Circumference, Area of A Circle and Circular Sector

The area of the circle

Given:

A₁A₂...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 = πR²

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 A₁A₂...A_n.

Knowing that r_n → R, P_n → 2πR, S_n → S as n→∞.

Then S = P_n • r_n = 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.

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.

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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 =

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Problem 131.

Given:

ABCD is the square,

the side of the square is AB = a

 

Calculate:

the area of the filled figure S_EFE1F1 = ?

Solution:

S =

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

S = S₁ = • 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= • = a²

Therefore, the area of the filled figure

S_EFE1F1 = S_ABCD – S _1+2+3+4 = a² – =

Answer: .

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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:

S_circle1 = πR² = π • (OH₁)² = π • 1 = π

S_circle2 = πR² = π • (OH₂)² = π • 4 = 4π

S_circle3 = πR² = π • (OH₃)² = π • 9 = 9π

S_circle4 = πR² = π • (OH₄)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.

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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) .

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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

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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.

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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 = 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: S_circle =

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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.

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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.

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