# Regular Polygons

• ### Корзина

Пусто

The contents of this webpage:

• – the examples of solutions for the lesson "The formula for solving the angles of a regular polygon" are presented in the problems 108 - 112;
• – the theme "The circumscribed circle" is explained in the answers for the worksheets under numbers 113 - 116 of the textbook;
• – the questions, how to find the radius of regular polygons, and the problems for the lesson "The inscribed and described regular polygons" are considered in the worksheets 117 - 123 of this workbook.

Definition of a regular polygon:

The regular polygon is a convex polygon if the following conditions hold: the angles and the sides of this polygon are equal to each other. The examples of the regular polygon are a square, and an equilateral triangle.

Theorem - Derivation of the formula for calculating the angles of a regular polygon.

Given:

the regular polygon,

the angles α1, α2, α3, α4 ...

Prove:

αn = • 180°

Proof:

The sum of the angles of this regular polygon (n - 2) • 180°.

By the condition α1 = α2 = α3 = α4 = ...

each angle is • 180°, i.e. true is

the formula for calculating the angles of a regular polygon:

αn = • 180°

***

Problem 108.

Given: the regular hexagon, i.e. n = 6

Find: the angle of the regular hexagon

Solution:

αn = • 180° = = 120°

***

Problem 109.

Given: the regular polygon, where n is the number of sides of the polygon

Find: n - how many sides are contained in the regular polygon

Solution:

 1) αn = 60° αn = • 180° 60° •n = (n - 2) • 180° 60° •n = 180°•n - 360° - 120° • n = - 360° n = 3 2) Since the angle of the regular polygon αn = 135° 135° •n = (n - 2) • 180° 135° •n - 180°•n = - 360° n = 8

***

Problem 110.

Given:

ABCDEF is the regular polygon,

external angles β1; β2; ... ; βn

Find:

the sum of the external angles β1 + β2 + ... + βn = ?

Solution:

Since the hexagon is regular, then by definition of a regular polygon

each angle in the regular polygon is αn = • 180°

αn = • 180° = 120°

Since the all angles in the regular polygon are equal, then the external angles will also be equal, namely β1 = β2 = β3 = β4 = β5 = β6 = 180° - FAB = 180° - 120° = 60°

Then β1 + β2 + β3 + β4 + β5 + β6 = 60° • 6 = 360°.

***

Problem 111.

Given:

the regular polygon,

 1) the regular triangle, n = 3 2) n = 5 3) n = 18

Find: the angle of the regular polygon

Solution:

 1) αn = • 180° 3 • αn = 180° αn = 60° 2) 5 • αn = 3 • 180° αn = 108° 3) 18 • αn = 16 • 180° αn = 160°

Answer: every angle of a regular polygon is 1) 60°; 2) 108°; 3) 160°.

***

Problem 112.

Given:

the regular polygon,

αn = 90°

Determine: how many sides does the regular polygon have, i.e. n = ?

Solution:

αn = • 180°

90° •n = (n - 2) • 180°

90° •n - 180°•n = - 360°

-90° •n = - 360°

n = 4

Answer: the number of sides of the regular polygon n = 4.

***

## Regular polygon and circumscribed circle

Definition:

A polygon is termed inscribed in a circle if all its vertices lie on a common circle.

Theorem:

About the regular polygon, it is possible to circumscribe a circle and, moreover, only one.

Given:

A1A2A3...An is the regular polygon

Prove:

there exists a unique circle of center at O and radius R such that the vertices of the regular polygon lie on this circle

! circle (O; R): A1; A2; A3;...An circle (O; R)

Proof:

1) We draw bisectors of the angle A1 and the angle A2.

Since the polygon is regular, we see that A1 = A2

1 =2 =3 =4.

From the condition 1 =3 it follows that the triangle ΔA1OA2 - is isosceles, so A1O = OA2

Consider the triangle ΔA2OA3:

1) A2O - common side

2) A1A2 = A2A3

3) 3 =4

Then, by the first theorem of the equality of the triangles

Δ A1OA2 = Δ A2OA3. Therefore, A2O = A3O.

By connecting each remaining vertex to the point O, it can be shown that all the triangles are equal.

