1. Chapter 10 Section 10.1 - Areas of Parallelograms and Triangles
Objectives:
To find the area of a parallelogram
To find the area of a triangle

2. h b A = bh Theorem 10.1 – Area of a Rectangle
The area of a rectangle is the product of its base and height. h b
A = bh

3. Base of a Parallelogram  any of its sides
Altitude  a segment perpendicular to the line containing that base, drawn from the side opposite the base.
Height  the length of an altitude.
Altitude
Base

4. A = bh h b Theorem 10.2 – Area of a Parallelogram
The area of a parallelogram is the product of a base and the corresponding height.
A = bh h b

5. Base of a Triangle  any of its sides
Height  the length of the altitude to the line containing that base

6. A = 1 2 bh h b Theorem 10.3 – Area of a Triangle
The area of a triangle is half the product of a base and the corresponding height.
A = 1 2 bh h b

7. Ex: Find the area of each Parallelogram
4.6 cm
3.5 cm
4 in
5 in
2 cm

8. Ex: Find the area of each Triangle.
13 cm
6.4 ft
5 cm
10 ft
4 ft
12 cm

9. When designing a building, you must be sure that the building can withstand hurricane-force winds, which have a velocity of 73 mi/h or more. The formula F = 0.004A 𝑣 2 gives the force F in pounds exerted by a wind blowing against a flat surface. A is the area of the surface in square feet, and v is the wind velocity in miles per hour. How much force is exerted by a 73 mi/h wind blowing against the side of the building shown below?
6 ft
12 ft
20 ft

10. Homework #18
Due Tuesday (March 12)
Page 536 – 538
# 1 – 27 odd

11. Section 10.2 – Areas of Trapezoids, Rhombuses, and Kites
Objectives:
To find the area of a trapezoid
To find the area of a rhombus or a kite

12. A = 1 2 h( 𝑏 1 + 𝑏 2 ) Theorem 10.4 – Area of a Trapezoid 𝑏 1 h 𝑏 2
The area of a trapezoid is half the product of the height and the sum of the bases.
A = 1 2 h( 𝑏 1 + 𝑏 2 )
𝑏 1 h
𝑏 2

13. A = 1 2 𝑑 1 𝑑 2 Theorem 10.5 – Area of a Rhombus or a Kite 𝑑 1 𝑑 2
The area of a rhombus or a kite is half the product of the lengths of its diagonals.
𝑑 1
𝑑 2
A = 𝑑 1 𝑑 2

14. Ex: Find the area of the trapezoid
12 cm
7 cm
15 cm

15. What is the area of trapezoid PQRS
What is the area of trapezoid PQRS? What would the area be if <P was changed to 45°?
5 m S R
60° P Q
7 m

16. Ex: Find the area of the kite.
3 m
5 m
3 m
2 m

17. Find the area of the rhombus.C D

18. Homework #19
Due Wednesday (March 13)
Page 542 – 543
# 1 – 29 odd

19. Section 10.3 – Areas of Regular Polygons
Objectives:
To find the area of a regular polygon

20. You can circumscribe a circle about any regular polygon.
The center of a regular polygon is the center or the circumscribed circle.
The radius is the distance from the center to a vertex.
The apothem is the perpendicular distance from the center to a side.
Center
Radius
Apothem

21. Ex: Finding Angle Measures
The figure below is a regular pentagon with radii and apothem drawn. Find the measure of each numbered angle.
m<1 = = 72 (Divide 360 by the number of sides)
m<2 = 1 2 m<1 = 36 (apothem bisects the vertex angle)
m< = 180 m<3 = 54 3 2 1

22. Find the measure of each angle of the half of an octagon.1 2 3

23. Suppose you have a regular n-gon with side s
Suppose you have a regular n-gon with side s. The radii divide the figure into n congruent isosceles triangles. Each isosceles triangle has area equal to 1 2 as (a being apothem/s being side).
Since there are n congruent triangles, the area of the n-gon is A = n · 1 2 as. The perimeter p of the n-gon is ns. Substituting p for ns results in a formula for the area of the polygon.

