Area of Figures – Chapter Problems Area of Rectangles Classwork 1. Find the area of a rectangle that has a length of 4.6 in. and a width of 3.4 in. 2. Annabelle wants to place new carpeting in her dining room. Her dining room is a rectangle with a length of 30 feet and a width of 15 feet. a. How much carpeting does she need to buy to cover her entire dining room? b. If the carpeting costs $7.50 per square foot, how much will it cost to carpet Annabelle’s dining room? 3. The diagonal of a rectangle is 26 feet, and its width is 10 feet. Find the length of the rectangle and its area. 4. The diagonal of a rectangle is 50 feet, and its length is 10 more feet than its width. Find the rectangle’s length, width, and area. PARCC-type Question 5. The population density is the amount of people living per square mile. If the town of Mathville is a rectangular town that has a length of 12 miles, a width of 6 miles, and a population of 3,982 people, what is the population density of the town? Round your answer to the nearest hundredth.

Area of Rectangles Homework 6. Find the area of a rectangle that has a length of 8.5 cm and a width of 3.7 cm. 7. Rick wants to place new carpeting in his living room. His living room is a rectangle with a length of 40 feet and a width of 20 feet. a. How much carpeting does he need to buy to cover his entire living room? b. If the carpeting costs $8.75 per square foot, how much will it cost to carpet Rick’s living room? 8. The diagonal of a rectangle is 30 feet, and its width is 6 feet less than its length. Find the rectangle’s length, width, and area. 9. A rectangle has a perimeter of 200 feet and its length is 4 times its width. Find the rectangle’s length, width, and area. PARCC-type Question 10. The population density is the amount of people living per square mile. If the town of Algebraville is a rectangular town that has a length of 13 miles, a width of 10 miles, and a population of 12,576 people, what is the population density of the town? Round your answer to the nearest hundredth.

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Area of Triangles Classwork Find the area of the triangle for #11-14. 11.

8

12.

55°

12

12

55° 47°

13

13.

14.

14

14

14

48° 70° 𝟕𝟑. 𝟒°

18

8

15. Derive the formula A = ½ ab sin(C) for the area of a triangle by drawing an altitude from vertex B perpendicular to the opposite side.

B c

a

C

b

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Area of Triangles Homework Find the area of the triangle for #16-19. 16.

17.

45°

18 16

23 37°

30°

10

18.

19. 22 100°

10

10

16

43°

𝟔𝟔. 𝟒𝟐°

8

20. Derive the formula A = ½ ac sin(B) for the area of a triangle by drawing an altitude from vertex A perpendicular to the opposite side.

B a

C

c

b

A

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Law of Sines Classwork Solve the triangle. Round your answers to the nearest hundredth. 21. B

7

5

37°

A

C

𝑚∠𝐴 =__________, 𝑚∠𝐵 = __________, AC = __________ 22.

E 17

8 D

126°

F 𝑚∠𝐸 =__________, 𝑚∠𝐹 = __________, DF = __________

23.

z 30°

65°

y x°

25 x = __________, y = __________, z = __________

24.

z

x°

y 60°

40° 10

x = __________, y = __________, z = __________

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Law of Sines Homework Solve the triangle. Round your answers to the nearest hundredth. 25.

I 8

5

32°

G

H

𝑚∠𝐻 =__________, 𝑚∠𝐼 = __________, GH = __________ 26.

L 12 105° K

J

10

𝑚∠𝐽 =__________, 𝑚∠𝐿 = __________, KL = __________ 27.

25

58°

z

x° y

23° x = __________, y = __________, z = __________

28.