Then A1O =A2O = A3O = ... = AnO

Since the point O is the center of the circle and the radius is R = A1O =A2O = A3O = ... = AnO , we see that,

! circle (O; R): A1; A2; A3;...An circle (O; R)

Uniqueness:

Take any three vertices of the regular polygon; they form a triangle such that only one circle can be circumscribed about this triangle, it follows that only one circle can be circumscribed about the given polygon.

***

Problem 113.

Given:

the regular polygon,

the arc AB= 60° ( AB = 60°)

AB is the side of the regular polygon

Find:

number of sides of the regular polygon, i.e. n = ?

Solution:

Since the degree measure AB = 60° < 180°, then the arc is equal to the angle, i.e.

AB = AOB.

ΔAOB - is isosceles, where OAB = OBA =

= (180° - 60°) : 2 = 60°.

Then ΔAOB is equilateral.

The radii of a circle circumscribed about the regular polygon are the bisectors of its angles, so each angle of the polygon is 60° • 2 = 120°.

Knowing that

αn = • 180°

120° •n = (n - 2) • 180°

120° •n - 180°•n = - 360°

-60° •n = - 360°

n = 6

Answer: the number of sides of the regular polygon is n = 6.

***

Problem 114.

Given:

the regular polygon,

 1) AB = 36° 2) AB = 18°

AB is the side of the regular polygon

Find:

the number of sides of the polygon, i.e. n = ?

Solution:

1)

Since the degree measure AB = 36° < 180°, we see that the arc is equal to the angle, i.e.

AB = AOB.

ΔAOB is isosceles, where OAB = OBA =

= (180° - 36°) : 2 = 72°.

The radii of the circle circumscribed about the polygon are the bisectors of its angles, so each angle of the polygon is 72° • 2 = 144°.

Knowing that

αn = • 180°

144° •n = (n - 2) • 180°

144° •n - 180°•n = - 360°

-36° •n = - 360°

n = 10

2)

Since the degree measure AB = 18° < 180°, then the arc is equal to the angle, i.e.

AB = AOB, where the angle AOB is central.

ΔAOB is isosceles (OA = OB = r), where OAB =

=OBA = (180° - 18°) : 2 = 81°.

The radii of the circle circumscribed about the polygon are the bisectors of its angles, so each angle of the polygon is 81° • 2 = 162°.

Knowing that

αn = • 180°

162° •n = (n - 2) • 180°

162° •n - 180°•n = - 360°

-18° •n = - 360°

n = 20

Answer: 1) n = 10; 2) n = 20.

***

## The inscribed and circumscribed circle in the regular polygon

Conclusion:

In regular polygons, the centers of the inscribed and circumscribed circles coincide.

Problem 115.

Given:

A1A2A3A4A5 is the regular pentagon

αn = • 180° = 108°

Prove:

A1A3 = A1A4

Proof:

By the definition of the regular polygon in a given pentagon, all sides and angles are equal.

Consider the triangles ΔA1A2A3 and ΔA1A4A5.

A1A2 = A1A5

A2A3 = A5A4

A2 =A5

Then, according to the first theorem of the equality of the triangles (ΔA1A2A3 = ΔA1A4A5) it follows that A1A3 = A1A4 as the corresponding sides.

***

Problem 116.

Given:

ABCD is the square

The ratio of the segments

AM : MK : KD = 1 : : 1

Prove:

MNOZLFEK is the regular polygon

Proof:

ΔAMN = ΔOBZ = ΔLCF = ΔEKD (by the first theorem of the equality of the triangles)

By the Pythagorean theorem:

MN = OZ = LF = EK =

By the condition NO = ZL = EF = MK =

Therefore, all sides are equal.

Since 1=2 = 45°, then NMK=MKE = KEF=EFL = FLZ=ZON = ... = 180° - 45° = 135°

It follows that MNOZLFEK is the regular octagon.

***

## The regular polygon and inscribed circle

Theorem:

In any regular polygon, it is possible to inscribe a circle, and moreover only one.

Given:

A1A2A3...An is the regular polygon

Prove:

There is a unique inscribed circle of center at O and radius R

! circle (O; R): H1; H2; H3;...Hn circle (O; R)

Proof:

1) We draw the heights of the triangles, i.e. OH1; OH2; ...; OHn

It is known that the triangles are equal, i.e. ΔOA1A2 = ΔOA2A3 = ... = ΔOAnA1.

Therefore, OH1 = OH2 = ... = OHn . Then H1; H2; H3;...Hn circle (O; R).

