Circumference, Area of A Circle and Circular Sector

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:

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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

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Problem 125.

Given:

a is the side of the regular triangle

 

Find: the length of the circumscribed circle

Solution:

Answer: 8π cm.

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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:

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

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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:

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:

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:

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:

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:

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:

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:

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:

Answer: ≈ 17.6 dm³.

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