12

45°

y

95°

x° z x = __________, y = __________, z = __________

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Area of Parallelograms Classwork 29. Find the area of a parallelogram that has a base of 3.6 inches and a height of 8.2 inches. 30. A parallelogram has a height of 5.2 cm and an area of 48.88 𝑐𝑚2 . Find the length of its base. 31. A parallelogram has a base length of 9.3 inches and an area of 56.73 𝑖𝑛2 . Find the height of the parallelogram. 32. A parallelogram has a base length of 7.5 cm. The length of its other side is 5 cm, and the obtuse angle between two of the sides is 133°. Find the height and area of the parallelogram. 33. A diagonal parking space creates a parallelogram. Its length is 21 feet. Its distance along the curb is 10.5 feet, and the acute angle that is made with the curb is 60°. Find the area of the parking space. 34. A window frame is in the shape of a parallelogram. Its base is 4 feet long and its other side length is 3 feet long. The acute angle between two of the sides is 48°. Find the height and the area of the window. 35. Mrs. Polygon is making a quilt for her granddaughter. The quilt is created by stitching together parallelograms that are different colors. Each parallelogram is 3 inches long, its other side is 2 inches long, and the obtuse angle between the two sides is 145°. a. What is the area of each parallelogram used to make the quilt? Round your answer to the nearest hundredth. b. The quilt requires 6 parallelograms horizontally and 12 parallelograms vertically. How much material will Mrs. Polygon need to make the quilt?

Area of Parallelograms Homework 36. Find the area of a parallelogram that has a base of 2.5 feet and a height of 7.9 feet 37. A parallelogram has a height of 4.8 cm and an area of 32.16 𝑐𝑚2 . Find the length of its base. 38. A parallelogram has a base length of 19.5 inches and an area of 232.05 𝑖𝑛2 . Find the height of the parallelogram.

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39. A parallelogram has a base length of 8.3 cm. The length of its other side is 6.2 cm, and the acute angle between two of the sides is 48°. Find the height and area of the parallelogram. 40. A diagonal parking space creates a parallelogram. Its length is 19 feet 10 inches. Its distance along the curb is 12 feet 9 inches, and the acute angle that is made with the curb is 45°. Find the area of the parking space. 41. A window frame is in the shape of a parallelogram. Its base is 3.5 feet long and its other side length is 2.5 feet long. The obtuse angle between two of the sides is 127°. Find the height and the area of the window.

42. In the Polygon household, the living room floor is shaped like a parallelogram. The length of the room is 30 meters and the length of the side adjacent to the base is 20 meters. The obtuse angle between the two sides is 125°. Mrs. Polygon wants to install hardwood flooring in her living room. a. What is the area of the living room? b. If hardwood flooring costs $8.25 per square foot, how much will it cost to get the hardwood floor installed?

Area of Regular Polygons Classwork Calculate the perimeter and area of each regular polygon. Round your answers to the nearest hundredth. 43.

12

44.

4

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45.

12

46.

10

Area of Regular Polygons Homework Calculate the perimeter and area of each regular polygon. Round your answers to the nearest hundredth. 47.

2 3

48.

12

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49.

12

50.

6

Area of Circles & Sectors Classwork Find the area of the minor sector. Round to the nearest hundredth or leave your answer in terms of 𝜋. 51. 52. 53. 54. 0 45

220o

R=6ft.

R=3in

100o r=10 cm

95o d=15 in

Find the area of the major sector. Round to the nearest hundredth or leave your answer in terms of 𝜋. 55. 56. 57. 58. o r=10 cm 0 45

R=3in

Geometry – Area of Figures

100

220o

R=6ft.

95o

d=15 in

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Area of Circles & Sectors Homework Find the area of the minor sector. Round to the nearest hundredth or leave your answer in terms of 𝜋. 59. 60. 61. 62. 115o

74o

105o

47o r=2.3 cm d=21in

d=10ft.

r=6.7 m

Find the area of the major sector. Round to the nearest hundredth or leave your answer in terms of 𝜋. 63. 64. 65. 66. 115o d=10ft.

74o

105o

47o r=2.3 cm d=21in

r=6.7 m

Area of Other Quadrilaterals Classwork Answer each question below. 67. A teacher has 15 desks in their classroom. Each desk was created using wood for the flat top surface and metal bars for the legs. The shape of all of the flat top surfaces is an isosceles trapezoid. The short and long bases have a length of 3 feet and 5 feet, respectively, and the acute angle formed by the legs and the long base is 68°. a. Find the height of the trapezoid.

b. Find the total amount of wood required to make all 15 desks in the classroom.

68. A ruby was cut into the shape of a kite to make the ring shown to the right. If its diagonals measure 20 mm and 11 mm, what is the area of the stone?