Uniqueness:

2) Let us suppose that as well as the circle (O; R) there is another circle inscribed in a given polygon.

Then its center O1 is equidistant from the sides of the polygon and coincides with the point O of the intersection of the bisectrixes lying on each angle of the polygon.

Therefore, the radius of this circle is equal to OH1 and from this it follows that the circles coincide.

***

Problem 117.

Given:

the regular polygon,

the arc AB= 72° ( AB = 72°)

AB is the side of the regular polygon

Find:

the number of sides of the polygon, i.e. n = ?

Solution:

Since the degree measure AB = 72° < 180°, then the arc is equal to the angle, i.e.

AB = AOB, where the angle AOB is central.

ΔAOB is isosceles (OA = OB = r), where

OAB = OBA = (180° - 72°) : 2 = 54°.

Then ΔAOB is equilateral.

The radii of the circle circumscribed about the polygon are the bisectors of its angles, so each angle of the polygon is 54° • 2 = 108°.

Knowing that

αn = • 180°

108° •n = (n - 2) • 180°

108° •n - 180°•n = - 360°

-72° •n = - 360°

n = 5

***

Formulas for calculating the area of the regular polygon, its side and the radius of the inscribed circle.

Given:

A1A2A3...An is the regular polygon

R is the radius of the circumscribed circle

r is the radius of the inscribed circle

an is the side of the polygon

Prove:

1) the area of the regular polygon is equal to half the product of the perimeter of the polygon by the radius of the inscribed circle

Sn = Pn • r

2) the side of the regular polygon is equal to twice the product of the radius of the circumscribed circle by the sine of the angle equal to the number from the division of 180° by n - the number of sides of the polygon

an = 2R • Sin ()

3) the radius of the inscribed circle is equal to the product of the radius of the circumscribed circle by the cosine of the angle equal to the number from the division of 180° by n - the number of sides of the regular polygon

r = R • Cos ()

Proof:

1) By connecting the point O to the vertices of the regular polygon, we obtain triangles Δ A1A2O = Δ A2A3O = ... = Δ A1AnO, where the number of all triangles in the polygon = n.

S (Δ A1A2O) = • A1A2 • OH1 = • an • r.

Then the area of the polygon is equal to the sum of the areas of all triangles

Sn = S (Δ A1A2O) • n = • an • r • n = (• an • n) • r = Pn • r

***

2)

Since the angle in the polygon is given by formula

αn = • 180°, we see that the angle A1 in the triangle A1H1O is half the angle of the polygon.

A1 = •(• 180°) = • 90° = = 90° -

Cos A1 = =

Then A1H1 = Cos A1 • R = Cos (90° - ) • R = R • Sin()

Knowing that an = A1H1 • 2, we obtain

an = 2R • Sin ()

***

Examples:

If n=3, then a3 = 2R • = R•

If n=4 (the square), then a4 = 2R • Sin45° = 2R • = R•

If n=6 (the regular hexagon), then a6 = 2R • 0,5 = R

***

3)

Sin A1 =

Then r = R • Sin A1 = R • (90° - ) = R • Cos ()

***

Problem 118.

Given:

The regular quadrilateral is the square

 R r a4 P S 1 - - 6 - - 2 - 2 - - - 3 - - - - 16

what are the radii of the inscribed and circumscribed circumference, the sides, perimeter and area of the regular polygon – the square

R, r, a4, P, S = ?

Solution:

1) If the side of the square a4 = 6 and since given is the square, we see that

the perimeter of the square P = 4 • a4 = 4 • 6 = 24

The area of the square S = a42 = 62 = 36

Since the area of the square can be found by the formula S4 = P4 • r, we see that the radius of the inscribed circle

r = = = 3

Since the number of sides n=4, we see that a4 = R•

The radius of the circumscribed circle R== ==

2) If r=2, then the radius of the circumscribed circle

r = R • Cos () = R • Cos 45° = R •

R = 2 • = 2

Since the number of sides n=4, then the side of the square

a4 = R•= 2= 4

Then the perimeter of the square P = 4 • a4 = 4 • 4 = 16

The area of the square S = a42 = 42 = 16

3) If the area of the square S = 16, then the side of the square

a42 = 16 a4 = ±. But a4 = - does not satisfy the condition of the problem.

