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Chapter 11

Chapter 11. Area of Polygons and Circles. Chapter 11 Objectives. Calculate the sum of the interior angles of any polygon Calculate the area of any regular polygon Compare the perimeters of similar polygons Compare the areas of similar polygons Define circumference Calculate arc length

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Chapter 11

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  1. Chapter 11 Area of Polygons and Circles

  2. Chapter 11 Objectives • Calculate the sum of the interior angles of any polygon • Calculate the area of any regular polygon • Compare the perimeters of similar polygons • Compare the areas of similar polygons • Define circumference • Calculate arc length • Find the area of a circle • Define a sector • Find the area of a sector • Utilize geometric probability

  3. Lesson 11.1 Angle Measure in Polygons

  4. Lesson 11.1 Objectives • Utilize the Polygon Interior Angles Theorem • Calculate the number of sides on a polygon know the interior angle sum • Find the sum of the exterior angles of a polygon

  5. Interior Angles of a Polygon • The sum of the interior angles of a triangle is • 180o • The sum of the interior angles of a quadrilateral is • 360o • The sum of the interior angles of a pentagon is • ??? • The sum of the interior angles of a hexagon is • ??? • By splitting the interior into triangles, it should be able to tell you the sum of the interior angles. • Just count up the number of triangles and multiply by 180o. 180o 360o 540o 720o

  6. Theorem 11.1:Polygon Interior Angles Theorem • The sum of the measure of the interior angles of a convexn-gon is • 180o(n – 2) • n is the number of sides • This basically states that for every side that you add, you add another 180o to the interior angles.

  7. Example 1 • First determine the number of sides of the polygon. • Plug n into Theorem 11.1 and solve. • 180(8–2) = • 1080o • Create an equation that has all interior angles equal to the answer from above. • 155 + 135 + 145 + 150 + 110 + 120 + 130 + x = 1080 • And solve for x • 945 + x = 1080 • x = 135o 145o 135o 155o 150o xo 110o 120o 130o

  8. Corollary to Theorem 11.1 • The measure of each interior angle of a regularn-gon is • 1/n(180)(n - 2) • It must be regular! • It basically states take the sum of the interior angles and divide by the number of sides to figure out how big each angle is.

  9. Example 2 Find the measure of each interior angle in the figure at right. • n = 5 • 1/5(180)(5 – 2) • 180/5(3) • (36)(3) • 108O 108O

  10. Example 3:Finding the Number of Sides = 120n = 180n – 360 -60n = – 360 n = 6 • Each interior angle is 120o, name the polygon. • Use Corollary 11.1 • 1/n(180)(n – 2) • Set it equal the measure of each interior angle and solve for n. • 1/n(180)(n – 2) = 120 • Multiply both sides by n • 180(n – 2) = 120n • Distribute • 180n – 360 = 120n • Subtract 180n from both sides • -360 = -60n • Divided by –60 • n = 6 • Hexagon Could be easier if set up as a proportion. 180(n – 2) 120 n 1 Dividing by n is the same as multiplying by 1/n 120n = 180(n – 2)

  11. Exterior Angles 4 3 5 2 1 • An exterior angle is formed by extending each side of a polygon in one direction. • Make sure they all extend either pointing clockwise or counter-clockwise.

  12. Theorem 11.2:Polygon Exterior Angles Theorem 4 3 5 2 1 • The sum of the measures of the exterior angles of a convexpolygon is 360o. • As if you were traveling in a circle!  1 +  2 +  3 +  4 +  5 = 360o

  13. Corollary to Theorem 11.2 • The measure of the each exterior angle of a regularn-gon is • 360/n • It must be regular! • This will be used to determine the number of sides in a polygon.

  14. Example 4 • The measure of an exterior angle of a regular polygon is 120o. Name it. • Plug into Corollary 11.2 • 360/120 = n • n = 3 • Triangle

  15. Homework 11.1 • In Class • 1-5 • p665-668 • HW • 6-41, 49-54, 58-61, 63-73 • Due Tomorrow

  16. Lesson 11.2 Areas of Regular Polygons

  17. Lesson 11.2 Objectives • Area of Equilateral Triangle Theorem • Use the Area of a Regular Polygon Theorem • Know parts of a polygon • Define central angle

  18. Theorem 11.3:Area of an Equilateral Triangle • Area of an equilateral triangle is • A = ¼ (√3) s2 • Take ¼ times the length of a side squared and write in front of √3. • Be sure to simplify if possible. s

  19. Parts of a Polygon • The centerof a polygon is the center of the polygon’s circumscribed circle. • A circumscribed circle is one in that is drawn to go through all the vertices of a polygon. • The radius of a polygon is the radius of its circumscribed circle. • Will go from the center to a vertex. r

  20. Apothem • The apothem is the distance from the center to any side of the polygon. • Not to the vertex, but to the center of the side. • The height of a triangle formed between the center and two consecutive vertices of the polygon. a

  21. Central Angle of a Polygon • The central angle of a polygon is the angle formed by drawing lines from the center to two consecutive vertices. • This is found by • 360/n • That is because the total degrees traveled around the center would be like a circle. • Then divide that by the number of sides because that determines how many central angles could be formed. 600

  22. In order to find the apothem, you must know one of the following length of a side length of radius central angle You will use trig to find the missing apothem If given the radius, you will use cosine and half the central angle. If given a side, you will use half the side, tangent, and half the central angle. In order to find the radius, you must know one of the following length of a side length of apothem central angle You will use trig to find the missing radius If given the apothem, you will use cosine and half the central angle If given a side, you will use half the side, sine, and half the central angle. Finding the Apothem or Radius r a s