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69. A rhombic triacontahedron is a convex 3-D solid with 30 rhombic faces. Each rhombic face has acute angles that measure 63.43°. The length of the small diagonal is 2 inches. a. Find the length of the long diagonal in the rhombus.

b. Find the area of one rhombic face.

c. Find the surface area (area of all of the faces) in this rhombic triacontahedron.

Find the area of each complex figure given below. 70. 71. The radius of each semicircle is 5 cm.

17 in 15 in

26 in

PARCC-type Questions Solve the following word problems based on the information below. The Bisect Building Company has created a building plan for the new patio for the Quadrilateral Family, shown in the figure. 72. The roof of the patio made from 2 isosceles trapezoids and 3 rectangles. a. What is the area of the entire roof? Explain your answer. b. Each bundle of shingles covers 36 𝑓𝑡 2 . Shingles cost $25.50 per bundle and must be purchased in full bundles. The builder has a budget of $125 for shingles. Did the Bisect Building Company budget enough money for the shingles? Explain your answer.

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73. The patio will cover the Quadrilateral’s new hot tub (or Jacuzzi tub) and possibly a sidewalk that is 2 feet wide on all four sides. The hot tub is 6’ 10” x 6’ 10”. a. How much concrete is needed to make the sidewalk surrounding the entire hot tub (or Jacuzzi tub)? Explain your answer.

b. If the price for sidewalk installation is $4.75 per square foot, how much will the Quadrilateral’s have to pay to get their sidewalk installed? Explain your answer.

c. Will the roof cover the hot tub (or Jacuzzi tub) and the sidewalk? Explain your answer.

Area of Other Quadrilaterals Homework Answer each question below. 74. A teacher has 12 desks in their classroom. Each desk was created using wood for the flat top surface and metal bars for the legs. The shape of all of the flat top surfaces is an isosceles trapezoid. The short and long bases have a length of 3.5 feet and 5.5 feet, respectively, and the acute angle formed by the legs and the long base is 57°. a. Find the height of the trapezoid. b. Find the total amount of wood required to make all 12 desks in the classroom. 75. The logo for Mitsubishi Motors is comprised of 3 congruent rhombi joined together at a center point to create a triangular figure. In each rhombus, the length of the long diagonal is 5 inches and the length of the short diagonal is 2.5 inches. a. What is the area of each rhombus in the logo?

b. What is the total rhombus area of the logo?

76. A blue sapphire stone was cut into the shape of a kite to make a necklace as shown to the right. If its diagonals measure 2.2 cm and 1.1 cm, what is the area of the stone?

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Find the area of each complex figure given below. 77. 78.

24 in 15 in 10 cm

62° 70°

12 in

2 in 30 in

PARCC-type Questions Solve the following word problems based on the information below. Mrs. Skew is going to redesign her bedroom, 30 ft shown in the picture to the right. 5 ft 79. The first thing that she needs to do is replace her carpet. If carpet is sold for $6.75 per square 3 ft foot, how much will it cost? Explain your answer.

15 ft

6 ft 80. Mrs. Skew is going to paint all of the walls, which 117° are 8 feet high, with a 2 coats of paint. The room contains a doorway that is 3 ft by 7 ft, 3 windows measuring 4 ft by 3.5 ft, and a closet doorway that is 6 ft by 7 ft. The doorway, windows and closet doorway will not be painted. a. What is the total amount of wall space needs to be covered with paint? Explain your answer.

b. If paint is sold in 1-gallon containers, and each gallon of paint covers 350 square feet and each can of paint costs $12.50, how much money does Mrs. Skew need to spend on paint? Explain your answer.

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1

81. Given the figure below, explain why the area formula for a kite is 𝐴 = 2 𝑑1 𝑑2 .

Area and Perimeter of Figures in the Coordinate Plane Classwork Calculate the perimeter & area of each figure below. 82. 83.

