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 α_{1}, α_{2}, α_{3}, α_{4} ...

Prove:

α_{n }= • 180°

Proof:

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

By the condition α_{1} = α_{2} = α_{3} = α_{4} = ...

each angle is • 180°, i.e. true is

the **formula for calculating the angles of a regular polygon**:

α_{n }= • 180°

***

**Problem 108.**

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

Find: the angle of the regular hexagon

Solution:

α_{n }= • 180° = = 120°

Answer: 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) α_{n }= 60° α 60° •n = (n - 2) • 180° 60° •n = 180°•n - 360° - 120° • n = - 360° n = 3 | 2) Since the angle of the regular polygon α_{n }= 135° 135° •n = (n - 2) • 180° 135° •n - 180°•n = - 360° n = 8 |

***

**Problem 110.**

Given:

ABCDEF is the regular polygon,

external angles β_{1}; β_{2}; ... ; β_{n}

Find:

the sum of the external angles β_{1} + β_{2} + ... + β_{n} = ?

Solution:

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

each angle in the regular polygon is α_{n }= • 180°

α_{n }= • 180° = 120°

Since the all angles in the regular polygon are equal, then the external angles will also be equal, namely β_{1} = β_{2} = β_{3} = β_{4} = β_{5} = β_{6} = 180° - FAB = 180° - 120° = 60°

Then β_{1} + β_{2} + β_{3} + β_{4} + β_{5} + β_{6} = 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) α_{n }= • 180° 3 • α α | 2) 5 • α α |
3) 18 • α α |

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

***

**Problem 112.**

Given:

the regular polygon,

α_{n }= 90°

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

Solution:

α_{n }= • 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:

A_{1}A_{2}A_{3}...A_{n} 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): A_{1}; A_{2}; A_{3};...A_{n} circle (O; R)

Proof:

1) We draw bisectors of the angle A_{1} and the angle A_{2}.

Since the polygon is regular, we see that A_{1} = A_{2}

1 =2 =3 =4.

From the condition 1 =3 it follows that the triangle ΔA_{1}OA_{2} - is isosceles, so A_{1}O = OA_{2}

Consider the triangle ΔA_{2}OA_{3}:

1) A_{2}O - common side

2) A_{1}A_{2} = A_{2}A_{3}

3) 3 =4

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

Δ A_{1}OA_{2} = Δ A_{2}OA_{3}. Therefore, A_{2}O = A_{3}O.

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

Then A_{1}O =A_{2}O = A_{3}O = ... = A_{n}O

Since the point O is the center of the circle and the radius is R = A_{1}O =A_{2}O = A_{3}O = ... = A_{n}O , we see that,

! circle (O; R): A_{1}; A_{2}; A_{3};...A_{n} 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

α_{n }= • 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

α_{n }= • 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

α_{n }= • 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:

A_{1}A_{2}A_{3}A_{4}A_{5} is the regular pentagon

α_{n }= • 180° = 108°

Prove:

A_{1}A_{3} = A_{1}A_{4}

Proof:

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

Consider the triangles ΔA_{1}A_{2}A_{3} and ΔA_{1}A_{4}A_{5}.

A_{1}A_{2 }= A_{1}A_{5}

A_{2}A_{3} = A_{5}A_{4}

A_{2 }=A_{5}

Then, according to the first theorem of the equality of the triangles (ΔA_{1}A_{2}A_{3} = ΔA_{1}A_{4}A_{5}) it follows that A_{1}A_{3} = A_{1}A_{4} 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:

A_{1}A_{2}A_{3}...A_{n} is the regular polygon

Prove:

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

! circle (O; R): H_{1}; H_{2}; H_{3};...H_{n} circle (O; R)

Proof:

1) We draw the heights of the triangles, i.e. OH_{1}; OH_{2}; ...; OH_{n}

It is known that the triangles are equal, i.e. ΔOA_{1}A_{2} = ΔOA_{2}A_{3} = ... = ΔOA_{n}A_{1}.

Therefore, OH_{1} = OH_{2} = ... = OH_{n} . Then H_{1}; H_{2}; H_{3};...H_{n} 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 O_{1} 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 OH_{1} 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

α_{n }= • 180°

108° •n = (n - 2) • 180°

108° •n - 180°•n = - 360°

-72° •n = - 360°

n = 5

Answer: n = 5.

***

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

Given:

A_{1}A_{2}A_{3}...A_{n} is the regular polygon

R is the radius of the circumscribed circle

r is the radius of the inscribed circle

a_{n} 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

S_{n} = P_{n} • 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

a_{n} = 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 A_{1}A_{2}O = Δ A_{2}A_{3}O = ... = Δ A_{1}A_{n}O, where the number of all triangles in the polygon = n.

S (Δ A_{1}A_{2}O) = • A_{1}A_{2} • OH_{1} = • a_{n} • r.

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

S_{n} = S (Δ A_{1}A_{2}O) • n = • a_{n} • r • n = (• a_{n} • n) • r = P_{n} • r

***

2)

Since the angle in the polygon is given by formula

α_{n }= • 180°, we see that the angle A_{1} in the triangle A_{1}H_{1}O is half the angle of the polygon.

