1. Geometry Marking Period Two

3. Application

4. Finding the Perimeter Find the distance around for the following irregular polygon Distance around = 5 inches + 4 inches + 2 inches + 3 inches + 6 inches = 20 inches

5. Finding the Perimeter Find the distance around for the following trapezoid Distance around = 5 inches + 8 inches + 4 inches + 3 inches = 20 inches

6. Finding the Perimeter Find the distance around the following triangle Distance around = 5 inches + 4 inches + 2 inches = 11 inches

7. Finding the Perimeter Find the distance around the following rectangle Distance around = 3 inches + 3 inches + 6 inches + 6 inches = 18 inches

8. Finding the Perimeter Find the distance around the following square Distance around = 5 inches + 5 inches + 5 inches + 5 inches = 20 inches You can get the same answer by doing 4 × 5 = 20

9. Perimeter and Inequalities The length of each side of a rectangular picture frame needs to be 15cm. You have only one 48cm piece of wood to use for this frame. Write and solve an inequality that describes the possible widths for this frame.

10. The Solution

11. Rooftop Garden A gardener is planning a rectangular garden area in a community garden. His garden will be next to an existing 12-ft fence. The gardener has a total of 44 ft. of fencing to build the other three sides of his garden. How long will the garden be if the width is 12 ft.?

12. The Solution

13. Kennel Construction You are fencing a rectangular puppy kennel with 25 m of fence. The side of the kennel against your house does not need a fence. This side is 9 m long. Find the dimensions of the kennel.

14. The Solution

15. Area of Shapes By definition, the area of shapes is the amount of space inside those shapes. You can find the area of any two-dimensional shape or any shape that has a width and a length.

16. Area of Shapes A tile, a tabletop, a rug, a ping pong table, a tennis court, a soccer field, and a football field are all examples of shapes that you can get the area for. To get the amount of space inside a figure, we use a square to represent 1 unit and we say that the area is measured in square units Take a look at the following rectangle. To get the area, we are going to draw squares of equal sizes inside of it.

17. Area of Shapes Take a look at the following rectangle. To get the area, we are going to draw squares of equal sizes inside of it. 1 square represents 1 square unit. The rectangle has 8 squares, so the area for this rectangle is 8 square units. We can also write 8 units 2 and it will mean the same thing

18. Area of Shapes Notice here the unit we are using is inch. That means we are going to use squares, which have a side of 1 inch to get the area for the rectangle. Area = length × width = 5 × 2 = 10 square inches or 10 inches 2 This also means that we can fit 10 squares with a side of 1 inch inside this rectangle.

19. Area of Shapes Find the area of the following rectangle Area = length × width = 10 × 2 = 20 square inches or 20 inches 2 Now,that you understand how to get the area for a rectangle, it is going to be easy to get the area of shapes such as squares, triangles, and trapezoid.

20. Finding Surface Area In this section we will calculate the surface area of solid figures such as the cube, the cylinder, cone, sphere, square pyramid and rectangular prism.

21. Surface Area of a Cube Start with a cube as shown below and call the length of one side a: In order to make a cube like the one shown above, you basically use the following cube template:

22. Surface Area of a Cube Looking at the cube template, it is easy to see that the cube has six sides and each side is a square The area of one square is a × a = a 2 Since there are six sides, the total surface area, call it SA is: SA = a 2 + a 2 + a 2 + a 2 + a 2 + a 2 SA = 6 × a 2

23. Surface Area of a Cube Find the surface area if the length of one side is 3 cm Surface area = 6 × a 2 Surface area = 6 × 3 2 Surface area = 6 × 3 × 3 Surface area = 54 cm 2

24. Surface Area of a Cube Find the surface area if the length of one side is 5 cm Surface area = 6 × a 2 Surface area = 6 × 5 2 Surface area = 6 × 5 × 5 Surface area = 150 cm 2

25. Surface Area of a Cube Find the surface area if the length of one side is 1/2 cm Surface area = 6 × a 2 Surface area = 6 × (1/2) 2 Surface area = 6 × 1/2 × 1/2 Surface area = 6 × 1/4 Surface area = 6/4 cm 2 Surface area = 3/2 cm 2 Surface area = 1.5 cm 2

26. Surface Area of a Rectangular Prism Start with a right rectangular prism as shown below and call the length l, the width w, and the height h: In order to make a rectangular prism like the one shown above, you basically use the following rectangular prism template:

27. Surface Area of a Rectangular Prism The bottom side and the top side are equal and have l and w as dimensions The area for the top and bottom side is l× w + l × w = 2 × l × w The front side (shown in sky blue) and the back side (not shown) are equal and have h and l as dimensions The area for the front and the back side is l× h + l × h = 2 × l × h Then, the last two sides have h and w as its dimensions. One side is shown in purple The area for the front and the back side is w× h + w × h = 2 × w × h The total surface area, call it SA is: SA = 2 × l × w + 2 × l × h + 2 × w × h

