The contents of this webpage:

- – the lesson "The circumference" is presented in the example of solving problems 124 - 129;
- – in worksheets under numbers 130 - 138 of this workbook for mathematics, we consider how to find the area of the circle and the sector of the circle.

## The circumference

**Derivation of the formula for the circumference.**

Let C and C′ - be the lengths of the circles with the radius R and R′. Let us inscribe the regular polygons in the circles.

P_{n} and P_{n}′ - are the perimeters of the regular polygons,

a_{n} and a_{n}′ are the sides of the regular polygons.

P_{n} = n • a_{n }= n • 2R • Sin

P_{n}′ = n • a_{n }= n • 2R′ • Sin

Then

Knowing that the perimeters P_{n} and P_{n}′ - are approximate values of the lengths of the circles C and C′, as n →∞, we obtain

But by the equality we obtain

By the property of the proportion

The value of the value of π (read "pi") is approximately 3.14.

**Formula for the circumference:**

***

**Problem 124. **

If the radius R = 4, then the circumference C = 2πR = 2 • 3.14 • 4 = 25.12

If C = 82, then the radius of the circle R == = 13.1

If C = 18π, then the radius of the circle R == = 9

***

**Problem 125. **

Given:

a is the side of the regular triangle

Find: the length of the circumscribed circle

Solution:

Since the side of a regular polygon

a_{n} = 2R • Sin (), then the side of the regular triangle

a = R R =

Then the length of the circle circumscribed about the regular triangle is C = 2πR =

***

**Derivation of the formula for calculating the arc L with the degree measure α.**

The degree measure of the circumference is 360°,

The circumference is C = 2πR

The arc length of 1° equals

Then **the length of the arc of the circle in α degrees:**

***

**Problem 126. **

Given:

the radius R= 6 cm,

the angle of the arc

1) α = 30° | 2) α = 45° | 3) α = 60° | 4) α = 90° |

Find: the length of the arc of the circle

Solution:

1) L = • 30° = • 30° = π (cm)

2) L = • 45° = • 3 = 1,5π (cm)

3) L = • 60° = 2π (cm)

4) L = • 90° = 3π (cm)

***

**Problem 127. **

Given:

ABCDEF is the regular hexagon,

the area of the hexagon S_{6} = 24 cm^{2}

Find: what is the length of the circumscribed circle C = ?

Solution:

C = 2πR

Therefore, we need to find the radius of the circumscribed circle.

The area of the hexagon is determined by the formula

S_{6} = • P_{6} • r_{6}

The radius of the inscribed circle is determined by the formula

r_{6} = R • Cos = • R

The side of the hexagon is equal to the radius of the circumscribed circle: a_{6} = R

Then the perimeter of the hexagon P_{6} = 6 • a_{6} = 6R (cm)

S_{6} = • P_{6} • r_{6} = • 6R • • R = 1.5•R^{2}

24= 1.5•R^{2}

R^{2} = = 16 We obtain the radius of the circumscribed circle

R = = 4 (cm)

Then the length of the circumscribed circle is

C = 2πR = 2π • 4 = 8π (cm)

Answer: 8π cm.

***

**Problem 128. **

Given:

ABCD is the square,

the side of the square is AB = a

Find: the length of the inscribed circle C = 2π • r = ?

Solution:

r_{4} = R • Cos = R • Cos 45° = R

C = 2π • r = 2π • R = π • R

AB = a = 2r = R. Therefore, C = π • R= π • a

Answer: the length of the circle inscribed in the square C = π • a

***

**Problem 129. **

Given: the circle (O; R) is circumscribed about the following figures

1) Δ ABC is the inscribed right triangle;

a, b are the legs of the triangle

2) Δ ABC is the inscribed isosceles triangle;

a is the base, b is the side

3) ABCD is the inscribed rectangle,

BC = a is the side of the rectangle,

α is the acute angle between the diagonals

Find: the length of the circumscribed circle C = 2πR = ?

Solution:

1)

2R = AB R = AB

AB =

Then the length of the circle circumscribed about the right triangle

C = 2π • • = π

2)

BH = =

The area of the triangle is equal to half the product of the base by the height

S_{Δ}_{ABC} = BH • AC = (1)

But the area of the triangle can also be found by dividing the product of its three sides by four radii of the circumscribed circle:

S_{Δ}_{ABC} = = (2)

Using equalities (1) and (2), we obtain

= R =

Then the length of the circle circumscribed about an isosceles triangle

C = 2π •

3)

OB = OC = OA = OD = R

We draw OH is the height and the bisector of the isosceles triangle ΔAOD.