  23. Example 4 • Find the length of the apothem and the side of a regular pentagon with a radius of 5. • Apothem • Using radius • cos (1/2CA) = a/r • cos (1/2(72) = a/5 • .8090 = a/5 • 5(.8090) = a • a = 4.045 • Side length • Using radius • sin (1/2 CA) = .5s/r • sin (36) = .5s/5 • .5879 = .5s/5 • 5(.5879) = .5s • 2.939 = .5s • s = 5.878 Remember that the angle used for calculation is half the central angle. And the bottom of the triangle is half the length of one entire side. CA = 72o

  24. Theorem 11.4:Area of a Regular Polygon • The area of a regular n-gon with side length s is half the product of the apothem and the perimeter. • A = 1/2aP • A stands for area • a stands for apothem • P stands for perimeter of the n-gon • Found by finding the side length and multiplying by the number of sides • A = 1/2a(ns)

  25. Homework 11.2 • In Class • 1-8 • p672-675 • HW • 9-34, 50-52, 54-64 • Due Tomorrow

  26. Lesson 11.3 Perimeters and Areas of Similar Figures

  27. Lesson 11.3 Objectives • Compare the perimeters of similar figures • Compare the areas of similar figures

  28. Theorem 11.5:Areas of Similar Polygons • If two polygons are similar with the lengths of corresponding sides in the ratio of a:b, then the ratio of their areas are a2:b2 • Remember that the ratio of side lengths a:b is the same as the ratio of the perimeters, a:b. • Theorem 8.1 Ratio of Sides 15/5 = 3 Ratio of Perimeters 15/5 = 3 Ratio of Areas 32 225/25 = 9 5 15

  29. Using Theorem 11.5 • First make sure the figures are similar. • They will tell you, or… • You need to use the Similarity Theorems from Chapter 8 • SSS Similarity • SAS Similarity • AA Similarity • Try to find the scale factor • Ratio of side lengths • The ratio of the areas is the square of the scale factor. • So the scale factor is the square root of the ratio of the areas.

  30. Homework 11.3 • In Class • 1-6 • p679-681 • HW • 7-28, 34-41 • Due Tomorrow • Quiz Wednesday • Lessons 11.1-11.3

  31. Lesson 11.4 Circumference and Arc Length

  32. Lesson 11.4 Objectives • Find the circumference of a circle. • Identify arc length • Define arc measure

  33. Circumference • The circumference of a circle is the distance around the circle. • For all circles, the ratio of circumference to the diameter is the same. •  , or pi C d

  34. Theorem 11.6:Circumference of a Circle • The circumference (C) of a circle is • C = d or… • C = 2r • where d is diameter • and r is radius C d r

  35. If asked to find the circumference. identify what you know diameter Use C =  d radius Use C = 2r If asked to find the diameter or radius, you must work backwards. diameter Divide by  radius Divide by 2 Using Theorem 11.6

  36. Example 5 Find the circumference of the circle • r = 5 • C = 2r • C = 2 (5) • C = 10 • Leave it! 5

  37. Example 6 • C = 32 • Find diameter • d = C/ • d = 32/ • d = 32 32

  38. Arc Length A AB mAB B • An arc length is a portion of the circumference. • denoted CD with an arc on top • It is determined by its arc measure • The measure of the angle made by joining the endpoints of the arc with the center of the circle. • denoted by placing an m in front to show we are finding the measure

  39. Corollary:Arc Length Corollary = • In a circle, the ratio of the length of the given arc to the entire circumference is equal to the ratio of the measure of the arc to the measure of the entire circle, 360o. • Set up a proportion using the smaller portions over the entire portions. Arc Length Arc Measure 360o Circumference

  40. Example 7 AB A Arc Length Arc Measure = 360o Circumference = B • Find the arc length for the following 60o 8 1 x 16 Arc Length = 6 16 Arc Length 60o Arc Length = 6 2 (8) 360o 16 2 r 8 60o Arc Length = Arc Length = x 16 3 360o

  41. Homework 11.4 • In Class • p686-689 • 1-14 • HW • 15-35, 39-41, 48-49 • Due Tomorrow

  42. Lesson 11.5 Areas of Circles and Sectors

  43. Lesson 11.5 Objectives • Find the area of a circle • Calculate the area of a sector of a circle • Apply the area of a circle and its sector to finding the area of complex figures

  44. Theorem 11.7:Area of a Circle • The area of a circle is  times the square of the radius. • A = r2 C r

  45. Sector A AB B • A sector of a circle is the region bounded by two radii of the circle and their arc. • Usually looks like a slice of pizza!

  46. Theorem 11.8:Area of a Sector = • The ratio of the area(A) of a sector of a circle to the area of the entire circle is equal to the ratio of the measure of the arc to the measure of the entire circle, 360o. Sector Area Arc Measure 360o Circle Area

  47. Example 8 A AB Sector Area Arc Measure = = B 360o Circle Area • Find the area of a sector with a radius 8 and an arc measure of 75o 4800 Sector Area = 360o 40 Sector Area = 3 75o Sector Area 360o  (8)2  r2 64 75o (64) Sector Area = 360o

  48. Homework 11.5 • In Class • 1-9 • p695-698 • HW • 10-37, 43, 44 • Due Tomorrow

  49. Lesson 11.6 Geometric Probability

  50. Lesson 11.6 Objectives • Recall probability • Apply probability to a line segment • Apply probability to a geometric area

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