Circumference, area of the circle and the circular sector

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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.

Pn and Pn′ - are the perimeters of the regular polygons,

an and an′ are the sides of the regular polygons.

Pn = n • an = n • 2R • Sin

Pn′ = n • an = n • 2R′ • Sin

Then

Knowing that the perimeters Pn and Pn′ - 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:

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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

an = 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 S6 = 24 cm2

 

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

S6 = • P6 • r6

The radius of the inscribed circle is determined by the formula

r6 = R • Cos = • R

The side of the hexagon is equal to the radius of the circumscribed circle: a6 = R

Then the perimeter of the hexagon P6 = 6 • a6 = 6R (cm)

S6 = • P6 • r6 = • 6R • • R = 1,5•R2

24= 1,5•R2

R2 = = 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:

r4 = 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

SΔABC = 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:

SΔABC = = (2)

 

Using the 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 • =

 

***

 

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The area of the circle

Given:

A1A2...An is the regular polygon

the circle (O; R)

the small circle′ (O; rn)

S is the area of the circle

Sn′ is the area of the small circle

 

Prove:

S = πR2

Proof:

Consider a regular polygon (see the figure).

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

Sn < Scircle

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

Sn′ < Sn

Then Sn′ < Sn < S (1)

 

The radius of a circle inscribed in the regular polygon

rn = R • Cos ()

 

As n→∞ the cosine Cos ()→1, therefore rn → R

Therefore, Sn′ → S as n→∞.

From the inequality (1) it follows that Sn → S as n→∞.

We know that the area of a regular polygon

Sn = Pn • rn, where Pn is the perimeter of the polygon A1A2...An.

Knowing that rn → R, Pn → 2πR, Sn → S as n→∞.

Then S = Pn • rn = 2πR • R = πR2

Formula of the area of the circle:

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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.

πR2 is the area of the circle.

The area of the circular sector, which measure is 1°, is equal to

The area of the circular sector, which measure is α degrees, 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 = πR2 • α

R2 = = 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 SEFE1F1 = ?

Solution:

S =

Consider in the figure the sector FAE1H3, where AF = AH3 = R =

S = S1 = • 90° =

The area of four sectors:

S1+2+3+4 = 4 • =

The area of the square

SABCD = AC • BD.

Consider ΔACD, where AD = CD = a.

By the Pythagorean theorem:

AC =

Then Sn= = a2

Therefore, the area of the filled figure

SEFE1F1 = SABCD – S 1+2+3+4 = a2=

Answer: .

***

 

Problem 132.

Given:

the circle (O;OH1)

the circle (O;OH2)

the circle (O;OH3)

the circle (O;OH4)

OH1 = 1, OH2 = 2

OH3 = 3; OH4 = 4

 

Find: the area of the circle (O;OH1),

the area of each of the three targets = ?

Solution:

Scircle1 = πR2 = π • (OH1)2 = π • 1 = π

Scircle2 = πR2 = π • (OH2)2 = π • 4 = 4π

Scircle3 = πR2 = π • (OH3)2 = π • 9 = 9π

Scircle4 = πR2 = π • (OH4)2 = π • 16 = 16π

 

Starget2 = Scircle2 – Scircle1 = 4π – π = 3π

Starget3 = Scircle3 – Scircle2 = 9π – 4π = 5π

Starget4 = Scircle4 – Scircle3 = 16π – 9π = 7π

Answer: Scircle1 = π; Starget2 = 3π; Starget3 = 5π; Starget4 = 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 = πR2 = ?

Solution:

1)

BD = 2BO = 2R

Consider the right triangle ΔABD.

By the Pythagorean theorem:

BD2 = AB2 + AD2

(2R)2 = a2 + b2

R2 =

Then the area of the circle S = πR2 = π •

2)

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

Sin α = AB =

2R = R =

 

Therefore,

S = πR2 = π • = π • =

Answer: 1) π •; 2) .

***

 

Problem 134.

Given:

ΔABC is the right triangle

circle1 (O1; AO1) on the hypotenuse AB

circle2 (O2; BO2) on the leg BC

circle3(O3; CO3) 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

S1 = S2 + S3

Proof:

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

The formula of the area of the circle sector

S =

S = =

S1 = • O1A2, where O1A = c S1 =•c2

S2 = • O2B2, where O2B = b S2 =•b2

S3 = • O3C2, where O3C = a S3 =•a2

Then S2 + S3 =•a2 +•b2 = • (a2 + b2)

By the Pythagorean theorem:

c2 = a2 + b2

Therefore,

S2 + S3 = • (a2 + b2) = • c2 = S1

Therefore S1 = S2 + S3

***

 

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 (cm2)

Answer: the area of the segment of the circle is S ≈ 262 cm2.

***

 

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

Scircle = ?

Solution:

S = πR2 = πr2

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 = πr2 = =

Answer: Scircle =

***

Problem 137.

Given:

the small circle (O; OD)

the area of a large circle

S circle(O; OC) = 314 mm2

the diameter of a small circle

D circle(O; OD) = 18,5 mm

 

Find: the difference in diameters

HC = ?

Solution:

S = πR2

314 = πR2

= = 10 (mm)

D2 = 2R = 2 • 9,25 = 18,5 (mm)

R2 = 9,25 mm

HC = R1 – R2 = 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 m2 → 0,8 dm3

 

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′ = πR2 = 9 • 3,14 ≈ 28,26 (m2)

Consider the circle′′ (O; OH+AB).

S′′ = πR2 = 16 • 3,14 ≈ 50,24 (m2)

Then the area between two pipes

S = S′′ – S′ = 50,24 – 28,26 ≈ 21,98 (m2)

Then the required amount of sand

21,98 • 0,8 ≈ 17,58 dm3 ≈ 17,6 dm3

Answer: ≈ 17,6 dm3.

***