Consider the right triangle ΔOHD.

AOD = 180° – α,

HOD = • (180° – α)

ODA = 180° – DHO – HOD

ODA =OAD = 180° – 90° – • (180° – α) =

AH = HD =

CosOAD = Cos = =

R =

Then the length of the circle circumscribed about the rectangle

C = 2π • =

***

## The area of the circle

Given:

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

the circle (O; R)

the small circle′ (O; r_{n})

S is the area of the circle

S_{n}′ is the area of the small circle

Prove:

S = πR^{2}

Proof:

Consider a regular polygon (see the figure).

The area of the circle is larger than the area of the polygon:

S_{n} < S_{circle}

The area of the polygon is larger than the area of the small circle:

S_{n}′ < S_{n}

Then S_{n}′ < S_{n} < S (1)

The radius of a circle inscribed in the regular polygon

r_{n} = R • Cos ()

As n→∞ the cosine Cos ()→1, therefore r_{n} → R

Therefore, S_{n}′ → S as n→∞.

From inequality (1) it follows that S_{n} → S as n→∞.

We know that the area of a regular polygon

S_{n} = P_{n} • r_{n}, where P_{n} is the perimeter of the polygon A_{1}A_{2}...A_{n}.

Knowing that r_{n} → R, P_{n} → 2πR, S_{n} → S as n→∞.

Then S = P_{n} • r_{n} = 2πR • R = πR^{2}

**Formula of the area of the circle:**

***

## The area of the circle sector

**Definition:**

**The sector of the circle **or simply the sector is a part of the circle bounded by an arc and two radii connecting the ends of the arc to the center of the circle.

πR^{2} is the area of the circle.

If the measure of the circular sector is equal to 1°, then the area of this sector is equal to

If the measure of the circular sector is equal to α degrees, then the area of this sector is equal to

**Formula of the sector of the circle**:

, where** **

α is the degree measure of the arc.

***

**Problem 130. **

Given:

AOB = 72°

S is the area of the circular sector

Find: R is the radius of the circle

Solution:

S =

360° • S = πR^{2} • α

R^{2} = = R =

Answer: the radius is R =

***

**Problem 131. **

Given:

ABCD is the square,

the side of the square is AB = a

Calculate:

the area of the filled figure S_{EFE}_{1}_{F1 }= ?

Solution:

S =

Consider in the figure the sector FAE_{1}H_{3}, where AF = AH_{3} = R =

S = S_{1} = • 90° =

The area of four sectors:

S_{1+2+3+4 }= 4 • =

The area of the square

S_{ABCD} = AC • BD.

Consider ΔACD, where AD = CD = a.

By the Pythagorean theorem:

AC =

Then S_{n}= • = a^{2}

Therefore, the area of the filled figure

S_{EFE}_{1}_{F1} = S_{ABCD} – S _{1+2+3+4} = a^{2} – =

Answer: .

***

**Problem 132. **

Given:

the circle (O;OH_{1})

the circle (O;OH_{2})

the circle (O;OH_{3})

the circle (O;OH_{4})

OH_{1} = 1, OH_{2} = 2

OH_{3} = 3; OH_{4} = 4

Find: the area of the circle (O;OH_{1}),

the area of each of the three targets = ?

Solution:

S_{circle1} = πR^{2} = π • (OH_{1})^{2} = π • 1 = π

S_{circle2} = πR^{2} = π • (OH_{2})^{2} = π • 4 = 4π

S_{circle3} = πR^{2} = π • (OH_{3})^{2} = π • 9 = 9π

S_{circle4} = πR^{2} = π • (OH_{4})^{2} = π • 16 = 16π

S_{target2 }= S_{circle2} – S_{circle1} = 4π – π = 3π

S_{target3 }= S_{circle3} – S_{circle2} = 9π – 4π = 5π

S_{target4 }= S_{circle4} – S_{circle3} = 16π – 9π = 7π

Answer: S_{circle1} = π; S_{target2 }= 3π; S_{target3 }= 5π; S_{target4 }= 7π.

***

**Problem 133. **

Given:

the circle (O;R), circumscribed about the quadrilateral and the triangle

1) ABCD is the rectangle,

a and b are the sides of the rectangle

2) Δ ABC is the right triangle,

a is the leg or the cathetus of the triangle,

α is the opposite angle

Find: the area of the circle shown in the figure.