PARCC-type Questions: 84. The figure shows polygon ABCDEF in the coordinate plane with point A at (0, 3.71), point B at (1.64, 3.71), point C at (1.64, 3), point D at (2.89, 3), point E at (2.89, 0), and point F at the origin. Polygon ABCDEF can be used to approximate the size of the state of Utah with x and y scales representing hundreds of miles. a. Based on the information given, how many miles is the perimeter of Utah? b. At the end of 2010, the population of Utah was 2,763,885 people. Based on the information given, what was the population density at the end of 2010? 85. The town of Geometryville has Polygon Park on a plot Geometry – Area of Figures

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of land. The figure represents a map of the Polygon Park showing the location of the Parking Area, Soccer Fields, and Playground. The coordinates represent points on a rectangular grid with the units in hundreds of feet. a. There is a walking trail around the entire park, including the Parking area. What is the length of the walking trail? Express your answer to the nearest foot.

b. What is the area of the plot of land that does not include the parking area? Express your answer to the nearest square foot.

c. The town is planning to put a fence around the Playground and Soccer Fields. What is the total length of fencing needed? Express your answer to the nearest foot.

d. In the future, Geometryville has plans to construct a circular pool centered at (35, 8) on the map. Which of the listed measurements could be the radius of the pool that will fit on the map and be at least 500 feet away from the Soccer Fields and edges of the park. Select all that apply. i. 150 feet iv. 600 feet ii. 300 feet v. 750 feet iii. 450 feet

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1 0

Area and 8 Perimeter of Figures in the Coordinate Plane Homework Calculate the perimeter & area of each figure below. 6 86. 87. 4

F2 G – 1 0

– 5

5 – 2

E

1 0

1 5

H

– 4

– 6

– 8

– 1 0

PARCC-type Questions: 88. The figure shows polygon ABCDEFGH in the coordinate plane with point A at (0, 3.92), point B at (3.02, 3.92), point C at (3.02, 0.48), point D at (1.42, 0.48), point E at (1.42, 0.35), point F at (0.46, 0.35), point G at (0.46, 0), and point H at the origin. Polygon ABCDEFGH can be used to approximate the size of the state of New Mexico with x and y scales representing hundreds of miles. a. Based on the information given, how many miles is the perimeter of New Mexico?

b. At the end of 2010, the population of New Mexico was 2,059,179 people. Based on the information given, what was the population density at the end of 2010?

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89. The town of Geometryville has a Pentagon Park on a plot of land. The figure represents a map of Pentagon Park showing the location of the Parking Area, Basketball Courts, and Playground. The coordinates represent points on a rectangular grid with the units in feet. a. There is a walking trail around the entire park, including the Parking area. What is the length of the walking trail? Express your answer to the nearest foot.

b. What is the area of the plot of land that does not include the parking area? Express your answer to the nearest square foot.

c. The town is planning to put a fence around the Playground and Basketball Courts. What is the total length of fencing needed? Express your answer to the nearest foot.

d. In the future, Geometryville has plans to construct a circular pool centered at (150, 165) on the map. Which of the listed measurements could be the diameter of the pool that will fit on the map and be at least 20 feet away from the Basketball Courts and edges of the park. Select all that apply. i. 50 feet iv. 20 feet ii. 40 feet v. 10 feet iii. 30 feet

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Area of Figures – Chapter Review 1. The diagonal of a rectangle is 39 feet, and its length is 21 more feet than its width. Find the rectangle’s length, width, and area.

2. The perimeter of a rectangle is 192 feet. The length of the rectangle is 1 more than quadruple its width. Find the length, width, and area of the rectangle.

3. The population density is the amount of people living per square mile. If the town of Calculusville is a rectangular town that has a length of 27 miles, a width of 13 miles, and a population of 13,479 people, what is the population density of the town? a. 38.4 people per square mile b. 130.1 people per square mile c. 168.5 people per square mile d. 537.6 people per square mile 4. Calculate the area of the triangle given below. Round your answer to the nearest hundredth.

Use the diagram of ⊙ 𝐶 to answer questions #5 & 6. ̂ , ̅̅̅̅ 5. What is the area of the sector formed by 𝐴𝐸 𝐸𝐶 , and ̅̅̅̅ 𝐴𝐶 , when 𝐶𝐷 = 6𝑚 and 𝑚∠𝐴𝐶𝐸 = 55°? a. 95.82 m2 b. 47.91 m2 c. 34.56 m2 d. 17.28 m2 ̂ , 𝐸𝐶 ̅̅̅̅ , and 𝐴𝐶 ̅̅̅̅ , 6. What is the area of the sector formed by 𝐴𝐷𝐸 when 𝐶𝐷 = 6𝑚 and 𝑚∠𝐴𝐶𝐸 = 55°? a. 95.82 m2 b. 47.91 m2 c. 34.56 m2 d. 17.28 m2

