Unit 6, Lesson.1 Circumference and Area of a Circle 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 can be proven. You know that the circumference of a specific circle divided by its diameter is the ratio pi, written as. Pi ( ) is an irrational number which is a number that cannot be written as a repeating decimal or as a fraction. It has an infinite number of non-repeating decimal places. You know that the circumference of a circle is equal to ( diameter ) or (radius ). Therefore, 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. He determined that each side of the hexagon equals the radius of the circle. AB = BD = DE = EF = FG = GA = AC Archimedes realized that if the perimeter of the hexagon (6r) were equal to the circumference ( r) 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. 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. MUnit6Lesson.1 1 1/3/019
He calculated the perimeter of the dodecagon to be approximately 6.1166. This means 3.1083. However, the circumference of the circle is still larger than the dodecagon, so 3.1083. must be greater than 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. Archimedes kept going with this process until he had inscribed a 48-sided polygon in the circle. As the number of sides of a polygon increases, the polygon looks more and more like a circle. As he added more sides to the polygon, the value of 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 the limit of the perimeter of a regular polygon inscribed in a circle divided by the radius of the circle, as the number of sides of the polygon approaches infinity, is equal to the value of. A limit is the value that a sequence approaches as a calculation becomes more and more accurate. This limit cannot be reached, because the number of sides can always be increased. Theoretically, if the polygon had an infinite number of sides, for the formula for finding the circumference of a circle. could be calculated. This is the basis Increasing the number of side lengths for the inscribed polygon causes the polygon s perimeter to get closer and closer to 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 below, 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 parallelogram. The height of the parallelogram equals the radius, r, of the original circle. The base is equal to half of the circumference, or (r ). The circle in the diagram to the right has been divided into 16 equal sections. MUnit6Lesson.1 1/3/019
You can arrange the 16 segments to form a new parallelogram. This figure looks more like a parallelogram. As the number of sections increases, the rounded bumps along its base become less and less distinct. The more sections, the closer the figure is to a parallelogram. This is another example of a limit; as the number of sections approaches infinity, the shape of the assembled sections approaches a parallelogram. The formula for the area of a rectangle is Area b(h). The base of the parallelogram made out of the circle segments is (r ). The height is r. Thus, the area of the circle is Area r ( r ) or A r. This proof is based on the dissection of the circle. Remember that a sector is a part of a circle that is determined 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. 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 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 meters. Compare the estimate with the circle s perimeter found by using the formula C r. 1. Draw a circle and inscribe a regular hexagon in the circle.. Create a triangle with a vertex at the center of the circle. Draw two line segments from the center of the circle to consecutive vertices on the hexagon. MUnit6Lesson.1 3 1/3/019
Example 1 (continued) 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, 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 a circle. 1 m CPB ( 360 ) 6 60 The measure of m CPB is 60.. 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. 6. Determine m BPD. DP bisects, or cuts in half, CPB. Since the measure of mcpb was found to be 60, divide 60 by to determine m BPD. 60 30 The measure of BPD is 30. 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 meters, and the measure of BPD is 30. The trigonometry ratio that uses the opposite and hypotenuse lengths is sine. sin 30 sin 30 BD BD BD 0. Evaluate the sine of 30 ( 0.) BD Multiply both sides of the equation by BD. The length of BD is. meters. Since BC is twice the length of BD, multiply. by. BC (.) BC The length of BC is meters. MUnit6Lesson.1 4 1/3/019
Example 1 (continued) 8. Find the perimeter of the hexagon. Perimeter BC( 6) ( 6 ) 30 The perimeter of the hexagon is 30 meters. 9. Compare the estimate with the calculated circumference of the circle. Example Calculate the circumference. C r Formula for circumference C () Substitute for r C 10 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 0.041 or 4.1% 31.416 From a proportional perspective, the circumference calculation is approximately 4.1% 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. 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 meters. Compare the estimate with the circle s area found by using the formula Area 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. MUnit6Lesson.1 1/3/019
Example (continued) First, determine the formula to use. 1 Area bh Area formula for a triangle 1 A PBC (BC )(PD ) Rewrite the formula with the base and height of the triangle whose area you are trying to determine. From Example 1, the following information is known: BC = meters BD =. meters m BPD 30 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 meters. Since the measure of BPD and the hypotenuse are known, use the cosine of 30º to find PD. PD cos30 PD cos30 PD 0.86604038 Evaluate the cosine of 30 ( 0.86604038) PD Multiply both sides by PD 4.3307019 meters Now that the length of PD is known, use that information to find the area of the formula determined earlier. 1 A PBC (BC )(PD ) Area formula for PBC 1 A PBC ( )(4.3307019 ) Substitute the values of BC and PD A 10.86748 m PBC 3. Find the area of the hexagon. Multiply the area of one triangle times 6, the number of triangles in the hexagon. 6(10.86748) 64.94088 m The area of the hexagon is about 64.9 m. 4. Compare the area of the hexagon with the area of the circle. Find the area of the circle. A circle P r Formula for the area of a circle circle P ) A ( Substitute the value for the radius A circle P circle P A 78.3981634 m The actual area of circle P is about 78.4 m. PBC using MUnit6Lesson.1 6 1/3/019
Example (continued) Example 3 Find the difference between the area of the hexagon and the area of the circle. 78.3981634 64.94088 13.876346 The actual area of the circle is approximately 13.9 m greater than the hexagon s area. Show the difference as a percentage. 13.9 0.1730 or 17.30% 78.4 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. 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 r 100 r 100 r r 0 r 1.91 meters. Calculate the area by using the formula for the area of a circle. Example 4 Area r A (1.91 ) A 3. 3031 A 79.77 m The area of a circle with a circumference of 100 meters is approximately What is the circumference of a circle that has an area of 1000 m? 796 m. 1. First, find the radius by solving for r in the formula for the area of the circle. Area r Formula for the area of a circle 1000 r Substitute Area into the equation 1000 r r 318.3099 Simplify Divide both sides by r 318.3099 Take the square root of both sides r 17.841 m Use only the positive result, as a distance is always positive The radius r is approximately 17.841 m. MUnit6Lesson.1 7 1/3/019
Example 4 (continued). Find the circumference using the formula C r. C r Formula for the circumference of a circle C (17.841) Substitute 17.841 for r C 11.1099 C 11.1 m The circumference of a circle with an area of 1000 m is approximately 11.1 meters. MUnit6Lesson.1 8 1/3/019