A dodecagon is a polygon with twelve sides and twelve angles. Dodecagon may be a convex or a concave polygon.
The above figure is a dodecagon with twelve vertices named from A through L. In the above dodecagon, the sides are of varying length. It is to be noted that the sides need not be of the same length.

## Regular Dodecagon

A Regular Dodecagon is a dodecagon in which all twelve sides are of equal lengths. As all the sides are of equal length, all the angles will be equal.
Each interior angle is equal to 150o. Sum of interior angles = 1800o.

The above polygon is a regular decagon with each side is of 2 unit length.

We can construct a dodecagon by following the steps given hereunder:
Step 1: Draw a straight line and take a point O on it. Draw a circle with center O. Let A and B be the points where the line intersects the circle. Then, AB is the diameter and hence OA=OB=r, the radius of the circle.

Step 2: Draw an arc of radius equal to the diameter 2r with centers at A and B. Let the intersecting points of the arcs be X and Y.

Step 3: Join X and Y with a straight edge. The line segment $\overline{XY}$ will be perpendicular to the line segment $\overline{AB}$. Suppose that the line segment $\overline{XY}$ intersects the circle at the points C and D.

Step 4: Draw arcs with radius equal to r with centers A, B, C, and D. Let the intersecting points of the arcs with the circle be V1, V2, V3, V4, V5, V6, V7 and V8. Join each point (including A, B, C, D) with its adjacent point using a straight edge.

By removing the arcs and circumscribed circle, we get a dodecagon.

## Irregular Dodecagon

A dodecagon in which sides are not equal is called an irregular dodecagon. As the sides are not equal, the angles in an irregular dodecagon are not equal.

## Dodecagon Angles

Each interior angle of a dodecagon will be 150o. Now, we are interested in finding the sum of angles of a dodecagon. Let us do the following construction to decide it and let us verify against the value obtained by the formula for the sum of angles in a polygon of n sides. Take any point X in the interior of the dodecagon, and join it to the vertices of the dodecagon. Then, we form twelve triangles as shown below. The triangles need not be similar, but it is known that the sum of the three angles in a triangle is always 180°.

As we have twelve triangles, the sum of the angles of the twelve triangles be 12×180°=2160°. But, the angles at the point X is not part of the sum of the angles of the dodecagon and the sum of angles at X is equal to one complete angle, 360°. Thus,
Sum of angles of a Dodecagon=2160°-360°=1800°For a polygon with n sides,
Sum of Angles= (n-2)180°
For a Dodecagon, n=12. So,
Sum of angles of a Dodecagon = (n-2) 180°
=(12-2) 180°
=10×180°
=1800°
But, in case of an irregular dodecagon, the measures of interior angles based on the lengths of sides. So, it is possible to find out the measure of each angle. In case of regular dodecagons, the measures of all interior angles are equal. As there are 12 angles in the regular dodecagon and sum of angles is 1800°. So,An interior angle in a Regular Dodecagon= $\frac{1800^o}{12}$=150°

## Area of a Dodecagon

The area of an irregular dodecagon can be found by following the three steps given hereunder:

1. Partition the dodecagon into the following triangles.
2. Find out the area of each triangle in the partition.
3. Add up the areas of all these triangles to get the area of the given dodecagon.
However, if we know the twelve vertices of an irregular dodecagon, it is possible to derive a formula for the area of the irregular dodecagon. The formula will be expressed entirely in terms of the coordinates of the vertices.

We can derive the formula for the area of a regular dodecagon in terms of the length of the size. For this we need to take apothem into consideration. Apothem of a regular polygon is a line segment drawn from the center to the midpoint of one of its sides. Let a units be the apothem of a regular dodecagon whose side is of s units. Join the center O with the vertices D and V3. Let M be the mid-point of DV3. Join the center O with M. Then, $\overline{OM}$ be the apothem. Let us denote |$\overline{OM}$| = a.
From the diagram, it is clear that $\angle$OV3 M=75° and MV3=$\frac{s}{2}$. Thus,

tan($\angle$OV3 M)= $\frac{OM}{MV_3}$

$\rightarrow$ tan(75o)= =$\frac{a}{s/2}$

$\rightarrow$ tan(30°+ 45°)= $\frac{2a}{s}$

$\rightarrow$ $\frac{(tan(30°)+tan(45°))}{(1-tan(30°) tan(45°))}$ = $\frac{2a}{s}$

$\rightarrow$ $\frac{\frac{\sqrt3}{3}+1}{1-(\frac{\sqrt3}{3})(1)}$ = $\frac{2a}{s}$

$\rightarrow$ $\frac{\sqrt{3}+3}{3- \sqrt{3}}$ = $\frac{2a}{s}$

$\rightarrow$ $\frac{(\sqrt{3}+3)^2}{6}$ = $\frac{2a}{s}$

$\rightarrow$ a = $\frac{1}{12}$($\sqrt{3}$+3)2 s

$\rightarrow$ a= $\frac{1}{12}$ (12+6$\sqrt{3}$)s

$\rightarrow$ a= $\frac{1}{2}$ (2+$\sqrt{3}$)s

It is clear that
Area of $\Delta$ODV3 = $\frac{1}{2}$as = $\frac{1}{4}$(2+$\sqrt{3}$) s2
Thus,
Area of the Regular Dodecagon A=12 × Area of $\Delta$ODV3

=12 x $\frac{1}{4}$ (2+$\sqrt{3}$) s2

=3 ($\sqrt{3}$+2) s2

### Solved Examples

Question 1: Given a regular dodecagon with side 4 cm. Find the apothem and area of the regular dodecagon.
Solution:

Given that
side s=4 cm
Then,
Apothem a=$\frac{1}{2}$ (2+√3)s=$\frac{1}{2}$ (2+√3)(4)=2 (2+√3)  cm
Hence,
Area A=12×$\frac{1}{2}$   as=12×$\frac{1}{2}$  [2(2+√3)]  (4)=48 (2+ √3)  cm2

Question 2: Given a regular dodecagon for which apothem is 4 cm. Find the length of the side and the area.
Solution:

Given that
Apothem  a=4 cm
Let s  cm be the side of the dodecagon. Now, consider the following relationship between the apothem and the side of a dodecagon.

a=$\frac{1}{2}$ (2+√3)s

Substituting a=4, we get

4=$\frac{1}{2}$  (2+ v3)s

$\rightarrow$ 8=(2+ √3)s

$\rightarrow$ s=8/(2+ √3)

$\rightarrow$ s= $\frac{8(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}$

$\rightarrow$ s=8(2-√3)  cm
Now,
Area A=3(2+√3) s2
=3 (2+√3)  [8(2-√3)]2
=3(64)(2+√3)(2-√3)(2-√3)
=192 (2-√3)  sq.cm

Question 3: The area of a regular dodecagon is 108 (2+√3) cm2. Find the length of the side and apothem.
Solution:

Let s cm be the length of a side and a be the apothem.  Given that

Area  A=108 (2+√3)   cm2

$\rightarrow$ 3 (2+√3) s2=108 (2+√3)

$\rightarrow$  s2=36

$\rightarrow$ s=6

So, the side is of length  6 cm.  Now, let us find out the apothem.

Apothem a=$\frac{1}{2}$ (2+√3)s

=$\frac{1}{2}$ (2+√3)(6)

=3 (2+√3)  cm

Thus, the apothem is of length 3(2+√3)  cm.