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7. A window frame is in the shape of a parallelogram. Its base is 2.5 feet long and its other side length is 4.5 feet long. The obtuse angle between two of the sides is 113°. What is the area of the window? a. 26.50 ft2 b. 12.21 ft2 c. 10.36 ft2 d. 4.14 ft2 8. Which formulas below are used to calculate the area of a shape? Select all that apply. 1 a. bh e. 2 𝑎𝑃 b. 𝜋𝑟 2 f. 2𝜋𝑟 1 c. 2ℓ + 2𝑤 g. 2 ℎ(𝑏1 + 𝑏2 ) d.

1 2

𝑑1 𝑑2

h. 4s

9. The head of a sprinkler has a radius of 15 ft and covers an arc region that measures 135°. How much grass can it water at one time?

10. The state flag of Delaware has a rhombus in its center. The length of the state flag is 38 feet and the width is 20 feet. If the longer diagonal of the rhombus is 2/3 the length of the rectangle and the shorter diagonal is 3/5 the width of the rectangle, what is the area of the rhombus in the flag?

11. The restaurant sign for McDonald’s, shown to the right has the 2 golden arches sitting on an isosceles trapezoid. If the bases are 35 feet and 27 feet respectively and the acute angle formed between the top base and its legs is 70°. What is the area of the red trapezoid in the sign?

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Extended Constructed Response – Solve the problems, showing all work. Partial credit may be given. 12. Solve the triangle. Round your answers to the nearest hundredth.

13. Calculate the perimeter and area of the regular polygon shown below. Round your answers to the nearest hundredth, or leave them in simplified radical form.

14. The picture below shows the yard of the Fraction family. The coordinates represent points on a rectangular grid with the units in feet. a. What is the perimeter of their yard?

b. Every spring, Mr. Fraction fertilizes the lawn. What is the area of it?

c. This year, the Fractions are going to have a fence installed, represented by the red dotted line. Determine the amount of fencing required.

d. In the future, the Fractions want to install a square-shaped pool centered at (150, 75). If the pool must be at least 10 feet away from the house and the fencing, which of the listed measurements could be the area of the pool. Select all that apply. i. 1,200 ft2 ii. 1050 ft2 iii. 900 ft2 iv. 750 ft2 v. 600 ft2

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15. Mrs. Octagon going to redesign her guest bedroom, shown in the picture to the right. a. The first thing that she needs to do is replace the carpet. If carpet is sold for $5.25 per square foot, how much will it cost? Explain your answer.

b. Mrs. Octagon is going to paint all of the walls, which are 8 feet high, with 2 coats of paint. The room contains a doorway that is 4 feet by 7 feet, 3 windows measuring 3 feet by 4.5 feet, and a closet doorway that is 7 feet by 7 feet. The doorway, windows, and closet doorway will not be painted. What is the total amount of wall space that needs to be covered with paint? Explain your answer.

c. If paint is sold in 1-gallon containers, each gallon of paint covers 350 square feet, and each gallon of paint costs $12.10, how much money does Mrs. Octagon need to spend on paint? Explain your answer.

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Answers 1. 15.64 𝑖𝑛2 2. a) 450 𝑓𝑡 2 b) $3,375 3. length = 24 ft Area = 240 𝑓𝑡 2 4. x = width = 30 ft length = 40 ft Area = 1,200 𝑓𝑡 2 5. 55.31 people per square mile 6. 31.45 𝑐𝑚2 7. a) 800 𝑓𝑡 2 b) $7,000 8. x = length = 24 ft width = 18 ft Area = 432 𝑓𝑡 2 9. x = width = 20 ft length = 80 ft Area = 1,600 𝑓𝑡 2 10. 96.74 people per square mile 11. 63.89 sq units 12. 46.95 sq units 13. 53.67 sq units 14. 93.64 sq units 15. sin C = h/a h = a sin C A = (1/2)(base)(height) A = (1/2)b (a sinC) A = (1/2)ab sin C 16. 103.5 sq units 17. 79.22 sq units 18. 36.66 sq units 19. 173.33 sq units 20. sin B = h/c h = c sin B A = (1/2)(base)(height) A = (1/2)a (c sinB) A = (1/2)ac sin B 21. 𝑚∠𝐴 = 57.41°, 𝑚∠𝐵 = 85.59°, AC = 8.28 units 22. 𝑚∠𝐸 = 31.62°, 𝑚∠𝐹 = 22.38°, DF = 11.02 units 23. x = 85°, y = 13.79 units, z = 27.48 units

