**A polygon** is a part of a plane enclosed by line segments that intersect at their endpoints.

The segments $\overline{A_1A_2}$, $\overline{A_2A_3}, \overline{A_3A_4}, \ldots , \overline{A_{n-1}A_n}$ are called **sides** of the polygon, and points $A_1, A_2, A_3, A_4, \ldots , A_{n-1}, A_n$ are called **vertices**.

A polygon with $n$ sides and $n$ vertices is called **$n$-sided polygon**.

Angles $\angle{A_nA_1A_2}, \angle{A_1A_2A_3}, \angle{A_2A_3A_4}, \ldots, \angle{A_{n-1}A_nA_1}$ are called **interior angles** of the $n$-sided polygon.

**An exterior angle** of a polygon is an adjacent interior angle. Interior and exterior angles are called supplementary angles, and the sum of their measures is equal to $180^{\circ}.$

**A diagonal** of a polygon is a segment line in which the ends are non-adjacent vertices of a polygon.

We already know that triangles which are three sided polygons and quadrilaterals are four sided polygons.

Polygons may be a convex set, however, not every polygon is a convex set. We can distinguish between *convex* and *concave* polygons.

**A concave polygon** is a polygon that has at least one interior angle whose measure is greater than $180^{\circ}$:

**A convex polygon** is a polygon in which every interior angle has a measure less than $180^{\circ}$.

In this lesson, we will observe only convex polygons.

Polygons are also divided into two special groups which are *regular* and *irregular polygons*.

**A regular polygon*** *is a polygon that* *has all sides of equal length and all interior angles of equal measure.

**An irregular polygon **is a polygon that has at least one set of unequal sides.

First, we will recall things we learned about quadrilaterals, and expand where required in order to conclude some useful points.

Quadrilaterals are polygons in a plane which contain four sides.

A regular polygon is a __square__, because only a square of a quadrilateral contains all sides of equal length and all angles of equal measure. The area of a square is equal to the square of the length of one side, and the perimeter to four lengths of any side. Each interior angle has a measure equal to $90^{\circ}$, and their sum is equal to $360^{\circ}$. It has two diagonals. These diagonals intersect at one point which is the center of an inscribed and circumscribed circle.

**An inscribed circle **in a regular polygon is a circle which touches all sides of a regular polygon, and **an** **circumscribed circle** runs through all vertices of a regular polygon.

To draw a circumscribed circle of a square we simply place the needle of the compass into the intersection of diagonals, extend it to one vertex, and draw. The circumscribed circle will then run through all vertices.

To draw an inscribed circle, we must first find the radius. To find the radius, we must draw a perpendicular line from the center to any side. When we inscribe the circle, it must touch all sides of the square.

**Pentagons** are polygons which contain five sides.

**A regular pentagon** has five congruent sides and five congruent angles. In addition it has five diagonals.

How would we know how many diagonals does a polygon has without having to draw and count it?

Let’s try to logically arrive at a formula for the number of diagonals of any convex polygon. Let’s say that polygon has $n$ vertices. From any vertex we can draw $n – 3$ diagonals and do that $n$ times (from any vertex), we can’t draw from that vertex and two adjacent’s. Because every two diagonals overlap we have to divide that number with two. Final formula is:

$$ D_{n} = \frac{n(n – 3)}{2}.$$

For the pentagon that would be: $ D_{5} = \frac{5 \cdot(5 – 3)}{2} = \frac{10}{2}=5$, i.e. pentagon has $5$ diagonals.

How can we determine what are the sum of the measures of all interior angles, and what is the measure of each angle in a regular polygon?

We will use a pentagon for example, however, everything else is the same.

In order to obtain the sum of the measures of all interior angles of a pentagon, we will draw diagonals of a pentagon from only one vertex. A pentagon is divided into three triangles. Since we know that the sum of the measures of all interior angles of a triangle is equal to $180^{\circ}$, which means that the sum of the measures of all interior angles of a pentagon is equal to $ 180^{\circ} \cdot 3 = 540^{\circ}$.

