

What is the diameter of the biggest regular pentagon that can be fitted inside a:

  1. circle, with diameter 25''
  2. square with side 25''
1. Fitting a regular pentagon in a circle

The biggest regular pentagon to fit inside a circle should touch the circle with all its vertices. In other words, the circle must be the circumcircle of the pentagon and as a result its radius should be the circumradius:

R_c=\frac{25''}{2}=12.5''

However, the circumradius is related to the side length a through the formula:

R_c=\frac{a}{2 \sin{\frac{\theta}{2}}}

Therefore:

a=R_c\ 2 \sin{\frac{\theta}{2}}

From the last equation we can calculate the required side length a , if we substitute the values of R_c=12.5'' and \theta=72^\circ :

a=12.5'' \times 2 \sin{\frac{72^\circ}{2}}=14.69''

2. Fitting a regular pentagon is a square

The height h and the width w of the regular polygon are approximated by the following expressions:

h\approx 1.539 a

w\approx 1.618 a

From these approximations, it is apparent, that the width is actually, the biggest of the two dimensions. Therefore, the biggest regular pentagon, to fit inside a square, should be limited by its width alone. In other words, the width of the pentagon must be equal to side of the square:

w=25''

However, the width of the regular pentagon is related to the side length a with the formula:

w=a+2a \cos\theta

Therefore:

a=\frac{w}{1+2\cos\theta}

From the last equation we can calculate the required side length a , if we substitute the values of w=25'' and \theta=72^\circ :

a=\frac{25''}{1+2\cos72^\circ}\approx 15.45''

Regular pentagon cheat-sheet

In the following table a concise list of the main formulas, related to the regular pentagon is included. Also some approximations that may prove handy for practical problems are listed too.

Regular pentagon quick reference

Circumradius: R_c=\frac{a}{2 \sin(36^\circ)}
Inradius: R_i=\frac{a}{2 \tan(36^\circ)}
Height: h=R_c+R_i
Width: w=a+2a \cos(72^\circ)
Area: A=\frac{5a^2}{4 \tan(36^\circ)}
Interior angle: \varphi=108^\circ
Central angle: \theta=72^\circ
Approximations:

R_c \approx 0.851a

R_i \approx 0.688a

h \approx 1.539a

w \approx 1.618a

A\approx 1.720 a^2

See also

Properties of Hexagon
Properties of Heptagon
Properties of Octagon
Properties of Decagon
Properties of a N-gon
All Geometric Shapes

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See also
Properties of Hexagon
Properties of Heptagon
Properties of Octagon
Properties of Decagon
Properties of a N-gon
All Geometric Shapes