Area of Figures – Chapter Problemscontent.njctl.org/courses/math/geometry-2015-16/area-of-figures/area... · 10/4/2015 · Area of Figures – Chapter Problems Area of Rectangles

Geometry – Area of Figures ~1~ NJCTL.org

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.


Area of Triangles Classwork Find the area of the triangle for #11-14. 11. 12.

13. 14.

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.

55°

13

12

8

14 14

b

ac

B

C A

70°48°

18

14

55°

47°

12

8

𝟕𝟑. 𝟒°


100°

43°16

22

Area of Triangles Homework Find the area of the triangle for #16-19. 16. 17.

18. 19. 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.

30°

23

18

45°

37°

10

16

ac

bC

B

A

8

10 10

𝟔𝟔. 𝟒𝟐°


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

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

22.

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

23.

x = __________, y = __________, z = __________

24.

x = __________, y = __________, z = __________

37°

75

B

A C

178

126°

E

DF

25

z y

x°

65°

30°

y

40° 60°

x°

10

z


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

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

26.

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

27.

x = __________, y = __________, z = __________

28.

x = __________, y = __________, z = __________

58

32°GH

I

105°

10

12

JK

L

25

y

z

x° 23°

58°

45°

95° x°

z

y12


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.


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.

44.

12

4


45.

46.

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

48.

12

10

2 3

12


49.

50.

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. Find the area of the major sector. Round to the nearest hundredth or leave your answer

in terms of 𝜋. 55. 56. 57. 58.

12

6

450

R=3in R=6ft.

220o

r=10 cm 100o

d=15 in

95o

450

R=3in R=6ft.

220o

d=15 in

95o r=10 cm

100o


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.

Find the area of the major sector. Round to the nearest hundredth or leave your answer

in terms of 𝜋. 63. 64. 65. 66. 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?

d=10ft.

115o 105o

r=6.7 m

r=2.3 cm 74o 47o

d=21in

d=10ft.

115o 105o

r=6.7 m

r=2.3 cm 74o 47o

d=21in


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.

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 𝑓𝑡2. 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.

26 in

17 in

15 in


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?


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

PARCC-type Questions Solve the following word problems based on the information below.

Mrs. Skew is going to redesign her bedroom, shown in the picture to the right.

79. The first thing that she needs to do is replace her

carpet. If carpet is sold for $6.75 per square foot, how much will it cost? Explain your answer.

80. Mrs. Skew is going to paint all of the walls, which 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.

70°

10 cm

24 in

2 in

62°

15 in

30 in

12 in

117°

5 ft

3 ft

6 ft

15 ft

30 ft


81. Given the figure below, explain why the area formula for a kite is 𝐴 =1

2𝑑1𝑑2.

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


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 ii. 300 feet iii. 450 feet

iv. 600 feet v. 750 feet


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

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?

10

8

6

4

2

–2

–4

–6

–8

–10

–10 –5 5 10 15

E

G

H

F


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 ii. 40 feet iii. 30 feet

iv. 20 feet v. 10 feet


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 𝐴�̂�, 𝐸𝐶̅̅ ̅̅ , and 𝐴𝐶̅̅ ̅̅ ,

when 𝐶𝐷 = 6𝑚 and 𝑚∠𝐴𝐶𝐸 = 55°? a. 95.82 m2 b. 47.91 m2 c. 34.56 m2 d. 17.28 m2

6. What is the area of the sector formed by 𝐴𝐷�̂�, 𝐸𝐶̅̅ ̅̅ , and 𝐴𝐶̅̅ ̅̅ , when 𝐶𝐷 = 6𝑚 and 𝑚∠𝐴𝐶𝐸 = 55°?

a. 95.82 m2 b. 47.91 m2 c. 34.56 m2 d. 17.28 m2


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.

a. bh e. 1

2𝑎𝑃

b. 𝜋𝑟2 f. 2𝜋𝑟

c. 2ℓ + 2𝑤 g. 1

2ℎ(𝑏1 + 𝑏2)

d. 1

2𝑑1𝑑2 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?


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


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.


Answers

1. 15.64 𝑖𝑛2

2. a) 450 𝑓𝑡2 b) $3,375

3. length = 24 ft

Area = 240 𝑓𝑡2 4. x = width = 30 ft

length = 40 ft

Area = 1,200 𝑓𝑡2 5. 55.31 people per square mile

6. 31.45 𝑐𝑚2

7. a) 800 𝑓𝑡2 b) $7,000

8. x = length = 24 ft width = 18 ft

Area = 432 𝑓𝑡2 9. x = width = 20 ft

length = 80 ft

Area = 1,600 𝑓𝑡2 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

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 𝑓𝑡2 ≈ 190.89 𝑓𝑡2 34. h = 2.23 ft

A = 8.92 𝑓𝑡2

35. a) A = 3.45 𝑖𝑛2

b) Total material = 248.4 𝑖𝑛2

36. 19.75 𝑓𝑡2 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 𝑓𝑡2 41. h = 2 ft