Therefore, a4 = = 4

Then the perimeter of the square P = 4 • a4 = 4 • 4 = 16

Since the number of sides n=4, then the radius of the circumscribed circle

a4 = R• 4 = R• R = = 2

Then the radius of the inscribed circle

r = R • Cos () = R • Cos 45° = 2= 2

***

Problem 119.

Given:

Δ ABC is equilateral,

the perimeter of the triangle

PΔ = 18 cm

Find: the side of a square inscribed in the same circle

a4 = ?

Solution:

Since Pn = n • an , we see that the side of the triangle

18 = 3 • a3 a3 = 6

Then the radius of the circumscribed circle

a3 = R• R = = 2

Then the side of the square a4 = • R = • 2= 2

***

Problem 120.

Given:

ABCD is the square,

the side of the square a4 = 6

Calculate: what is the doubled radius of the inscribed circle 2r = ?

Solution:

Since the area of the square can be found by the formula S4 = P4 • r, then the radius of the inscribed circle

r = = = 3 (cm)

Therefore, 2r = 2 • 3 = 6 (cm).

***

Problem 121.

Given:

the square, the regular hexagon are inscribed in the circle (O;r) of the center at the point O and the radius r ;

the perimeter of the hexagon P6 = 48 cm

Find: the perimeter of the square P4 = ?

Solution:

Since the formula of the perimeter of the hexagon P6 = 6 • a6 , then the side of the hexagon a6 = 48 : 6 = 8 (cm)

Since given is a regular hexagon for n=6, then

a6 = 2R • 0.5 = R

Since given is one circle, the radii of the circumscribed circle are equal, i.e. R4 = R6 = 8 cm

Then the side of the square a4 = R•= 8cm

Therefore, the perimeter of the square P4 = 4 • 8= 32(cm)

***

Problem 122.

Given:

ABCDEF is the regular hexagon

BF = 1.5 cm

Find: the area of the hexagon

SABCDEF = ?

Solution:

We draw the diagonal AD and consider the isosceles triangle ΔABF.

By the property of an isosceles triangle AO - is the median.

Therefore, FO = OB = BF : 2 = 1.5 : 2 = 0.75 (cm)

Since the hexagon is regular, we see that using the formula to calculate the angles of the regular polygon

αn = • 180°, we obtain

α6 = • 180° = 120°

Then the angle BAO = 60°

Consider the triangle ΔAOB, where the angle O = 90°.

Sin 60° = = AB = = (cm)

Then the perimeter of the hexagon

P6 = 6 • = 3(cm)

Then the radius of the inscribed circle

r6 = OF = cm

Therefore, the area of the hexagon

S6 = • P6 • r6 = = (см2)

***

Problem 123.

Given:

A1A2A3A4A5A6 is a regular hexagon inscribed in a circle(O; R)

B1B2B3B4B5B6 is a regular hexagon, circumscribed about the circle(O; R)

OH = r

OH1 = R

Find: the proportion of areas of hexagons

= ?

Solution:

1)

Consider the hexagon A1A2A3A4A5A6

The sides of the hexagon are equal to the radii of the circumscribed circle A1A2 = A2A3 = A6A1 = R

Knowing that S (A1A2A3A4A5A6) = P • r = 6Rr = 3R•r

Consider the triangle ΔOHA6 is right, since the radius of the inscribed circle is perpendicular to the side of the hexagon r A1A6

Since the point O - is the point of intersection of bisectors, we see that R - is the bisector of the angle A1A6A5

Knowing that OA6A1 = 60°, we obtain

Sin 60° = r = Sin 60° • R =

Therefore, the area of the hexagon

S (A1A2A3A4A5A6) = 3R •=R2

2)

Consider the hexagon B1B2B3B4B5B6

B1B2 = ... = B6B1 = OB5 , where OB5 - is the radius of the circumscribed circle about a given regular polygon, namely the hexagon B1B2B3B4B5B6

S (B1B2B3B4B5B6) = • (6 • OB5) • R

Consider ΔH1OB5 - is a right triangle, since R - of the inscribed circle in the polygon B1B2B3B4B5B6:

R B5B6

Then OB5H1= 60°. Therefore

Sin 60° = OB5 = 2R

Then the area of the hexagon is B1B2B3B4B5B6

S (B1B2B3B4B5B6) = • 6 • 2R • R = 2R2

Then the proportion of the areas of the hexagons

= 2R2 =