A_{1} = •(• 180°) = • 90° = = 90° -

Cos A_{1} = =

Then A_{1}H_{1} = Cos A_{1} • R = Cos (90° - ) • R = R • Sin()

Knowing that a_{n} = A_{1}H_{1} • 2, we obtain

a_{n} = 2R • Sin ()

***

Examples:

If n=3, then a_{3} = 2R • = R•

If n=4 (the square), then a_{4} = 2R • Sin45° = 2R • = R•

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

***

3)

Sin A_{1 }=

Then r = R • Sin A_{1} = R • (90° - ) = R • Cos ()

***

**Problem 118. **

Given:

The regular quadrilateral is the square

R | r | a_{4} | 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, a_{4}, P, S = ?

Solution:

1) If the side of the square a_{4} = 6 and since given is the square, we see that

the perimeter of the square P = 4 • a_{4} = 4 • 6 = 24

The area of the square S = a_{4}^{2} = 6^{2} = 36

Since the area of the square can be found by the formula S_{4} = P_{4} • r, we see that the radius of the inscribed circle

r = = = 3

Since the number of sides n=4, we see that a_{4} = 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

a_{4} = R•= 2• = 4

Then the perimeter of the square P = 4 • a_{4} = 4 • 4 = 16

The area of the square S = a_{4}^{2} = 4^{2} = 16

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

a_{4}^{2} = 16 a_{4} = ±. But a_{4} = - does not satisfy the condition of the problem.

Therefore, a_{4} = = 4

Then the perimeter of the square P = 4 • a_{4} = 4 • 4 = 16

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

a_{4} = 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

P_{Δ} = 18 cm

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

a_{4} = ?

Solution:

Since P_{n} = n • a_{n} , we see that the side of the triangle

18 = 3 • a_{3} a_{3} = 6

Then the radius of the circumscribed circle

a_{3} = R• R = = 2

Then the side of the square a_{4} = • R = • 2= 2

Answer: 2.

***

**Problem 120. **

Given:

ABCD is the square,

the side of the square a_{4} = 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 S_{4} = P_{4} • r, then the radius of the inscribed circle

r = = = 3 (cm)

Therefore, 2r = 2 • 3 = 6 (cm).

Answer: 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 P_{6} = 48 cm

Find: the perimeter of the square P_{4} = ?

Solution:

Since the formula of the perimeter of the hexagon P_{6} = 6 • a_{6} , then the side of the hexagon a_{6} = 48 : 6 = 8 (cm)

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

a_{6} = 2R • 0,5 = R

Since given is one circle, the radii of the circumscribed circle are equal, i.e. R_{4} = R_{6} = 8 cm

Then the side of the square a_{4} = R•= 8cm

Therefore, the perimeter of the square P_{4} = 4 • 8= 32(cm)

Answer: P_{4} = 32cm.

***

**Problem 122. **

Given:

ABCDEF is the regular hexagon

BF = 1,5 cm

Find: the area of the hexagon

S_{ABCDEF} = ?

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

α_{n }= • 180°, we obtain

α_{6 }= • 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

P_{6} = 6 • = 3(cm)

Then the radius of the inscribed circle

r_{6} = OF = cm

Therefore, the area of the hexagon

S_{6} = • P_{6} • r_{6} = = (см^{2})

Answer: S_{ABCDEF} = cm^{2}.

***

**Problem 123. **

Given:

A_{1}A_{2}A_{3}A_{4}A_{5}A_{6} is a regular hexagon inscribed in a circle(O; R)

B_{1}B_{2}B_{3}B_{4}B_{5}B_{6} is a regular hexagon, circumscribed about the circle(O; R)

OH = r

OH_{1} = R

Find: the proportion of areas of hexagons

= ?

Solution:

1)

Consider the hexagon A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}

The sides of the hexagon are equal to the radii of the circumscribed circle A_{1}A_{2 }= A_{2}A_{3 }= A_{6}A_{1} = R

Knowing that S (A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}) = P • r = 6Rr = 3R•r

Consider the triangle ΔOHA_{6 }is right, since the radius of the inscribed circle is perpendicular to the side of the hexagon r A_{1}A_{6}

Since the point O - is the point of intersection of bisectors, we see that R - is the bisector of the angle A_{1}A_{6}A_{5}

Knowing that OA_{6}A_{1} = 60°, we obtain

Sin 60° = r = Sin 60° • R =

Therefore, the area of the hexagon

S (A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}) = 3R •=R^{2}

2)

Consider the hexagon B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}

B_{1}B_{2 }= ... = B_{6}B_{1} = OB_{5} , where OB_{5} - is the radius of the circumscribed circle about a given regular polygon, namely the hexagon B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}

S (B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}) = • (6 • OB_{5}) • R

Consider ΔH_{1}OB_{5} - is a right triangle, since R - of the inscribed circle in the polygon B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}:

R B_{5}B_{6}

Then OB_{5}H_{1}= 60°. Therefore

Sin 60° = OB_{5} = 2R

Then the area of the hexagon is B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}

S (B_{1}B_{2}B_{3}B_{4}B_{5}B_{6}) = • 6 • 2R • R = 2R^{2}

Then the proportion of the areas of the hexagons

= 2R^{2} • =

Answer:

***