Geometry – Area of Figures

24. x = 80°, y = 8.79 units, z = 6.53 units 25. 𝑚∠𝐻 = 57.98°, 𝑚∠𝐼 = 90.02°, GH = 9.44 26. 𝑚∠𝐽 = 21.4°, 𝑚∠𝐿 = 53.6°, KL = 4.53 27. x = 99 o, y = 54.26 units, z = 63.19 units 28. x = 40 o, y = 18.60 units, z = 13.20 units 29. 29.52 𝑖𝑛2 30. 9.4 cm 31. 6.1 in 32. h = 3.66 cm A = 27.45 𝑐𝑚2 33. ℎ = 5.25√3 𝑓𝑡 ≈ 9.09 𝑓𝑡 𝐴 = 110.25√3 𝑓𝑡 2 ≈ 190.89 𝑓𝑡 2 34. h = 2.23 ft A = 8.92 𝑓𝑡 2 35. a) A = 3.45 𝑖𝑛2 b) Total material = 248.4 𝑖𝑛2 36. 19.75 𝑓𝑡 2 37. 6.7 cm 38. 11.9 in. 39. h = 4.61 cm A = 38.26 𝑐𝑚2 40. h = 9.02 ft A = 178.90 𝑓𝑡 2 41. h = 2 ft A = 7 𝑓𝑡 2 42. A = 491.4 𝑓𝑡 2 Cost = $4,054.05 43. P = 72 units 𝐴 = 216√3 𝑢𝑛𝑖𝑡𝑠 2 ≈ 374.12 𝑢𝑛𝑖𝑡𝑠 2 44. 𝑃 = 24√3 𝑢𝑛𝑖𝑡𝑠 ≈ 41.57 𝑢𝑛𝑖𝑡𝑠 𝐴 = 48√3 𝑢𝑛𝑖𝑡𝑠 2 ≈ 83.14 𝑢𝑛𝑖𝑡𝑠 2 45. P = 73.47 units A = 407.29 𝑢𝑛𝑖𝑡𝑠 2 46. P = 72.65 units A = 363.25 𝑢𝑛𝑖𝑡𝑠 2 47. P = 24 units 𝐴 = 24√3 𝑢𝑛𝑖𝑡𝑠 2 ≈ 41.57 𝑢𝑛𝑖𝑡𝑠 2 ~22~

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48. P = 74.16 units A = 423.82 𝑢𝑛𝑖𝑡𝑠 2 49. P = 60 units A = 247.74 𝑢𝑛𝑖𝑡𝑠 2 50. 𝑃 = 24√2 𝑢𝑛𝑖𝑡𝑠 ≈ 33.94 𝑢𝑛𝑖𝑡𝑠 A = 72 𝑢𝑛𝑖𝑡𝑠 2 51. A = 1.125𝜋 in2/3.53 in2 52. A = 14𝜋 / 43.98 ft2 53. A = 27.78𝜋 / 87.27 cm2 54. A = 53.13𝜋 / 166.9 in2 55. A = 7.77𝜋 / 24.4 in2 56. A = 22𝜋 / 69.12 ft2 57. A = 72.22𝜋 / 226.89 cm2 58. A = 112.5𝜋 / 353.43 in2 59. A = 18.06𝜋 / 56.72 ft2 60. A = 13.09𝜋 / 41.13 m2 61. A = 1.09𝜋 / 3.42 cm2 62. A = 57.58𝜋 / 180.88 in2 63. A = 50𝜋 / 157.08 ft2 64. A = 31.8𝜋 / 99.89 m2 65. A = 4.2𝜋 / 13.2 cm2 66. A = 220.5𝜋 / 692.72 in2 67. a) h = 2.48 ft b) A = 148.8 𝑓𝑡 2 68. A = 110 𝑚𝑚2 69. a) Long diagonal = 3.24 in. b) A = 3.24 𝑖𝑛2 c) Total: A = 97.2 𝑖𝑛2 70. A = 161.25 𝑖𝑛2 71. A = (100 + 50𝜋) 𝑐𝑚2 = 257.08 𝑐𝑚2 72. a) Sample Answer: Before finding the area of the roof, you need to determine the length of the legs and height in the isosceles trapezoids using trigonometry and the segment addition postulate. Since the top base is 6’ 2.5”, or 74.5”, the bottom base, which is 9’ or 108” in length, can be divided into 3 sections, measuring 16.75”, 74.5” and 16.75” from left to right. Using