24. A = 1 2 ap p a Theorem 10.6 – Area of a Regular Polygon
The area of a regular polygon is half the product of the apothem and the perimeter.
A = 1 2 ap p a

25. Ex: Find the area of each regular polygon.
A regular decagon with a 12.3 in. apothem and 8 in. sides
A regular pentagon with 11.6 cm sides and an 8 cm apothem

26. Ex: Find the area of the hexagon.
10 mm
5 mm

27. Homework #20
Due Thurs/Fri (March 14/15)
Page 548
# 1 – 23 all

28. Section 10.4 – Perimeters and Areas of Similar Figures
Objectives:
To find the perimeters and areas of similar figures

29. Theorem 10.7 – Perimeters and Areas of Similar Figures
If the similarity ratio of two similar figures is 𝑎 𝑏 , then
1. the ratio of their perimeters is 𝑎 𝑏
2. the ratio of their areas is 𝑎 2 𝑏 2

30. Ex: Finding Ratios in Similar Figures
The trapezoids below are similar. Find the ratio of their perimeters and ratio of their areas.
9 m
6 m

31. Ex: Two similar polygons have corresponding sides in the ratio 5 : 7.
a. Find the ratio of their perimeters
b. Find the ratio of their areas

32. Ex: Finding Areas Using Similar Figures
The area of the smaller regular pentagon is 𝑐𝑚 2 . What is the area of the larger pentagon?
4 cm
10 cm

33. Ex:
The corresponding sides of two similar parallelograms are in the ratio 3 : 4. The area of the larger parallelogram is 96 𝒊𝒏 𝟐 . Find the area of the smaller parallelogram.

34. Ex: Finding Similarity and Perimeter Ratios
The areas of two similar triangles are 50 𝑐𝑚 2 and 98 𝑐𝑚 2 . What is the similarity ratio? What is the ratio of their perimeters?
The areas of two similar rectangles are 𝑓𝑡 2 and 135 𝑓𝑡 2 . Find the ratio of their perimeters.

35. Homework #21
Due Monday (March 18)
Page 555 – 556
# 1 – 23 all

36. Section 10.5 – Trigonometry and Area
Objectives:
To find the area of a regular polygon using trigonometry
To find the area of a triangle using trigonometry

37. In the last lesson, we learned how to find the area of a regular polygon by using the formula A = 1 2 ap. By using this formula and trigonometric ratios, you can solve other types of problems.

38. Trigonometry Review Sine = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 SOH
Cosine = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 CAH
Tangent = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 TOA

39. Ex: Find the area of the regular pentagon with 8cm sides.

40. Ex: Finding the Area and Perimeter
Find the area and perimeter of a regular octagon with radius 16m.

41. Suppose we want to find the area of triangle ABC (below), but you are only given m<A and lengths b and c. To use the formula A = 1 2 bh, you need to find the height. This can be found by using the sine ratio:
sin A = ℎ 𝑐 therefore h = c(sin A) B
Area = 1 2 bc(sin A) c a h A C b

42. Area ΔABC = 1 2 bc(sin A) Theorem 10.8 – Area of a Triangle Given SAS
The area of a triangle is one half the product of the lengths of two sides and the sine of the included angle. B
Area ΔABC = 1 2 bc(sin A) a c A C b

43. Ex: Finding the area of a Triangle
Two sides of a triangular building plot are 120 ft and 85 ft long. They include an angle of 85°. Find the area of the building plot to the nearest square foot.

44. Quiz Tuesday (10.1 – 10.5) Homework #22 Due Monday (March 18)
Page 561 – 562
# 1 – 27 odd
Quiz Tuesday (10.1 – 10.5)

45. Section 10.6 – Circles and Arcs
Objectives:
To find the measures of central angles and arcs
To find circumference and arc length

46. Circle  the set of all points equidistant from a given point called the center.
Radius  a segment that has one endpoint at the center and the other endpoint on the circle.
Congruent Circles  have congruent radii
Diameter  a segment that contains the center of a circle and has both endpoints on the circle
Central Angle  an angle whose vertex is the center of the circle.

