51 basic shapes and formulas

basic shapes and formulas

1. Basic Geometrical Shapes and Formulas

2. If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. Basic Geometrical Shapes and Formulas

3. If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. Basic Geometrical Shapes and Formulas

4. If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. Basic Geometrical Shapes and Formulas

5. If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. Basic Geometrical Shapes and Formulas

6. If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses a flat area, or a plane-shape. Basic Geometrical Shapes and Formulas

7. If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses a flat area, or a plane-shape. The length of the border, i.e. the length of the rope, is also referred to as the perimeter of the area. Basic Geometrical Shapes and Formulas

8. If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses a flat area, or a plane-shape. The length of the border, i.e. the length of the rope, is also referred to as the perimeter of the area. All the areas above are enclosed by the same rope, so they have equal perimeters. Basic Geometrical Shapes and Formulas

9. If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses a flat area, or a plane-shape. The length of the border, i.e. the length of the rope, is also referred to as the perimeter of the area. All the areas above are enclosed by the same rope, so they have equal perimeters. Following shapes are polygons: A plane-shape is a polygon if it is formed by straight lines. Basic Geometrical Shapes and Formulas

10. If we connect the two ends of a rope that’s resting flat in a plane, we obtain a loop. The loop forms a perimeter or border that encloses a flat area, or a plane-shape. The length of the border, i.e. the length of the rope, is also referred to as the perimeter of the area. All the areas above are enclosed by the same rope, so they have equal perimeters. Following shapes are polygons: These are not polygons: A plane-shape is a polygon if it is formed by straight lines. Basic Geometrical Shapes and Formulas

11. Three sided polygons are triangles . Basic Geometrical Shapes and Formulas

12. Three sided polygons are triangles . Basic Geometrical Shapes and Formulas If the sides of a triangle are labeled as a, b , and c, then the perimeter is P = a + b + c. a b c

13. Triangles with three equal sides are equilateral triangles . s Three sided polygons are triangles . Basic Geometrical Shapes and Formulas If the sides of a triangle are labeled as a, b , and c, then the perimeter is P = a + b + c. a b c s s

14. Triangles with three equal sides are equilateral triangles . s Three sided polygons are triangles . Basic Geometrical Shapes and Formulas If the sides of a triangle are labeled as a, b , and c, then the perimeter is P = a + b + c. a b c The perimeter of an equilateral triangle is P = 3s. s s

15. Triangles with three equal sides are equilateral triangles . s Three sided polygons are triangles . Basic Geometrical Shapes and Formulas If the sides of a triangle are labeled as a, b , and c, then the perimeter is P = a + b + c. a b c The perimeter of an equilateral triangle is P = 3s. s s Rectangles are 4-sided polygons where the sides are joint at a right angle as shown.

16. Triangles with three equal sides are equilateral triangles . s Three sided polygons are triangles . Basic Geometrical Shapes and Formulas If the sides of a triangle are labeled as a, b , and c, then the perimeter is P = a + b + c. a b c The perimeter of an equilateral triangle is P = 3s. s s Rectangles are 4-sided polygons where the sides are joint at a right angle as shown. s s s s A square Rectangle with four equal sides are squares.

17. Triangles with three equal sides are equilateral triangles . s Three sided polygons are triangles . Basic Geometrical Shapes and Formulas If the sides of a triangle are labeled as a, b , and c, then the perimeter is P = a + b + c. a b c The perimeter of an equilateral triangle is P = 3s. s s Rectangles are 4-sided polygons where the sides are joint at a right angle as shown. s s s s A square Rectangle with four equal sides are squares. The perimeter of a squares is P = s + s + s + s = 4s

18. Basic Geometrical Shapes and Formulas Example A. We have to fence in an area with two squares and an equilateral triangle as shown. How many feet of fences do we need? 20 ft

19. Basic Geometrical Shapes and Formulas Example A. We have to fence in an area with two squares and an equilateral triangle as shown. How many feet of fences do we need? 20 ft The area consists of two squares and an equilateral triangle so all the sides, measured from corner to corner, are equal.

20. Basic Geometrical Shapes and Formulas Example A. We have to fence in an area with two squares and an equilateral triangle as shown. How many feet of fences do we need? 20 ft The area consists of two squares and an equilateral triangle so all the sides, measured from corner to corner, are equal. There are 9 sections where each is 20 ft hence we need 9 x 20 = 180 ft of fence.

