Introduction You have used the formulas for finding the circumference and area of a circle. In this...

IntroductionYou have used the formulas for finding the circumference and area of a circle. In this lesson, you will prove why the formulas for circumference and area work. You will see how the ratio of can be proven.

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3.5.1: Circumference and Area of a Circle

Key Concepts• You know that the circumference of a specific circle

divided by its diameter is the ratio pi, written as .

• Pi is an irrational number that cannot be written as a repeating decimal or as a fraction. It has an infinite number of non-repeating decimal places.

• We know that the circumference of a

circle = • diameter or 2 • radius.

• Therefore,

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Key Concepts, continued• Long ago, mathematicians didn’t yet know the value

of pi. Archimedes, a great mathematician from ancient Greece, used inscribed polygons to determine the value of pi. He started by inscribing a regular hexagon in a circle.

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Key Concepts, continued

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Key Concepts, continued• He determined that each side of the hexagon equals

the radius of the circle.

AB = BD = DE = EF = FG = GA = CE

• Archimedes realized that if the perimeter of the hexagon were equal to the circumference of the circle, then both would equal 6r. This would mean that = 3. However, the circumference is larger than the hexagon; therefore, Archimedes thought, must be larger than 3.

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Key Concepts, continued• Next, Archimedes

inscribed a regular dodecagon—a 12-sided polygon—in the circle. The perimeter of the dodecagon wasmuch closer to the actual perimeter ofthe circle.

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Key Concepts, continued• He calculated the perimeter of the dodecagon to be

approximately 6.21166. This means ≈ 3.10583. However, the circumference of the circle is still larger than the dodecagon, so must be greater than 3.10583.

• Next, Archimedes inscribed a 24-sided regular polygon and calculated its perimeter. This polygon’s perimeter is even closer to the circumference of a circle. Archimedes found that the ratio of the perimeter to the diameter is closer to the value of .

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Key Concepts, continued• Archimedes kept going with this process until he had

inscribed a 48-sided polygon. As the number of sides of a polygon increases, the polygon looks more and more like a circle. As he worked, the number for the ratio of became more and more accurate. The more sides an inscribed polygon has, the closer its perimeter is to the actual circumference of the circle.

• Therefore, Archimedes determined that as the number of sides of a polygon inside a circle increases, the calculation approaches the limit for the value of .

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Key Concepts, continued• A limit is the value that a sequence approaches as a

calculation becomes more and more accurate. This limit cannot be reached.

• Theoretically, if the polygon had an infinite number of sides, could be calculated. This is the basis for the formula for finding the circumference of a circle.

• Increasing the number of side lengths for the inscribed polygon causes the polygon’s perimeter to get closer and closer to the length of the circumference of the circle.

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Key Concepts, continued• The area of the circle can be derived similarly using

dissection principles. Dissection involves breaking a figure down into its components.

• In the diagram below, a circle has been divided into four equal sections.

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Key Concepts, continued• If you cut the four sections from the circle apart, you

can arrange them to resemble a rectangle.

• The width of the “rectangle” equals the radius, r, of the original circle. The length is equal to half of the circumference, or • r. 11


Key Concepts, continued• The circle in the

diagram to the right has been divided into 16 equal sections.

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Key Concepts, continued• You can arrange the 16 segments to form a new

“rectangle.”

• This figure looks more like a rectangle.

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Key Concepts, continued• As the number of sections increases, the rounded

“bumps” along its length and the “slant” of its width become less and less distinct. The figure will approach the limit of being a rectangle.

• The formula for the area of a rectangle is l • w = a. The length of the rectangle made out of the circle segments is • r. The width is r. Thus, the area of the circle is . This proof is a dissection of the circle.

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Key Concepts, continued• Remember that a sector is the part of a circle that is

enclosed by a central angle.

• A central angle has its vertex on the center of the circle. A sector will have an angular measure greater than 0° and less than 360°.

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Common Errors/Misconceptions• not realizing that there is more than one way to

prove a formula • using the wrong formula for area, circumference, or

the area of a sector • using the diameter in a formula instead of the radius

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Guided Practice

Example 1Show how the perimeter of a hexagon can be used to find an estimate for the circumference of a circle that has a radius of 5 meters. Compare the estimate with the circle’s perimeter found by using the formula C = 2 r.

