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 –180 o The sum of the interior angles of a quadrilateral is –360 o 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 180 o. 360 o 180 o 540 o 720 o

6. Theorem 11.1: Polygon Interior Angles Theorem The sum of the measure of the interior angles of a convex n-gon is – 180 o (n – 2) n is the number of sides This basically states that for every side that you add, you add another 180 o 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) = – 1080 o 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 = 135 o 130 o 120 o 110 o 150 o 145 o 135 o 155 o xoxo

8. Corollary to Theorem 11.1 The measure of each interior angle of a regular n-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) 108 O

10. Example 3: Finding the Number of Sides Each interior angle is 120 o, 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) n 120 = 1 Dividing by n is the same as multiplying by 1 / n 120n = 180(n – 2) 120n = 180n – 360 -60n = – 360 n = 6

11. Exterior Angles 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. 1 2 3 4 5

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

13. Corollary to Theorem 11.2 The measure of the each exterior angle of a regular n-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 120 o. 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) s 2 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 center of 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. 60 0

22. Finding the Apothem or Radius 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. 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 / 2 CA) = 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 CA = 72 o 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.

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 / 2 aP –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 / 2 a(ns) n stands for the number of sides s stands for the length of one side

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 a 2 :b 2 –Remember that the ratio of side lengths a:b is the same as the ratio of the perimeters, a:b. Theorem 8.1 515 Ratio of Sides 15 / 5 = 3 Ratio of Perimeters 15 / 5 = 3 Ratio of Areas 225 / 25 = 9 3232

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. Using Theorem 11.6 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 

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 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 A B AB mAB

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, 360 o. –Set up a proportion using the smaller portions over the entire portions. = Arc Length Circumference Arc Measure 360 o

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

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 =  r 2 C r

45. Sector 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! AB A B

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, 360 o. = Sector Area Circle Area Arc Measure 360 o

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

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

51. Probability Recall that probability is a number that represents the chance that an event will occur. That number is a decimal or fraction from 0 to 1 –0 means the event cannot occur –1 means the event will always occur The probability is calculated by taking the number of favorable outcomes and dividing by the total number of possible outcomes.

52. Geometric Probability The probability of finding point K on a line segment is determined by divided the length of the target segment divided by the length of the entire segment. The probability of finding point K in a given area is determined by finding the target area and dividing by the entire area of the surface. Any time that probability is calculated using geometric measures such as length and area, you are finding the geometric probability of the event occurring. If K is on segment CD P(K is on segment CD) = CD AZ If K is in area M P(K is in area M) = Area M Area J P stands for probability of the inside of parentheses happening.

53. Example 9 Find the probability that a point randomly chosen is on line segment WX. 01 VWXYZ P(K is on segment WX) = WX VZ P(K is on segment WX) = 3 11

54. Example 10 Find the probability that a point randomly chosen lies inside the circle. s = 10 Area Circle P(K is in the circle) = Area Square  r 2 = s2s2 =  (5) 2 10 2 25  100 ≈ 78.5%

55. Homework 11.6 2-19, 31-34, 37-39 –skip 18 –p701-704 In Class – 8, 11, 31, 37 Due Tomorrow Quiz Thursday –Lessons 11.4-11.6