21. Basic Geometrical Shapes and Formulas Example A. We have to fence in an area with two squares and an equilateral triangle as shown. How many feet of fences do we need? 20 ft The area consists of two squares and an equilateral triangle so all the sides, measured from corner to corner, are equal. There are 9 sections where each is 20 ft hence we need 9 x 20 = 180 ft of fence. If we know two adjacent sides of a rectangle, then we know all four sides because their opposites sides are identical.

22. Basic Geometrical Shapes and Formulas Example A. We have to fence in an area with two squares and an equilateral triangle as shown. How many feet of fences do we need? 20 ft The area consists of two squares and an equilateral triangle so all the sides, measured from corner to corner, are equal. There are 9 sections where each is 20 ft hence we need 9 x 20 = 180 ft of fence. If we know two adjacent sides of a rectangle, then we know all four sides because their opposites sides are identical. We will use the word “height” for the vertical side and “width” for the horizontal side . width (w) height (h)

23. Basic Geometrical Shapes and Formulas Example A. We have to fence in an area with two squares and an equilateral triangle as shown. How many feet of fences do we need? 20 ft The area consists of two squares and an equilateral triangle so all the sides, measured from corner to corner, are equal. There are 9 sections where each is 20 ft hence we need 9 x 20 = 180 ft of fence. If we know two adjacent sides of a rectangle, then we know all four sides because their opposites sides are identical. We will use the word “height” for the vertical side and “width” for the horizontal side . The perimeter of a rectangle is h + h + w + w or that width (w) height (h) P = 2h + 2w

24. Example B. a. We want to rope off a 50-meter by 70-meter rectangular area and also rope off sections of area as shown. How many meters of rope do we need ? Basic Geometrical Shapes and Formulas 50 m 70 m

25. Example B. a. We want to rope off a 50-meter by 70-meter rectangular area and also rope off sections of area as shown. How many meters of rope do we need ? Basic Geometrical Shapes and Formulas 50 m 70 m We have three heights where each requires 50 meters of rope,

26. Example B. a. We want to rope off a 50-meter by 70-meter rectangular area and also rope off sections of area as shown. How many meters of rope do we need ? Basic Geometrical Shapes and Formulas and three widths where each requires 70 meters of rope. 50 m 70 m We have three heights where each requires 50 meters of rope,

27. Example B. a. We want to rope off a 50-meter by 70-meter rectangular area and also rope off sections of area as shown. How many meters of rope do we need ? Basic Geometrical Shapes and Formulas and three widths where each requires 70 meters of rope. Hence it requires 3(50) + 3(70) = 150 + 210 = 360 meters of rope. 50 m 70 m We have three heights where each requires 50 meters of rope,

28. Example B. a. We want to rope off a 50-meter by 70-meter rectangular area and also rope off sections of area as shown. How many meters of rope do we need ? Basic Geometrical Shapes and Formulas and three widths where each requires 70 meters of rope. Hence it requires 3(50) + 3(70) = 150 + 210 = 360 meters of rope. 50 m 70 m We have three heights where each requires 50 meters of rope, b. What is the perimeter of the following step-shape if all the short segments are 2 feet? 2 ft The perimeter of the step-shape is the same as the perimeter of the rectangle that boxes it in as shown.

29. Example B. a. We want to rope off a 50-meter by 70-meter rectangular area and also rope off sections of area as shown. How many meters of rope do we need ? Basic Geometrical Shapes and Formulas and three widths where each requires 70 meters of rope. Hence it requires 3(50) + 3(70) = 150 + 210 = 360 meters of rope. 50 m 70 m We have three heights where each requires 50 meters of rope, b. What is the perimeter of the following step-shape if all the short segments are 2 feet? 2 ft

30. Example B. a. We want to rope off a 50-meter by 70-meter rectangular area and also rope off sections of area as shown. How many meters of rope do we need ? Basic Geometrical Shapes and Formulas and three widths where each requires 70 meters of rope. Hence it requires 3(50) + 3(70) = 150 + 210 = 360 meters of rope. 50 m 70 m We have three heights where each requires 50 meters of rope, b. What is the perimeter of the following step-shape if all the short segments are 2 feet? 2 ft The perimeter of the step-shape is the same as the perimeter of the rectangle that boxes it in as shown. 2 ft

31. Example B. a. We want to rope off a 50-meter by 70-meter rectangular area and also rope off sections of area as shown. How many meters of rope do we need ? Basic Geometrical Shapes and Formulas and three widths where each requires 70 meters of rope. Hence it requires 3(50) + 3(70) = 150 + 210 = 360 meters of rope. 50 m 70 m We have three heights where each requires 50 meters of rope, b. What is the perimeter of the following step-shape if all the short segments are 2 feet? 2 ft The perimeter of the step-shape is the same as the perimeter of the rectangle that boxes it in as shown. 2 ft The height of the rectangle is 6 ft and the width is 10 ft , so the perimeter P = 2(6 ) + 2(10) = 32 ft.