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Guided Practice: Example 1, continued

1. Draw a circle and inscribe a regular hexagon in the circle. Find the length of one side of the hexagon and multiply that length by 6 to find the hexagon’s perimeter.

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2. Create a triangle with a vertex at the center of the circle. Draw two line segments from the center of the circle to vertices that are next to each other on the hexagon.

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3. To find the length of , first determine the known lengths of and Both lengths are equal to the radius of circle P,

5 meters.

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4. DetermineThe hexagon has 6 sides. A central angle drawn from P will be equal to one-sixth of the number of degrees in circle P.

The measure of is 60°.

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5. Use trigonometry to find the length of Make a right triangle

inside of by

drawing a

perpendicular line,

or altitude, from P

to .

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6. Determine bisects, or cuts in half, . Since the measure of was found to be 60°, divide 60 by 2 to determine

The measure of is 30°.

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7. Use trigonometry to find the length of

and multiply that value by 2 to find the

length of is opposite .

The length of the hypotenuse, , is 5 meters.

The trigonometry ratio that uses the opposite and hypotenuse lengths is sine.

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The length of is 2.5 meters.

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Substitute the sine of 30°.

Multiply both sides of the equation by 5.


Since is twice the length of , multiply 2.5 by 2.

The length of is 5 meters.

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8. Find the perimeter of the hexagon.

The perimeter of the hexagon is 30 meters.

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9. Compare the estimate with the calculated circumference of the circle. Calculate the circumference.

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Formula for circumference

Substitute 5 for r.

Guided Practice: Example 1, continuedFind the difference between the perimeter of the hexagon and the circumference of the circle.

The formula for circumference gives a calculation that is 1.416 meters longer than the perimeter of the hexagon. You can show this as a percentage difference between the two values.

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From a proportional perspective, the circumference calculation is approximately 4.51% larger than the estimate that came from using the perimeter of the hexagon.

If you inscribed a regular polygon with more side lengths than a hexagon, the perimeter of the polygon would be closer in value to the circumference of the circle.

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Guided Practice

Example 2Show how the area of a hexagon can be used to find an estimate for the area of a circle that has a radius of 5 meters. Compare the estimate with the circle’s area found by using the formula

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1. Inscribe a hexagon into a circle and divide it into 6 equal triangles.

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2. Use the measurements from Example 1 to find the area of one of the six triangles.

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Guided Practice: Example 2, continuedFirst, determine the formula to use.

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Area formula for a triangle

Rewrite the formula with the base and height of the triangle whose area you are trying to determine.

Guided Practice: Example 2, continuedFrom Example 1, the following information is known:

You need to find the height, . In , the height, , is the adjacent side length.

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Guided Practice: Example 2, continuedSince the hypotenuse, , is a radius of the circle, it is 5 meters. Since the measure of and the hypotenuse are known, use the cosine of 30º to find

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Substitute the cosine of 30°.

Multiply both sides by 5.

Guided Practice: Example 2, continuedNow that the length of is known, use that information to find the area of using the formula determined earlier.

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Area formula for

Substitute the values of BC and PD.


3. Find the area of the hexagon. Multiply the area of one triangle times 6, the number of triangles in the hexagon.

The area of hexagon is about 64.95 m2.

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4. Compare the area of the hexagon with the area of the circle. Find the area of the circle.

The actual area of circle P is about 78.54 m2.42


Formula for the area of a circle

Substitute the value for the radius.

Guided Practice: Example 2, continuedFind the difference between the area of the hexagon and the area of the circle.

The actual area of the circle is approximately13.59 m2 greater than the hexagon’s area.

Show the difference as a percentage.

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Guided Practice: Example 2, continuedThe actual area of the circle is about 17.30% larger than the estimate found by using the area of the hexagon.

The estimate of a circle’s area calculated by using an inscribed polygon can be made closer to the actual area of the circle by increasing the number of side lengths of the polygon.

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Introduction You have used the formulas for finding the circumference and area of a circle. In this...

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Transcript of Introduction You have used the formulas for finding the circumference and area of a circle. In this...