S = πR^{2} = ?

Solution:

1)

BD = 2BO = 2R

Consider the right triangle ΔABD.

By the Pythagorean theorem:

BD^{2} = AB^{2} + AD^{2}

(2R)^{2} = a^{2} + b^{2}

R^{2} =

Then the area of the circle S = πR^{2} = π •

2)

Find the area of a circle having used its diameter AB = 2AO = R

Sin α = AB =

2R = R =

Therefore,

S = πR^{2} = π • = π • =

Answer: 1) π •; 2) .

***

**Problem 134. **

Given:

ΔABC is the right triangle

circle_{1} (O_{1}; AO_{1}) on the hypotenuse AB

circle_{2} (O_{2}; BO_{2}) on the leg BC

circle_{3}(O_{3}; CO_{3}) on the leg AC

Prove:

The sum of the area of the semicircle on the hypotenuse is equal to the sum of the areas of the semicircles on the legs

S_{1} = S_{2} + S_{3}

Proof:

Suppose that AB = c; AC = a; BC = b.

The formula of the area of the circle sector

S =

S = =

S_{1} = • O_{1}A^{2}, where O_{1}A = c S_{1} =•c^{2}

S_{2} = • O_{2}B^{2}, where O_{2}B = b S_{2} =•b^{2}

S_{3} = • O_{3}C^{2}, where O_{3}C = a S_{3} =•a^{2}

Then S_{2} + S_{3} =•a^{2 }+•b^{2 }= • (a^{2} + b^{2})

By the Pythagorean theorem:

c^{2} = a^{2} + b^{2}

Therefore,

S_{2} + S_{3}^{ }= • (a^{2} + b^{2}) = • c^{2} = S_{1}

Therefore S_{1} = S_{2} + S_{3}

***

**Problem 135. **

Given:

the circle (O; AO)

AO is the radius

AO = 10 cm

AMB = AOB = 60°

Find:

what is the area of the circle sector with the arc ALB = ?

Solution:

The degree measure of the arc

ALB = 360° – 60° = 300°

Then the area of the circle sector

S = = = ≈ 261.67 ≈ 262 (cm^{2})

Answer: the area of the segment of the circle is S ≈ 262 cm^{2}.

***

**Problem 136. **

Given:

circle (O; OH) is inscribed in ΔABC

the triangle ΔABC is equilateral

AB = a

Find: what is the area of the circle

S_{circle} = ?

Solution:

S = πR^{2} = πr^{2}

Consider the triangle ΔABH – is the right triangle.

AO is the bisector of the angle A.

Therefore, OAH = 60° : 2 = 30°

OH = AO

r =

AOH = 180° – (30° + 90°) = 60°

Sin 60° = the radius of the circumscribed circle R =

Then the radius of the inscribed circle r = : 2 =

Therefore, the area of the circle S = πr^{2} = =

Answer: S_{circle} =

***

**Problem 137. **

Given:

the small circle (O; OD)

the area of a large circle

S _{circle(}_{O; }_{OC)} = 314 mm^{2}

the diameter of a small circle

D_{ circle(}_{O; }_{OD)} = 18.5 mm

Find: the difference in diameters

HC = ?

Solution:

S = πR^{2}

314 = πR^{2}

= = 10 (mm)

D_{2} = 2R = 2 • 9.25 = 18.5 (mm)

R_{2} = 9.25 mm

HC = R_{1} – R_{2} = 10 – 9.25 = 0.75 (mm).

Answer: 0.75 mm.

***

**Problem 138. **

Given:

the small circle (O; OH) is the hole of the pipe

OH = 3m is the radius of a small pipe

AB = 1m is the difference in diameters between the two pipes

A square meter requires 0.8 cubic decimeters of sand

1 m^{2} → 0.8 dm^{3}

Find: how much sand you need to fill the space between two pipes

Solution:

Consider the circle′ (O; OH).

The area of this circle:

S′ = πR^{2} = 9 • 3.14 ≈ 28.26 (m^{2})

Consider the circle′′ (O; OH+AB).

S′′ = πR^{2} = 16 • 3.14 ≈ 50.24 (m^{2})

Then the area between two pipes

S = S′′ – S′ = 50.24 – 28.26 ≈ 21.98 (m^{2})

Then the required amount of sand

21.98 • 0.8 ≈ 17.58 dm^{3} ≈ 17.6 dm^{3}

Answer: ≈ 17.6 dm^{3}.

***