32. Area If we connect the two ends of a rope that’s resting flat in a plane, the rope form a loop that encloses an area.

33. Area If we connect the two ends of a rope that’s resting flat in a plane, the rope form a loop that encloses an area.

34. Area If we connect the two ends of a rope that’s resting flat in a plane, the rope form a loop that encloses an area.

35. Area If we connect the two ends of a rope that’s resting flat in a plane, the rope form a loop that encloses an area. The word “ area ” also denotes the amount of surface enclosed.

36. Area If we connect the two ends of a rope that’s resting flat in a plane, the rope form a loop that encloses an area. The word “ area ” also denotes the amount of surface enclosed. If each side of a square is 1 unit, we define the area of the square to be 1 unit x 1 unit = 1 unit 2 , i.e. 1 square-unit.

37. Area If we connect the two ends of a rope that’s resting flat in a plane, the rope form a loop that encloses an area. The word “ area ” also denotes the amount of surface enclosed. 1 in 1 in 1 in 2 If each side of a square is 1 unit, we define the area of the square to be 1 unit x 1 unit = 1 unit 2 , i.e. 1 square-unit. Hence the areas of the following squares are: 1 square-inch

38. Area If we connect the two ends of a rope that’s resting flat in a plane, the rope form a loop that encloses an area. The word “ area ” also denotes the amount of surface enclosed. 1 in 1 in 1 in 2 If each side of a square is 1 unit, we define the area of the square to be 1 unit x 1 unit = 1 unit 2 , i.e. 1 square-unit. Hence the areas of the following squares are: 1 m 1 m 1 m 2 1 square-inch 1 square-meter

39. Area If we connect the two ends of a rope that’s resting flat in a plane, the rope form a loop that encloses an area. The word “ area ” also denotes the amount of surface enclosed. 1 in 1 in 1 in 2 If each side of a square is 1 unit, we define the area of the square to be 1 unit x 1 unit = 1 unit 2 , i.e. 1 square-unit. Hence the areas of the following squares are: 1 m 1 m 1 mi 1 mi 1 m 2 1 mi 2 1 square-inch 1 square-meter 1 square-mile

40. Area If we connect the two ends of a rope that’s resting flat in a plane, the rope form a loop that encloses an area. The word “ area ” also denotes the amount of surface enclosed. 1 in 1 in 1 in 2 If each side of a square is 1 unit, we define the area of the square to be 1 unit x 1 unit = 1 unit 2 , i.e. 1 square-unit. Hence the areas of the following squares are: 1 m 1 m 1 mi 1 mi 1 m 2 1 mi 2 1 square-inch 1 square-meter 1 square-mile We find the area of rectangles by cutting them into squares.

41. 2 mi 3 mi Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi 2 (square miles). w = 6 mi 2 2 x 3

42. 2 mi 3 mi Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi 2 (square miles). In general, given the rectangle with height = h (units) width* = w (units), h w = 6 mi 2 2 x 3 A = h x w (unit 2 ) then its area A = h x w (unit 2 ).

43. 2 mi 3 mi Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi 2 (square miles). In general, given the rectangle with height = h (units) width* = w (units), h w = 6 mi 2 2 x 3 A = h x w (unit 2 ) then its area A = h x w (unit 2 ). The area of a square is s*s = s 2 .

44. 2 mi 3 mi Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi 2 (square miles). In general, given the rectangle with height = h (units) width* = w (units), h w = 6 mi 2 2 x 3 A = h x w (unit 2 ) then its area A = h x w (unit 2 ). The area of a square is s*s = s 2 . By cutting and pasting, we may find areas of other shapes.