Geometry – Area of Figures

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the 16.75” as your adjacent side to the 65 degree angle, you can use tangent to calculate the height of the trapezoid h = 16.75tan(65) = 35.92 in. Using the same adjacent side measurement and cosine, you can find the length of each leg in the isosceles trapezoid leg = 16.75/tan(65) = 39.63 in. Using these measurements, you can now determine the total area of the roof. A = 108(39.63)(2) + 74.5(108) + 2(½)(35.92)(74.5 + 108) = 23,161.48 𝑖𝑛2 ≈ 160.84 𝑓𝑡 2 b) Sample Answer: Since 1 bundle of shingles covers and area of 36 𝑓𝑡 2 , the total bundles that need to be purchased is 160.84/36 = 4.47 ≈ 5 bundles, making the cost $127.50. Since their budget was $125, which is less than $125, they did not budget enough money for the shingles. 73. a) Sample Answer: The dimensions of the square formed by the hot tub and sidewalk are 10’10” by 10’10”, or 130” by 130”, making its area 16,900 𝑖𝑛2 ≈ 117.36 𝑓𝑡 2 . The dimensions of the hot tub are 6’10” by 6’10” or 82” by 82” making its area 6,724 𝑖𝑛2 ≈ 46.69 𝑓𝑡 2 . By subtracting these two areas, we calculate that the area of the sidewalk is 10,176 𝑖𝑛2 ≈ 70.67 𝑓𝑡 2 b) Sample Answer: Since the unit price is $4.75 per square foot, the total cost for the sidewalk installation is 4.75(70.67) = $335.68 c) Sample Answer: No, the total

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length of the hot tub & side walk is 10’10” or 130”, which exceeds the 9’ or 108” measurement from the roof. 74. a) h = 1.54 ft b) A = 83.16 𝑓𝑡 2 75. a) A = 6.25 𝑖𝑛2 b) A = 18.75 𝑖𝑛2 76. A = 1.21 𝑐𝑚2 77. A = 46.40 𝑐𝑚2 78. A = 632.6 𝑖𝑛2 79. Sample Answer: The total area of the floor is 5(30) + 27(10) + ½(20.61)(27+6) 150 + 270 + 340.065 760.065 𝑓𝑡 2 If you take the total area of the floor and multiply it by the unit price of $6.75, then the total cost is $5,130.44. 80. a) Sample Answer: Since all of the walls are rectangular, you can find the complete area of the walls by multiplying the perimeter of the room by the height of the wall, and then subtract the areas that will not be painted (door, windows & closet). Using this method, the total amount of wall space that will be painted is Area of walls: 81(8) = 648 𝑓𝑡 2 Area of door: 7(3) = 21 𝑓𝑡 2 Area of windows: 3(4)(3.5) = 42 𝑓𝑡 2 Area of closet: 6(7) = 42 𝑓𝑡 2 Area to be painted: 2(648 – 21 – 42 – 42) = 2(543) = 1,086 𝑓𝑡 2 b) Sample Answer: 1,086/350 = 3.10 ≈ 4 𝑐𝑎𝑛𝑠 of paint are required. Since each can of paint costs $12.50, the total cost for Mrs. Skew will be 4(12.50) = $50. 81. Sample Answer: If you take the bottom right triangle from the kite shown and reflect it vertically, it