47. Circle = ΘP Central Angle = <CPA Radius = CP Diameter = AB

48. Semicircle  half of a circle (180°)
Minor Arc  smaller than a semicircle (< 180°). Its measure is the measure of its corresponding angle.
Major Arc  greater than a semicircle (> 180°). Its measure is 360 minus the measure of its related minor arc.
Adjacent Arcs  arcs of the same circle that have exactly one point in common.

49. Ex: Identify the following in Θ O:
The minor arcs (4)
The semicircles (4)
The major arcs that contain point A (4) A C O D E

50. mABC = mAB + mBC Postulate 10.1 – Arc Addition Postulate B C A
The measure of the arc formed by two adjacent arcs is the sum of the measures of the two arcs. B C A
mABC = mAB + mBC

51. C = Πd C = 2Πr Circumference  the distance around a circle
Pi (Π)  the ratio or the circumference of a circle to its diameter
Theorem 10.9 – Circumference of a Circle
The circumference of a circle is Π (pi) times the diameter d
C = Πd
C = 2Πr r C O

52. Arc Length  a fraction of a circle’s circumference
Theorem  Arc Length
The length of an arc of a circle is the product of the ratio 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑐 360 and the circumference of the circle. A r
Length of AB = 𝑚𝐴𝐵 360 · 2Πr O B

53. Ex: Find the measure of each arc. a. BC b. BD c. ABC d. AB
58° D O
32° A

54. Ex:
The diameter of a bicycle wheel is 22in. To the nearest whole number, how many revolutions does the wheel make when the bicycle travels 100 feet?

55. Homework #23
Due Thurs/Fri (March 21/22)
Page 569 – 570
# 1 – 39 odd

56. Section 10.7 – Areas of Circles and Sectors
Objectives:
To find the areas of circles, sectors, and segments of circles.

57. A = Π 𝑟 2 Theorem 10.11 – Area of a Circle r O
The area of a circle is the product of Π (pi) and the square of the radius.
A = Π 𝑟 2 r O

58. Sector of a Circle  a region bounded by an arc of the circle and the two radii to the arc’s endpoints. A sector is named using one arc endpoint, the center of the circle, and the other arc endpoint.

59. Theorem 10-12 – Area of a Sector of a Circle
The area of a sector of a circle is the product of the ratio 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑐 360 and the area of the circle. A
Area Sector AOB = 𝑚𝐴𝐵 360 · Π 𝑟 2 O r B

60. Segment of a circle  a part of a circle bounded by an arc and the segment joining its endpoints. To find the area of a segment, draw radii to form a sector. The area of the segment equals the area of the sector minus the area of the triangle formed. T

61. Ex:
You’re hungry one day and decide to go to Warehouse Pizza to get some food. When you arrive, you check the menu and notice they are having deals on 14-in and 12-in pizzas. If a 14-in pizza costs $20.00 and a 12-in pizza costs $16.00, which price gives you the most pizza for your dollar?

62. Ex: Find the Area of a Segment of a Circle
Find the area of the circle segment if the radius is 10-in and central angle ATB forms a right angle. B T

63. Section 10.8 – Geometric Probability
Objectives:
To use segment and area models to find the probabilities of events.

64. P(event) = 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
You may remember that the probability of an event is the ratio of the number of favorable outcomes to the number of possible outcomes
P(event) = 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

65. P(event) = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑒𝑛𝑡𝑖𝑟𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡
Geometric Probability  a model in which points represent outcomes. We find probabilities by comparing measurements of sets of points.
P(event) = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑓𝑎𝑣𝑜𝑟𝑎𝑏𝑙𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑒𝑛𝑡𝑖𝑟𝑒 𝑠𝑒𝑔𝑚𝑒𝑛𝑡

66. Chapter 10 Test Tuesday/Wednesday (Notebooks also due)
Homework #24
Due Monday (March 25)
Page 577 – 578
# 1 – 27 odd
Chapter 10 Test Tuesday/Wednesday
(Notebooks also due)