45. 2 mi 3 mi Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi 2 (square miles). In general, given the rectangle with height = h (units) width* = w (units), h w = 6 mi 2 2 x 3 A = h x w (unit 2 ) then its area A = h x w (unit 2 ). The area of a square is s*s = s 2 . By cutting and pasting, we may find areas of other shapes. Example C. a. Find the area of R as shown. Assume the unit is meter . 4 4 R 12 12

46. 2 mi 3 mi Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi 2 (square miles). In general, given the rectangle with height = h (units) width* = w (units), h w = 6 mi 2 2 x 3 A = h x w (unit 2 ) then its area A = h x w (unit 2 ). The area of a square is s*s = s 2 . By cutting and pasting, we may find areas of other shapes. Example C. a. Find the area of R as shown. Assume the unit is meter . 4 4 There are two basic approaches. R 12 12

47. 2 mi 3 mi Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi 2 (square miles). In general, given the rectangle with height = h (units) width* = w (units), h w = 6 mi 2 2 x 3 A = h x w (unit 2 ) then its area A = h x w (unit 2 ). The area of a square is s*s = s 2 . By cutting and pasting, we may find areas of other shapes. Example C. a. Find the area of R as shown. Assume the unit is meter . 4 4 There are two basic approaches. R I . We may view R as a 12 x 12 square with a 4 x 8 corner removed. 12 8 12 4 4 R 12 12

48. 2 mi 3 mi Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi 2 (square miles). In general, given the rectangle with height = h (units) width* = w (units), h w = 6 mi 2 2 x 3 A = h x w (unit 2 ) then its area A = h x w (unit 2 ). The area of a square is s*s = s 2 . By cutting and pasting, we may find areas of other shapes. Example C. a. Find the area of R as shown. Assume the unit is meter . 4 4 There are two basic approaches. R I . We may view R as a 12 x 12 square with a 4 x 8 corner removed. 12 8 12 4 4 R 12 12 Hence the area of R is 12 x 12 – 4 x 8

49. 2 mi 3 mi Area A 2 mi x 3 mi rectangle may be cut into six 1 x 1 squares so it covers an area of 2 x 3 = 6 mi 2 (square miles). In general, given the rectangle with height = h (units) width* = w (units), h w = 6 mi 2 2 x 3 A = h x w (unit 2 ) then its area A = h x w (unit 2 ). The area of a square is s*s = s 2 . By cutting and pasting, we may find areas of other shapes. Example C. a. Find the area of R as shown. Assume the unit is meter . 4 4 There are two basic approaches. R I . We may view R as a 12 x 12 square with a 4 x 8 corner removed. 12 8 12 4 4 R 12 12 Hence the area of R is 12 x 12 – 4 x 8 = 144 – 32 = 112 m 2 .

50. Area Il. We may dissect R into two rectangles labeled I and II. 12 12 4 4 I II

51. Area Il. We may dissect R into two rectangles labeled I and II. 12 12 4 4 8 I II Area of I is 12 x 8 = 96,

52. Area Il. We may dissect R into two rectangles labeled I and II. 12 12 4 4 8 I II Area of I is 12 x 8 = 96, area of II is 4 x 4 = 16.

53. Area Il. We may dissect R into two rectangles labeled I and II. 12 12 4 4 8 I II Area of I is 12 x 8 = 96, area of II is 4 x 4 = 16. The area of R is the sum of the two or 96 + 16 = 112 m 2 .

54. Area Il. We may dissect R into two rectangles labeled I and II. 12 12 4 4 8 I II Area of I is 12 x 8 = 96, area of II is 4 x 4 = 16. The area of R is the sum of the two or 96 + 16 = 112 m 2 . b. Find the area of the following shape R where all the short segments are 2 ft. 2 ft

55. Area Il. We may dissect R into two rectangles labeled I and II. 12 12 4 4 8 I II Area of I is 12 x 8 = 96, area of II is 4 x 4 = 16. The area of R is the sum of the two or 96 + 16 = 112 m 2 . b. Find the area of the following shape R where all the short segments are 2 ft. Let’s cut R into three rectangles as shown. 2 ft

56. Area Il. We may dissect R into two rectangles labeled I and II. 12 12 4 4 8 I II Area of I is 12 x 8 = 96, area of II is 4 x 4 = 16. The area of R is the sum of the two or 96 + 16 = 112 m 2 . b. Find the area of the following shape R where all the short segments are 2 ft. Let’s cut R into three rectangles as shown. I II III 2 ft

57. Area Il. We may dissect R into two rectangles labeled I and II. 12 12 4 4 8 I II Area of I is 12 x 8 = 96, area of II is 4 x 4 = 16. The area of R is the sum of the two or 96 + 16 = 112 m 2 . b. Find the area of the following shape R where all the short segments are 2 ft. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, I II III 2 ft