Geometry – Area of Figures

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will fit on the other side of the bottom left triangle. The same can be done to the upper right triangle: move it and reflect it vertically; then it will fit on the other side of the top left triangle. When this occurs, a rectangle is formed, and the length would be diagonal 2 and the width will be ½ the length of diagonal 1. Therefore the formula for the 1 area of a kite is 𝐴 = 2 𝑑1 𝑑2 . Visual Sample Answer: d2

d1

Figure 1

Figure 2 d2

d1

Figure 3 Using the area of a rectangle formula and the dimensions given above in Figure 3, we can conclude that the area of a kite is 1 𝐴 = 2 𝑑1 𝑑2 . 82. P = 6√41 𝑢𝑛𝑖𝑡𝑠 ≈ 38.42 𝑢𝑛𝑖𝑡𝑠 A = 82 𝑢𝑛𝑖𝑡𝑠 2 83. P = (4√5 + 10√2)𝑢𝑛𝑖𝑡𝑠 P ≈ 23.09 𝑢𝑛𝑖𝑡𝑠 A = 30 𝑢𝑛𝑖𝑡𝑠 2 84. a) 1,320 miles b) 28.10 people per square mile 85. a) 18,110 ft b) 14,250,000 𝑓𝑡 2 c) 12,781 ft d) i & ii

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86. P = (3√17 + √85 + √34) 𝑢𝑛𝑖𝑡𝑠 P ≈ 27.42 𝑢𝑛𝑖𝑡𝑠 A = 25.5 𝑢𝑛𝑖𝑡𝑠 2 87. P = (7√13 + √26 + √85) 𝑢𝑛𝑖𝑡𝑠 P ≈ 38.40 𝑢𝑛𝑖𝑡𝑠 A = 45.5 𝑢𝑛𝑖𝑡𝑠 2 88. a) 1,388 miles b) 19.18 people per square mile 89. a) 857 ft b) A = 35,375 𝑓𝑡 2 c) 667 ft d) iii, iv & v

Answers: Chapter Review 1. ℓ = 36 𝑓𝑡, 𝑤 = 15 𝑓𝑡, 𝐴 = 540 𝑓𝑡 2 2. ℓ = 77 𝑓𝑡, 𝑤 = 19 𝑓𝑡 𝐴 = 1,463 𝑓𝑡 2 3. A 4. A = 78.14 m2 5. D 6. A 7. C 8. A, B, D, E, G 675𝜋 9. 8 𝑓𝑡 2 = 265.07 𝑓𝑡 2 10. A = 152 ft2 11. A = 340.69 ft2 12. 𝑚∠𝐶 = 54°, 𝐵𝐶 = 13 𝑐𝑚 𝐴𝐶 = 4.43 𝑐𝑚 13. 𝑃 = 72√3 𝑢𝑛𝑖𝑡𝑠 ≈ 124.71 𝑢𝑛𝑖𝑡𝑠 𝐴 = 648√3 𝑢𝑛𝑖𝑡𝑠 2 ≈ 1,122.37 𝑢𝑛𝑖𝑡𝑠 2 14. a) P = 660 ft b) Area of the lawn = 20,200 ft2 c) 525 ft of fencing d) iii, iv, and v

Geometry – Area of Figures

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15. a) Sample Answer: The total area of the floor is 1 (10√3)(10 + 30) + 25(37) 2 346.41 + 925 1,271.41 𝑓𝑡 2 If you take the total area of the floor and multiply it by the unit price of $5.25, then the total cost is $6,674.90 b) Sample Answer: Since all of the walls are rectangular, you can find the complete area of the walls by multiplying the perimeter of the room by the height of the wall, and then subtract the areas that will not be painted (door, windows & closet). Using this method, the total amount of wall space that will be painted is Area of walls: 144(8) = 1,152 𝑓𝑡 2 Area of door: 7(4) = 28 𝑓𝑡 2 Area of windows: 3(4.5)(3) = 40.5 𝑓𝑡 2 Area of closet: 7(7) = 49 𝑓𝑡 2 Area to be painted: = 1,152 – 28 – 40.5 – 49 = 2,069 𝑓𝑡 2 c) Sample Answer: 2,069/350 = 5.91 ≈ 6 𝑐𝑎𝑛𝑠 of paint are required. Since each can of paint costs $12.10, the total cost for Mrs. Octagon will be 6(12.10) = $72.60

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