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Angles, areas, diagonals, inscribed and circumscribed circles of regular polygons

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.

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 everyinterior angle has a measure less than  $180^{\circ}$.

In this lesson, we will observe only convex polygons.

 

Diagonals

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

How many diagonals does n-polygon have? Let’s see for the first few polygons.

For $n=3$ we have a triangle. We can see triangle has no diagonals because each vertex has only adjacent vertices.

For $n=4$ we have quadrilateral.  It has $2$ diagonals.

For $n=5$, we have pentagon with $5$ diagonals.

For $n=6$, $n$-polygon is called hexagon and it has $9$ diagonals.

Since $n$ was a lower number we could easily draw the diagonals of $n$-polygons and then count them.

How would we know the number of diagonals without having to draw all of them? Let’s try to logically come up with 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) since we can’t draw from that vertex and two adjacent’s. Every two diagonals overlap and because we don’t want to count each diagonal twice , we have to divide that number with two. Final formula is:

$$ D_{n} = \frac{n(n – 3)}{2}.$$

Angles

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. On the picture above, they are colored green.

An exterior angle of a polygon is an adjacent interior angle, colored red on picture. Interior and exterior angles are supplementary angles, meaning that the sum of their measures is equal to $180^{\circ}.$

We can see on the picture that the sum of interior angle $\alpha$ and exterior angle on the same vertex $\alpha^{‘}$ is

$$\alpha+ \alpha^{‘} =127.72^{\circ} + 52.28^{\circ} = 180^{\circ}$$

The same thing can be applied to all the pairs of angles on the same vertex, $\beta+\beta^{‘}=180^{\circ}$, $\gamma + \gamma^{‘}=180^{\circ}$ and so on.

 

How can we determine what are the sum of the measures of all interior angles?

We will use a pentagon for example, however, we can use the same process for every other polygon.

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. 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}$.

Let’s find the appropriate formula for the sum of the measures of all interior angles of a convex polygon of $n$ vertices. We know that the sum of the measures of all interior angles of a triangle is equal to $ (3 – 2) \cdot 180^{\circ} = 180^{\circ}$, of a quadrilateral is equal to $ (4 – 2) \cdot 180^{\circ} = 360^{\circ}$, of a pentagon is equal to $ (5 – 2) \cdot 180^{\circ}= 540^{\circ}$ and so fourth.

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:

$$ (n – 2)\cdot180^{\circ}.$$

 

Polygons are also divided into two special groups:

  • regular
  • irregular

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.

Regular polygons have both an inscribed circle (circle that touches all sides of a regular polygon), and an circumscribed circle (circle that runs through all vertices of a regular polygon). They are also called incircle and circumcircle. The center of both of these circles is the same and is also called the center of a polygon.

 

 

Center of regular polygon

Center of a polygon is a point inside a regular polygon that is equidistant from each vertex.

How do we find it?

In some regular polygons, the center of polygon is intersection of diagonals. For example in quadrilaterals and hexagons.

However, that’s not the case with all the polygons. We will find universal strategy for finding the center.

Let’s use a pentagon for example.

We will try to find the center by bisecting the angles. By doing this we obtain $5$ triangles. These triangles are called characteristic triangles of regular polygon.

 

These $5$ tringles are congruent. Since this is a regular polygon, all sides have equal lengths and interior angles have equal measures. Therefore, angle bisectors give us the same angles in triangles.

This means that $ |AS|= |BS| = |CS| = |DS| = |ES|$, and the point $S$ is the center of an inscribed and circumscribed circles.

 

Area of regular polygons

The triangles we got by dividing our $n$-sided regular polygons will also be useful for finding its area.

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

Once again, let’s take pentagon as an example.

We know that all triangles that we have divided into a regular pentagon are congruent and isosceles. 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 $$ A_t = \displaystyle{\frac{a \cdot h_a}{2}}.$$

$h_a$ is also called apothem of a regular polygon.

Using the area of characteristic triangle we can get the area of a regular pentagon. It will be $ A_p = 5\cdot P_t$.

In general, the area of a regular polygon with $n$ vertices is equal to:

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

 

Quadrilaterals

Quadrilaterals are polygons in a plane with $4$ sides and $4$ vertices.

the quadrilatelar

A regular quadrilateral is a square, because square is the only quadrilateral with 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.

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.

 

Pentagon

Pentagons are polygons which contain five sides.

A regular pentagon has five congruent sides and five congruent angles.

How many diagonals does pentagon have?

We can use the formula from above.

$$ D_{5} = \frac{5 \cdot(5 – 3)}{2} = \frac{10}{2}=5$$

What’s the sum of the measures of all interior angles?

By using the formula we had earlier,$ (n – 2)\cdot180^{\circ}$, we know that the sum of all interior angles in pentagon is

$$ (n – 2)\cdot180^{\circ}=(5 – 2)\cdot180^{\circ}=3\cdot180^{\circ}=540^{\circ}$$

 

What is the measure of each angle in a regular polygon?

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. To draw an  inscribed and circumscribed circle we need to find their center by the process we described before – 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.

Area of pentagon is

$$A_t=5\cdot\displaystyle{\frac{a\cdot h_a}{2}}.$$

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

$$ tan (54^{\circ}) = \displaystyle{\frac{h_a}{\displaystyle{\frac{a}{2}}}}$$

$$h_a=\frac{a \cdot tan (54^{\circ})} {2} $$

Now we can calculate the area of a regular pentagon:

$$A_t=5\cdot\displaystyle{\frac{a\cdot \frac{a \cdot tan (54^{\circ})} {2}}{2}}.$$

$$A_t=\displaystyle{\frac{5}{4}}\cdot a^{2}\cdot tan (54^{\circ}).$$

 

Hexagon

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.

Polygons worksheets

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