58. Area Il. We may dissect R into two rectangles labeled I and II. 12 12 4 4 8 I II Area of I is 12 x 8 = 96, area of II is 4 x 4 = 16. The area of R is the sum of the two or 96 + 16 = 112 m 2 . b. Find the area of the following shape R where all the short segments are 2 ft. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, I II III 2 ft

59. Area Il. We may dissect R into two rectangles labeled I and II. 12 12 4 4 8 I II Area of I is 12 x 8 = 96, area of II is 4 x 4 = 16. The area of R is the sum of the two or 96 + 16 = 112 m 2 . b. Find the area of the following shape R where all the short segments are 2 ft. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. I II III 2 ft

60. Area Il. We may dissect R into two rectangles labeled I and II. 12 12 4 4 8 I II Area of I is 12 x 8 = 96, area of II is 4 x 4 = 16. The area of R is the sum of the two or 96 + 16 = 112 m 2 . b. Find the area of the following shape R where all the short segments are 2 ft. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft 2 . I II III 2 ft

61. Area Il. We may dissect R into two rectangles labeled I and II. 12 12 4 4 8 I II Area of I is 12 x 8 = 96, area of II is 4 x 4 = 16. The area of R is the sum of the two or 96 + 16 = 112 m 2 . b. Find the area of the following shape R where all the short segments are 2 ft. Let’s cut R into three rectangles as shown. Area of I is 2 x 2 = 4, area of II is 2 x 6 = 12, and area of III is 2 x 5 = 10. Hence the total area is 4 + 12 + 10 = 16 ft 2 . I II III 2 ft By cutting and pasting we obtain the following area formulas.

62. Area A parallelogram is a shape enclosed by two sets of parallel lines. h b

63. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h b

64. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h b

65. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. h b h b

66. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. Hence the area of the parallelogram is A = h x b where h = height and b = base. h b h b

67. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. Hence the area of the parallelogram is A = h x b where h = height and b = base. h b h b For example, the area of all the parallelograms shown here is 8 x 12 = 96 ft 2 , so they are the same size. 12 ft 8 ft

68. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. Hence the area of the parallelogram is A = h x b where h = height and b = base. h b h b For example, the area of all the parallelograms shown here is 8 x 12 = 96 ft 2 , so they are the same size. 12 ft 8 ft 12 ft 8 ft

69. Area A parallelogram is a shape enclosed by two sets of parallel lines. By cutting and pasting, we may arrange a parallelogram into a rectangle. Hence the area of the parallelogram is A = h x b where h = height and b = base. h b h b For example, the area of all the parallelograms shown here is 8 x 12 = 96 ft 2 , so they are the same size. 12 ft 8 ft 8 ft 8 ft 12 ft 12 ft 12 ft 8 ft

70. Area A triangle is half of a parallelogram.

71. Area A triangle is half of a parallelogram.

72. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, h b h b

73. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. h b h b

74. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. h b h b Therefore the area of a triangle is h x b 2 A = ( h x b) ÷ 2 or A = where h = height and b = base.

75. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. h b h b Therefore the area of a triangle is h x b 2 A = ( h x b) ÷ 2 or A = where h = height and b = base. For example, the area of all the triangles shown here is (8 x 12) ÷ 2 = 48 ft 2 , i.e. they are the same size. 12 ft 8 ft 8 ft

76. Area A triangle is half of a parallelogram. Given a triangle, we may copy it and paste to itself to make a parallelogram, and the area of the triangle is half of the area of the parallelogram formed. h b h b Therefore the area of a triangle is h x b 2 A = ( h x b) ÷ 2 or A = where h = height and b = base. For example, the area of all the triangles shown here is (8 x 12) ÷ 2 = 48 ft 2 , i.e. they are the same size. 12 ft 8 ft 8 ft 8 ft 12 ft 12 ft 12 ft 8 ft

77. Area A trapezoid is a 4-sided figure with one set of opposite sides parallel.

78. Area Example D. Find the area of the following trapezoid R. Assume the unit is meter. A trapezoid is a 4-sided figure with one set of opposite sides parallel. 12 5 8 R

79. Area By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. 12 8 8 4 5 Example D. Find the area of the following trapezoid R. Assume the unit is meter. A trapezoid is a 4-sided figure with one set of opposite sides parallel. R

80. Area By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. 12 8 8 4 The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m 2 . 5 Example D. Find the area of the following trapezoid R. Assume the unit is meter. A trapezoid is a 4-sided figure with one set of opposite sides parallel. R

