Prerequisite Skills This lesson requires the use of the following skills: using formulas for the surface areas of polygons and circles performing calculations with the angles in circles using the Pythagorean Theorem using ratios of trigonometry understanding how to bisect angles and side lengths Introduction You 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 π can be proven. Key Concepts You know that the circumference of a specific circle divided by its diameter is the ratio pi, written as π. Pi (π) 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 = π diameter or π radius. Therefore, π = circumference diameter = circumference radius 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.. U3-03
A B G C D F E 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 π = 3. However, the circumference is larger than the hexagon; therefore, Archimedes thought, π must be larger than 3. Next, Archimedes inscribed a regular dodecagon a 1-sided polygon in the circle. The perimeter of the dodecagon was much closer to the actual perimeter of the circle. U3-04
He calculated the perimeter of the dodecagon to be approximately 6.1166. This means π 3.10583. However, the circumference of the circle is still larger than the dodecagon, so π must be greater than 3.10583. Next, Archimedes inscribed a 4-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 π. 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 π 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 π. 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, π 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. The area of the circle can be derived similarly using dissection principles. Dissection involves breaking a figure down into its components. In the diagram that follows, a circle has been divided into four equal sections. If you cut the four sections from the circle apart, you can arrange them to resemble a rectangle. r r U3-05
The width of the rectangle equals the radius, r, of the original circle. The length is equal to half of the circumference, or π r. The circle in the diagram below has been divided into 16 equal sections. You can arrange the 16 segments to form a new rectangle. This figure looks more like a rectangle. r r 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 π r. The width is r. Thus, the area of the circle is a = r π r = π r. This proof is a dissection of the circle. 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 0º and less than 360º. 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 U3-06
Guided Practice 3.5.1 Example 1 Show 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 = πr. 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.. 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. C B P U3-07
3. To find the length of BC, first determine the known lengths of PB and PC. Both lengths are equal to the radius of circle P, 5 meters. 4. Determine m CPB. The hexagon has 6 sides. A central angle drawn from P will be equal to one-sixth of the number of degrees in circle P. 1 m CPB = 360 = 60 6 The measure of CPB is 60º. 5. Use trigonometry to find the length of BC. Make a right triangle inside of PBC by drawing a perpendicular line, or altitude, from P to BC. C D B P U3-08
6. Determine m BPD. DP bisects, or cuts in half, CPB. Since the measure of CPB was found to be 60º, divide 60 by to determine m BPD. 60 = 30 The measure of BPD is 30º. 7. Use trigonometry to find the length of BD and multiply that value by to find the length of BC. BD is opposite BPD. The length of the hypotenuse, PB, is 5 meters. The trigonometry ratio that uses the opposite and hypotenuse lengths is sine. BD sinbpd = sin 30º = 5 05. = BD Substitute the sine of 30º. 5 5 0.5 = BD Multiply both sides of the equation by 5. BD =.5 The length of BD is.5 meters. Since BC is twice the length of BD, multiply.5 by. BC =.5 = 5 The length of BC is 5 meters. U3-09
8. Find the perimeter of the hexagon. Perimeter = BC 6 = 5 6 = 30 The perimeter of the hexagon is 30 meters. 9. Compare the estimate with the calculated circumference of the circle. Calculate the circumference. C = πr Formula for circumference C = π 5 Substitute 5 for r. C 31.416 meters Find the difference between the perimeter of the hexagon and the circumference of the circle. 31.416 30 = 1.416 meters 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. 1. 416 = 0. 0451 = 451. % 31. 416 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. U3-10
Example Show 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 A = π r. 1. Inscribe a hexagon into a circle and divide it into 6 equal triangles.. Use the measurements from Example 1 to find the area of one of the six triangles. C D B P First, determine the formula to use. A= 1 bh A PBC = 1 BC PD Area formula for a triangle Rewrite the formula with the base and height of the triangle whose area you are trying to determine. (continued) U3-11
From Example 1, the following information is known: BC = 5 meters BD =.5 meters m BPD =30º You need to find the height, PD. In BPD, the height, PD, is the adjacent side length. Since the hypotenuse, BP, is a radius of the circle, it is 5 meters. Since the measure of BPD and the hypotenuse are known, use the cosine of 30º to find PD. cos30º = PD 5 0. 866054038 = PD Substitute the cosine of 30º. 5 5 0.866054038 = PD Multiply both sides by 5. PD 4.3307019 meters Now that the length of PD is known, use that information to find the area of PBC using the formula determined earlier. A PBC = 1 BC PD Area formula for PBC A PBC A PBC = 1 5 4.3307019 ( ) ( ) 10. 8567548 m 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. 6 10.8567548 = 64.9540588 m The area of hexagon is about 64.95 m. U3-1
4. Compare the area of the hexagon with the area of the circle. Find the area of the circle. Acircle P =πr Formula for the area of a circle A circle P =π 5 Substitute the value for the radius. A circle P 78. 53981634 m The actual area of circle P is about 78.54 m. Find the difference between the area of the hexagon and the area of the circle. 78.53981634 64.9540588 = 13.58576346 The actual area of the circle is approximately 13.59 m greater than the hexagon s area. Show the difference as a percentage. 13. 59 0. 1730 17. 30% 78. 54 The 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. U3-13
Example 3 Find the area of a circle that has a circumference of 100 meters. 1. First, find the measure of the radius by using the formula for circumference. C = πr 100 = πr r = 100 π r 15.9155 m. Calculate the area by using the formula for the area of a circle. A = π r A π( 15. 9155) A 53.3031π m A 795.775 m The area of a circle with a circumference of 100 meters is approximately 796 m. U3-14
Example 4 What is the circumference of a circle that has an area of 1,000 m? 1. First, find the radius by solving for r in the formula for the area of the circle. A = π r Formula for the area of a circle 1000 = π r Substitute A into the equation. 1000 = r π Divide both sides by π. r 318.3099 Simplify. r =± 318. 3099 Take the square root of both sides. Use only the positive result, as distance is r 17.841 m always positive. The radius is approximately 17.841 meters.. Find the circumference using the formula C = π r. C = πr Formula for circumference of a circle C = π(17.841) Substitute 17.841 for r. C 11.1 m The circumference of a circle with an area of 1,000 m is approximately 11.1 meters. U3-15