By incomplete induction we can therefore conclude that the formula for the sum of the measures of all interior angles of a convex polygon of $n$ vertices is equal to:

$$ (n – 2)\cdot180^{\circ}.$$

Since all regular polygons have all angles of equal measure, to obtain to the measure of each angle in a polygon with $n$ vertices we can simply divide the sum of the measures of all interior angles by $n$.

For a regular pentagon that will be: $ 540 : 5 =108^{\circ}$. This means that the measure of each angle in a regular pentagon will be $ 108^{\circ}$.

As we already noticed, diagonals in a regular polygon do not intersect at one point. That means that we have no candidates for the center of an inscribed and circumscribed circle. We will try to find it by bisecting the angles. Through doing this we obtained five congruent triangles.

This means that $ |AS|= |BS| = |CS| = |DS| = |ES|$ and the point $S$ is the center of an inscribed and circumscribed circles. To draw a circumscribed circle we simply place the needle of the compass on point $S$ and extend it to any vertex of the regular pentagon. To obtain the radius of an inscribed circle, we must draw a perpendicular line to any side from the center. This will constitute our radius. Again, a circumscribed circle must run through all vertices and an inscribed circle must touch all sides.

The triangles we divided in our regular pentagon will also be useful for finding the area of our regular pentagon.

Since we already know how to calculate area of a triangle, we simply multiply that area by $ 5$ to get our whole area of a regular pentagon. All these triangles are thus isosceles triangles, whose angles we know. Then it is fairly simple to calculate area.

We know that all triangles that we have divided into a regular pentagon are congruent. This also means that their areas are equal. If $|AB|=|BC|=|CD|=|DE|=|EA|=a$ and $h_a$ is the height of a characteristic triangle of a regular pentagon then the area of a characteristic triangle of a regular pentagon is equal to $ A_t = \frac{a \cdot h_a}{2}$ and then the area of a regular pentagon will be $ A_p = 5\cdot P_t$. In general, the area of a regular polygon with $n$ vertices is equal to:

$$A_t=n\cdot\frac{a\cdot h_a}{2}.$$

Let’s observe a triangle $P_1BS$. This triangle is a right angled triangle. We know that the measure of each interior angle of a regular pentagon is equal to $ 108^{\circ}$. That means that $\angle{P_1BS}=54^{\circ}$, because segment $\overline{BS}$ divides an interior angle $\angle{ABC}$ of a regular pentagon into two angles both of equal measures. According to this, $\angle{BSP_1}$ is equal to $36^{\circ}$. Since triangle $ABS$ is an isosceles triangle and $|P_1S|=h_a$ is a height of that triangle then $|P_1B|=\frac{a}{2}$. By knowing this, we can use trigonometry of a right angled triangle $P_1BS$:

$$ tan (54^{\circ}) = \frac{h_a}{2}\Rightarrow h_a = 2.75 cm.$$

Now we can calculate the area of a regular pentagon:

$$ A_t= 4 \cdot 2.75 = 11 cm^2 \Rightarrow A_p = 5 \cdot11 = 55 cm^2.$$

**A hexagon** is a polygon which contains six sides.

**A regular hexagon** contains six congruent sides and six congruent angles.

Let’s use what we know to determine other properties.

A number of diagonals is:

$$ d = \frac{n (n – 3)}{2} = \frac{6 (6 – 3)}{2} = 9.$$

The sum of the measures of all interior angles is:

$$ (n – 2) \cdot 180^{\circ}= 4 \cdot 180^{\circ}= 720^{\circ}.$$

The measure of each interior angle:

$$ 720^{\circ} : 6 = 120^{\circ}.$$

The center of an inscribed and an circumscribed circle is in the intersection of opposite vertices. If we are unsure at which point to use as the center for an inscribed and circumscribed circle, the best way is to bisect the angles and then their intersection will be the point we are looking for.

These diagonals divide a hexagon into six congruent equilateral triangles, which means that their sides are all congruent and each of their angles are $ 60^{\circ}$. For the area, we must again calculate the area of one triangle and multiply it by $6$.

The same rules and formulas apply to other regular polygons.

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