81. Area By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. 12 8 8 4 The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m 2 . The triangle has base = 4 and height = 5, hence its area is (4 x 5) ÷ 2 = 10 m 2 . 5 Example D. Find the area of the following trapezoid R. Assume the unit is meter. A trapezoid is a 4-sided figure with one set of opposite sides parallel. R

82. Area By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. 12 Therefore the area of the trapezoid is 40 + 10 = 50 m 2 . 8 8 4 The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m 2 . The triangle has base = 4 and height = 5, hence its area is (4 x 5) ÷ 2 = 10 m 2 . 5 Example D. Find the area of the following trapezoid R. Assume the unit is meter. A trapezoid is a 4-sided figure with one set of opposite sides parallel. R

83. Area By cutting R parallel to one side as shown, we split R into two areas, one parallelogram and one triangle. 12 Therefore the area of the trapezoid is 40 + 10 = 50 m 2 . 8 8 4 The parallelogram has base = 8 and height = 5, hence its area is 8 x 5 = 40 m 2 . The triangle has base = 4 and height = 5, hence its area is (4 x 5) ÷ 2 = 10 m 2 . We may find the area of any trapezoid by cutting it into one parallelogram and one triangle. 5 Example D. Find the area of the following trapezoid R. Assume the unit is meter. A trapezoid is a 4-sided figure with one set of opposite sides parallel. R

84. Circumference and Area of Circles A circle has a center x, x

85. Circumference and Area of Circles A circle has a center x, and the distance from any location on the circle to C is a fixed number r, r is called the radius of the circle . r r x

86. Circumference and Area of Circles A circle has a center x, and the distance from any location on the circle to C is a fixed number r, r is called the radius of the circle . r r x The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points.

87. Circumference and Area of Circles A circle has a center x, and the distance from any location on the circle to C is a fixed number r, r is called the radius of the circle . r r x r r x d (diameter) The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points.

88. Circumference and Area of Circles A circle has a center x, and the distance from any location on the circle to C is a fixed number r, r is called the radius of the circle . r r x r r x d (diameter) The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points. Hence d = 2r.

89. Circumference and Area of Circles A circle has a center x, and the distance from any location on the circle to C is a fixed number r, r is called the radius of the circle . r r x r r x d (diameter) The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points. Hence d = 2r. The perimeter C of a circle is called the circumference and C = π d or C = 2 π r where π ≈ 3.14 …

90. Circumference and Area of Circles A circle has a center x, and the distance from any location on the circle to C is a fixed number r, r is called the radius of the circle . r r x r r x d (diameter) The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points. Hence d = 2r. The perimeter C of a circle is called the circumference and C = π d or C = 2 π r where π ≈ 3.14 … We may use 3 as an under–estimation for π .

91. Circumference and Area of Circles A circle has a center x, and the distance from any location on the circle to C is a fixed number r, r is called the radius of the circle . r r x r r x d (diameter) The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points. Hence d = 2r. The perimeter C of a circle is called the circumference and C = π d or C = 2 π r where π ≈ 3.14 … We may use 3 as an under–estimation for π . Example D. Is 25 feet of rope enough to mark off a circle of radius r = 9 ft on the ground?

92. Circumference and Area of Circles A circle has a center x, and the distance from any location on the circle to C is a fixed number r, r is called the radius of the circle . r r x r r x d (diameter) The diameter d of the circle is the length any straight line that goes through the center x connecting two opposite points. Hence d = 2r. The perimeter C of a circle is called the circumference and C = π d or C = 2 π r where π ≈ 3.14 … We may use 3 as an under–estimation for π . Example D. Is 25 feet of rope enough to mark off a circle of radius r = 9 ft on the ground? No, 25 ft is not enough since the circumference C is at least 3 x 9 = 27 ft.

93. Circumference and Area of Circles r x The area A (enclosed by) of a circle is A = π r 2 where π ≈ 3.14 … A

94. Circumference and Area of Circles r x The area A (enclosed by) of a circle is A = π r 2 where π ≈ 3.14 … Example E. a. Approximate the area of the circle with a 5–meter radius using 3 as the estimated value of π . A

95. Circumference and Area of Circles r x The area A (enclosed by) of a circle is A = π r 2 where π ≈ 3.14 … Example E. a. Approximate the area of the circle with a 5–meter radius using 3 as the estimated value of π . Estimating using 3 in stead of 3.14 … we have the area A to be at least 3 x 5 2 A