28. Surface Area of a Rectangular Prism Find the surface area of a rectangular prism with a length of 11 cm, a width of 3 cm, and a height of 5 cm SA = 2 × l × w + 2 × l × h + 2 × w × h SA = 2 × 11 × 3 + 2 × 11 × 5 + 2 × 3 × 5 SA = 66 + 110 + 30 SA = 206 cm 2

29. Surface Area of a Rectangular Prism Find the surface area of a rectangular prism with a length of 10 inches, a width of 4 inches, and a height of 5 inches. SA = 2 × l × w + 2 × l × h + 2 × w × h SA = 2 × 10 × 4 + 2 × 10 × 5 + 2 × 4 × 5 SA = 80 + 100 + 40 SA = 220 in 2

30. Surface Area of a Rectangular Prism Find the surface area with a length of 8 cm, a width of 3 cm, and a height of 4 cm SA = 2 × l × w + 2 × l × h + 2 × w × h SA = 2 × 8 × 3 + 2 × 8 × 4 + 2 × 3 × 4 SA = 48 + 64 + 24 SA = 136 cm 2

33. Surface Area of a Cylinder To derive the formula of the surface area of a cylinder, we will start by showing you how you can make a cylinder. Start with a rectangle and two circles.

34. Surface Area of a Cylinder Fold the rectangle until you make an open cylinder with it. An open cylinder is a cylinder that has no bases. A good real life example of an open cylinder is a water pipe. Next, using the two circles as bases for the cylinder, put one on top of the cylinder and put one beneath it.

35. Surface Area of a Cylinder Now you have a completed cylinder. And if you can make the cylinder with the rectangle and the two circles, you can use them to derive the surface area of the cylinder.

36. Surface Area of a Cylinder The area of the two circles is straightforward. The area of one circle is pi × r 2, so for two circles, you get 2 × pi × r 2 To get the area of the rectangle, multiply h by 2 × pi × r and that is equal to 2 × pi × r × h Therefore, the total surface area of the cylinder is: SA = 2 × pi × r 2 + 2 × pi × r × h

37. Surface Area of a Cylinder Example #1 Find the surface area of a cylinder with a radius of 3 cm, and a height of 7 cm SA = 2 × pi × r 2 + 2 × pi × r × h SA = 2 × 3.14 × 3 2 + 2 × 3.14 × 3 × 7 SA = 6.28 × 9 + 6.28 × 21 SA = 56.52 + 131.88 Surface area = 188.4 cm 2

38. Surface Area of a Cylinder Example #2 Find the surface area of a cylinder with a radius of 2 cm, and a height of 5 cm SA = 2 × pi × r 2 + 2 × pi × r × h SA = 2 × 3.14 × 2 2 + 2 × 3.14 × 2 × 5 SA = 6.28 × 4 + 6.28 × 10 SA = 25.12 + 62.8 Surface area = 87.92 cm 2

39. Surface Area of a Square Pyramid

44. NOTEBOOK TEST 1.How many feet of fence do you need to enclose your new garden? 2.What was the perimeter for the irregular polygon? 3.What is the perimeter for the trapezoid? 4.Recopy the inequality that describes the possible widths for hanging a picture frame. 5.What is the length of the rooftop garden? 6.What are the dimensions of the amazing dog kennel? 7.What is the area of the rectangle? 8.What is the formula for the surface area of a cube? 9.What is the formula for the surface area of a rectangular prism? 10.What is the formula for the surface area of a cylinder? 11.What is the answer to example #2 of finding the surface area of a cylinder? 12.List the answers for the surface areas of the Square Pyramids for examples #3 and #4.

45. Surface Area of a Sphere To derive the formula of the surface area of a sphere, we imagine a sphere with many pyramids inside of it until the base of all the pyramids cover the entire surface area of the sphere. In the figure below, only one of such pyramids is shown. Next, complete a ratio of the area of the pyramid to the volume of the pyramid. The area of the pyramid is A The volume of the pyramid is V = (1/3) × A × r = (A × r) / 3 So, the ratio of area to volume is A / V = A ÷ (A × r) / 3 = (3 × A) / (A × r ) = 3 / r

46. Surface Area of a Sphere For a large number of pyramids, let say that n is such large number, the ratio of the surface area of the sphere to the volume of the sphere is the same as 3 / r For n pyramids, the total area is n × A Also for n pyramids, the total volume is n × V Therefore, ratio of total area to total volume is n × A / n × V = A / V and we already saw before that A / V = 3 / r

47. Surface Area of a Sphere

48. Find the surface area of a sphere with a radius of 6 m SA = 4 × pi × r 2 SA = 4 × 3.14 × 6 2 SA = 12.56 × 36 SA = 452.16 Surface area = 452.16 m 2

49. Surface Area of a Sphere Find the surface area of a sphere with a radius of 5 cm SA = 4 × pi × r 2 SA = 4 × 3.14 × 5 2 SA = 12.56 × 25 SA = 50.24 Surface area = 50.24 cm 2

50. Surface Area of a Cone The surface area of a cone can be derived from the surface area of a square pyramid. Start with a square pyramid and just keep increasing the number of sides of the base. After a very large number of sides, you can see that the figure will eventually look like a cone.