A = 7 𝑓𝑡2

42. A = 491.4 𝑓𝑡2 Cost = $4,054.05

43. P = 72 units

𝐴 = 216√3 𝑢𝑛𝑖𝑡𝑠2 ≈374.12 𝑢𝑛𝑖𝑡𝑠2

44. 𝑃 = 24√3 𝑢𝑛𝑖𝑡𝑠 ≈ 41.57 𝑢𝑛𝑖𝑡𝑠

𝐴 = 48√3 𝑢𝑛𝑖𝑡𝑠2 ≈ 83.14 𝑢𝑛𝑖𝑡𝑠2 45. P = 73.47 units

A = 407.29 𝑢𝑛𝑖𝑡𝑠2 46. P = 72.65 units

A = 363.25 𝑢𝑛𝑖𝑡𝑠2 47. P = 24 units

𝐴 = 24√3 𝑢𝑛𝑖𝑡𝑠2 ≈ 41.57 𝑢𝑛𝑖𝑡𝑠2


48. P = 74.16 units

A = 423.82 𝑢𝑛𝑖𝑡𝑠2 49. P = 60 units

A = 247.74 𝑢𝑛𝑖𝑡𝑠2

50. 𝑃 = 24√2 𝑢𝑛𝑖𝑡𝑠 ≈ 33.94 𝑢𝑛𝑖𝑡𝑠

A = 72 𝑢𝑛𝑖𝑡𝑠2 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 𝑓𝑡2

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

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 𝑓𝑡2 b) Sample Answer: Since 1 bundle of shingles covers and

area of 36 𝑓𝑡2, 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 𝑓𝑡2. The dimensions of the hot tub are 6’10” by 6’10” or 82” by 82” making its area

6,724 𝑖𝑛2 ≈ 46.69 𝑓𝑡2. By subtracting these two areas, we calculate that the area of the

sidewalk is 10,176 𝑖𝑛2 ≈

70.67 𝑓𝑡2 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


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 𝑓𝑡2

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 𝑓𝑡2 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 𝑓𝑡2

Area of door: 7(3) = 21 𝑓𝑡2 Area of windows:

3(4)(3.5) = 42 𝑓𝑡2

Area of closet: 6(7) = 42 𝑓𝑡2 Area to be painted: 2(648 – 21 – 42 – 42)

= 2(543) = 1,086 𝑓𝑡2 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

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

area of a kite is 𝐴 =1

2𝑑1𝑑2.

Visual Sample Answer:

Figure 1 Figure 2

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

𝐴 =1

2𝑑1𝑑2.

82. P = 6√41 𝑢𝑛𝑖𝑡𝑠 ≈ 38.42 𝑢𝑛𝑖𝑡𝑠

A = 82 𝑢𝑛𝑖𝑡𝑠2

83. P = (4√5 + 10√2)𝑢𝑛𝑖𝑡𝑠

P ≈ 23.09 𝑢𝑛𝑖𝑡𝑠

A = 30 𝑢𝑛𝑖𝑡𝑠2 84. a) 1,320 miles

b) 28.10 people per square mile 85. a) 18,110 ft

b) 14,250,000 𝑓𝑡2 c) 12,781 ft d) i & ii

d2

d1

d1

d2


86. P = (3√17 + √85 + √34) 𝑢𝑛𝑖𝑡𝑠


A = 25.5 𝑢𝑛𝑖𝑡𝑠2

87. P = (7√13 + √26 + √85) 𝑢𝑛𝑖𝑡𝑠


A = 45.5 𝑢𝑛𝑖𝑡𝑠2 88. a) 1,388 miles

b) 19.18 people per square mile 89. a) 857 ft

b) A = 35,375 𝑓𝑡2 c) 667 ft d) iii, iv & v

Answers: Chapter Review

1. ℓ = 36 𝑓𝑡, 𝑤 = 15 𝑓𝑡, 𝐴 = 540 𝑓𝑡2

2. ℓ = 77 𝑓𝑡, 𝑤 = 19 𝑓𝑡 𝐴 = 1,463 𝑓𝑡2

3. A 4. A = 78.14 m2 5. D 6. A 7. C 8. A, B, D, E, G

9. 675𝜋

8 𝑓𝑡2 = 265.07 𝑓𝑡2

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 𝑢𝑛𝑖𝑡𝑠2 ≈ 1,122.37 𝑢𝑛𝑖𝑡𝑠2

14. a) P = 660 ft b) Area of the lawn = 20,200 ft2 c) 525 ft of fencing d) iii, iv, and v

15. a) Sample Answer: The total area of the floor is

1

2(10√3)(10 + 30) + 25(37)

346.41 + 925 1,271.41 𝑓𝑡2

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 𝑓𝑡2

Area of door: 7(4) = 28 𝑓𝑡2 Area of windows:

3(4.5)(3) = 40.5 𝑓𝑡2

Area of closet: 7(7) = 49 𝑓𝑡2 Area to be painted: = 1,152 – 28 – 40.5 – 49

= 2,069 𝑓𝑡2 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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Area of Figures – Chapter Problemscontent.njctl.org/courses/math/geometry-2015-16/area-of-figures/area... · 10/4/2015 · Area of Figures – Chapter Problems Area of Rectangles