96. Circumference and Area of Circles r x The area A (enclosed by) of a circle is A = π r 2 where π ≈ 3.14 … Example E. a. Approximate the area of the circle with a 5–meter radius using 3 as the estimated value of π . Estimating using 3 in stead of 3.14 … we have the area A to be at least 3 x 5 2 = 3 x 25 = 75 m 2 . A

97. Circumference and Area of Circles r x The area A (enclosed by) of a circle is A = π r 2 where π ≈ 3.14 … Example E. a. Approximate the area of the circle with a 5–meter radius using 3 as the estimated value of π . Estimating using 3 in stead of 3.14 … we have the area A to be at least 3 x 5 2 = 3 x 25 = 75 m 2 . b. Approximate the area of the circle with a 5–meter radius using π = 3.14 as the estimated value of π . A

98. Circumference and Area of Circles r x The area A (enclosed by ) of a circle is A = π r 2 where π ≈ 3.14 … Example E. a. Approximate the area of the circle with a 5–meter radius using 3 as the estimated value of π . The better approximate answer using π = 3.14 is 3.14 x 5 2 = 3.14 x 75 = 78.5 m 2 Estimating using 3 in stead of 3.14 … we have the area A to be at least 3 x 5 2 = 3 x 25 = 75 m 2 . b. Approximate the area of the circle with a 5–meter radius using π = 3.14 as the estimated value of π . A

99. Volume The volume of a solid is the measurement of the amount of “room ” or “space” the solid occupies.

100. Volume s The volume of a solid is the measurement of the amount of “room ” or “space” the solid occupies. A cube is a square–box, i.e. a box whose edges are the same. s s A cube

101. Volume s The volume of a solid is the measurement of the amount of “room ” or “space” the solid occupies. We define the volume of a cube whose sides are 1 unit to be 1 unit x 1 unit x 1 unit = 1 unit 3 , i.e. 1 cubic unit. A cube is a square–box, i.e. a box whose edges are the same. s s A cube

102. Volume s The volume of a solid is the measurement of the amount of “room ” or “space” the solid occupies. We define the volume of a cube whose sides are 1 unit to be 1 unit x 1 unit x 1 unit = 1 unit 3 , i.e. 1 cubic unit. A cube is a square–box, i.e. a box whose edges are the same. s s A cube 1 in 1 in 1 in 3 1 cubic inch 1 in

103. Volume s The volume of a solid is the measurement of the amount of “room ” or “space” the solid occupies. We define the volume of a cube whose sides are 1 unit to be 1 unit x 1 unit x 1 unit = 1 unit 3 , i.e. 1 cubic unit. A cube is a square–box, i.e. a box whose edges are the same. s s A cube 1 in 1 in 1 in 3 1 cubic inch 1 in 1 m 1 m 1 m 3 1 cubic meter 1 m

104. Volume s The volume of a solid is the measurement of the amount of “room ” or “space” the solid occupies. We define the volume of a cube whose sides are 1 unit to be 1 unit x 1 unit x 1 unit = 1 unit 3 , i.e. 1 cubic unit. A cube is a square–box, i.e. a box whose edges are the same. s s A cube 1 in 1 in 1 in 3 1 cubic inch 1 in 1 m 1 m 1 m 3 1 cubic meter 1 m 1 mi 1 mi 1 mi 3 1 mi 1 cubic mile

105. Volume s The volume of a solid is the measurement of the amount of “room ” or “space” the solid occupies. We define the volume of a cube whose sides are 1 unit to be 1 unit x 1 unit x 1 unit = 1 unit 3 , i.e. 1 cubic unit. A cube is a square–box, i.e. a box whose edges are the same. s s A cube 1 in 1 in 1 in 3 1 cubic inch 1 in 1 m 1 m 1 m 3 1 cubic meter 1 m 1 mi 1 mi 1 mi 3 1 mi 1 cubic mile We can cut larger cubes into smaller cubes to calculate their volume.

106. Volume A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 2 3 = 8, a 3 x 3 x 3 cube has volume 3 3 = 27, a 4 x 4 x 4 cube has volume 4 3 = 64 (unit 3 ).

107. Volume w = width A rectangular box is specified by three sides: the length, the width, and the height. We say that the dimension of the box is “l by w by h”. l = length h = height A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 2 3 = 8, a 3 x 3 x 3 cube has volume 3 3 = 27, a 4 x 4 x 4 cube has volume 4 3 = 64 (unit 3 ).