51. Surface Area of a Cone It is important to understand this, because we can use the formula of the surface area of a square pyramid to find that of a cone. l is the slant height. The area of the square is s 2 The area of one triangle is (s × l)/2 Since there are 4 triangles, the area is 4 × (s × l)/2 = 2 × s × l Therefore, the surface area, call it SA is: SA = s 2 + 2 × s × l :

52. The Pythagorean Theorem The Pythagorean Theorem was named after famous Greek mathematician Pythagoras.

54. Pythagorean Theorem It is an important formula that states that in any right triangle. Let c be the length of the longest side, called hypotenuse Let a and b be the length of the other two sides, called legs The theorem states that the length of the hypotenuse squared is equal to the length of side a squared and the length of side b squared.

55. Pythagorean Theorem

56. Skills You Need Two important skills you will need to solve the Pythagorean Theorem. 1 – How to deal with Square Roots 2 – How to solve equations with subtraction

57. Square Root of a Number Before understanding what the square root of a number is, it is important to understand the meaning of root of a number. The root of a number is an equal factor of the number

58. Square Root of a Number For example, find the root of 16 First, we need to factor 16 16 = 1 × 16 16 = 2 × 8 16 = 4 × 4 The root of 16 is 4 because 4 is the equal factor for 16 We call 4 the square root of 16 4 is called square root because we have to square 4 or raise 4 to a power of 2 to get 16

59. Square Root of a Number For the previous example, the square was equal to a whole number. It is not always possible to get the square root as a whole number. Sometimes, you may get a real number when finding the square root. For example, use the square root calculator below to find the square root of 5 The result includes lots of numbers after the decimal point.

60. Solving Equations Using Subtraction When solving equations using subtraction, we will use this form: x + b = c When solving one step equation of the form x + b = c, subtract b from both sides of the equation.

61. Example No. 1 Solve x + 3 = 8 for x

62. Example No. 2 Solve x + 5 = 10 for x

63. Example No. 3 Solve x + 5 = - 10 for x

64. Example No. 4

65. Example No. 5 Let a = 3 and b = 4. Find c, or the longest side c 2 = a 2 + b 2 c 2 = 3 2 + 4 2 c 2 = 9 + 16 c 2 = 25 c = √25

66. Example No. 6 Let c = 10 and a = 8. Find b, or the other leg. c 2 = a 2 + b 2 10 2 = 8 2 + b 2 100 = 64 + b 2 100 - 64 = 64 - 64 + b 2 (minus 64 from both sides to isolate b 2 ) 36 = 0 + b 2 36 = b 2 b = √36 = 6

67. Example No. 7 Example #3 Let c = 13 and b = 5. Find a c 2 = a 2 + b 2 13 2 = a 2 + 5 2 169 = a 2 + 25 169 - 25 = a 2 + 25-25 144 = a 2 + 0 144 = a 2 a = √144 = 12

68. Pythagorean Theorem – Zip Line

69. Zip Line Solution

70. Pythagorean Theorem – Corner Shot

71. Corner Shot Solution

72. Pythagorean Theorem – Long Throw

73. Long Throw Solution

74. Pythagorean Theorem – Classic Ladder

75. Classic Ladder Solution

76. James Garfield November 19, 1831 – September 19, 1881 served as a major general in the Union Army during the American Civil War. was the 20 th President of the United States, serving from March 4, 1881 until his assassination later that year. James A. Garfield had no intention of becoming America’s 20th president, but was unanimously elected to run as the Republican candidate after giving an inspiring speech in support of the nominee. Garfield advocated agricultural technology, education, and civil rights for African-Americans.

77. James Garfield Quotes “Right reason is stronger than force.” “Man cannot live by bread alone; he must have peanut butter.” “There are men and women who make the world better just by being the kind of people they are. They have the gift of kindness or courage or loyalty or integrity. It really matters very little whether they are behind the wheel of a truck or running a business or bringing up a family. They teach the truth by living it.”

78. Assassination Attempt Garfield’s presidency lasted only 200 days, 80 of which he fought heroically in his bed attempting to recover from an assassin’s bullet, shot into his back and resting behind his pancreas. He died not from the bullet wound though, but of heart failure, caused by massive infection.

79. Garfield’s Pythagorean Proof President Garfield studied math at Williams College (in Williamstown, MA) and taught in the public school in Pownal, Vermont. Garfield was fascinated by geometry, like President Lincoln, but was not a professional mathematician. President James Garfield developed his own proof in The Journal of Education (Volume 3 issue161) in 1876. Completed his proof of the Pythagorean Theorem while he was a member of Congress in 1876. About his proof he joked that "we think it something on which the members of both houses can unite without distinction of the party.“ A nice feature of mathematical proofs is that they are not subject to political opinion.

80. Garfield’s Proof

81. Notebook Test