108. Volume w = width A rectangular box is specified by three sides: the length, the width, and the height. We say that the dimension of the box is “l by w by h”. l = length h = height A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 2 3 = 8, Here is a “4 by 3 by 2” box. 4 3 2 a 3 x 3 x 3 cube has volume 3 3 = 27, a 4 x 4 x 4 cube has volume 4 3 = 64 (unit 3 ).

109. Volume w = width A rectangular box is specified by three sides: the length, the width, and the height. We say that the dimension of the box is “l by w by h”. l = length h = height A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 2 3 = 8, Here is a “4 by 3 by 2” box. Assuming the unit is inch, then the box may be cut into 2 x 3 x 4 = 24 1–inch cubes so its volume is 24 in 3 . 4 3 2 a 3 x 3 x 3 cube has volume 3 3 = 27, a 4 x 4 x 4 cube has volume 4 3 = 64 (unit 3 ).

110. Volume w = width A rectangular box is specified by three sides: the length, the width, and the height. We say that the dimension of the box is “l by w by h”. l = length h = height A 2 x 2 x 2 cube has volume 2 x 2 x 2 = 2 3 = 8, Here is a “4 by 3 by 2” box. Assuming the unit is inch, then the box may be cut into 2 x 3 x 4 = 24 1–inch cubes so its volume is 24 in 3 . We define the volume V of a box whose sides are l, w, and h to be V = l x w x h unit 3 . In particular the volume of a cube whose sides equal to s is V = s x s x s = s 3 unit 3 . 4 3 2 a 3 x 3 x 3 cube has volume 3 3 = 27, a 4 x 4 x 4 cube has volume 4 3 = 64 (unit 3 ).

111. Volume Example F. a. How many cubic inches are there in a cubic foot? (There are 12 inches in 1 foot.)

112. Volume Example F. a. How many cubic inches are there in a cubic foot? (There are 12 inches in 1 foot.) One cubic foot is 1 ft x 1 ft x 1ft

113. Volume Example F. a. How many cubic inches are there in a cubic foot? (There are 12 inches in 1 foot.) One cubic foot is 1 ft x 1 ft x 1ft or 12 in x 12 in x 12 in = 1728 in 3 .

114. Volume Example F. a. How many cubic inches are there in a cubic foot? (There are 12 inches in 1 foot.) Example b. How many cubic feet are there in the following solid? One cubic foot is 1 ft x 1 ft x 1ft or 12 in x 12 in x 12 in = 1728 in 3 .

115. Volume Example F. a. How many cubic inches are there in a cubic foot? (There are 12 inches in 1 foot.) Example b. How many cubic feet are there in the following solid? One cubic foot is 1 ft x 1 ft x 1ft or 12 in x 12 in x 12 in = 1728 in 3 . We may view the solid is consisted of two solids I and II as shown. II I Method 1.

116. Volume Example F. a. How many cubic inches are there in a cubic foot? (There are 12 inches in 1 foot.) Example b. How many cubic feet are there in the following solid? One cubic foot is 1 ft x 1 ft x 1ft or 12 in x 12 in x 12 in = 1728 in 3 . We may view the solid is consisted of two solids I and II as shown. II I The volume of I is 3 x 3 x 3 = 27, the volume of II is 10 x 3 x 6 = 180. Method 1.

117. Volume Example F. a. How many cubic inches are there in a cubic foot? (There are 12 inches in 1 foot.) Example b. How many cubic feet are there in the following solid? One cubic foot is 1 ft x 1 ft x 1ft or 12 in x 12 in x 12 in = 1728 in 3 . We may view the solid is consisted of two solids I and II as shown. II I The volume of I is 3 x 3 x 3 = 27, the volume of II is 10 x 3 x 6 = 180. Hence the volume of the entire solid is 180 + 27 = 207 ft 3 . Method 1.

118. Volume The solid may be viewed as a box with volume 3 x 10 x 9 = 270 with a top portion removed. Method 2. 9 ft 10 ft 3 ft

119. Volume The solid may be viewed as a box with volume 3 x 10 x 9 = 270 with a top portion removed. The dimension of the removed portion is also a box with volume 3 x 3 x 7 = 63. Method 2. 9 ft 10 ft 3 ft 3 ft 3 ft 7 ft

120. Volume The solid may be viewed as a box with volume 3 x 10 x 9 = 270 with a top portion removed. The dimension of the removed portion is also a box with volume 3 x 3 x 7 = 63. Hence the volume of the given solid is 270 – 63 = 207 ft 3 . Method 2. 9 ft 10 ft 3 ft 3 ft 3 ft 7 ft