ID: CH 3 Test Review
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CH 3 Test Review
Boundary Lines: Area of Parallelograms and Triangles
Calculate the area of each figure. Each square on the grid represents a square that is one meter long
and one meter wide.
1. You are making a kite out of nylon fabric. The height of the kite will be 36 inches and the widest part of the
kite will be 24 inches as shown in the diagram. How much nylon fabric will you need to make the kite? Write
the answer in square inches and square feet.
The Keystone Effect: Area of a Trapezoid
Calculate the area of each trapezoid. Each square on the grid represents a square that is one inch long
and one inch wide.
2. The area of a trapezoid is 209 square yards and the bases are 15 yards and 23 yards. What is the height of the
trapezoid?
.
3. The area of a trapezoid is 150 square meters. The height is 10 meters and one base is two meters longer than
the other base. What is each base?
ID: CH 3 Test Review
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Signs, Signs, Every Place There Are Signs!: Area of Regular Polygons
Calculate the area of each regular polygon.
4.
.
5.
.
6. A regular heptagon has a side length of 24 inches and an apothem of 24.9 inches. What is the area of the
regular heptagon?
.
7. A stop sign has a perimeter of 160 inches and an apothem of 24.1 inches. What is the area of the stop sign?
.
8. A regular nonagon has an area of 378 square yards and an apothem of 10.5 yards. What is the length of a side
of the regular nonagon
.
ID: CH 3 Test Review
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9. A regular polygon has an area of 10,080 square meters. The length of a side of the polygon is 30 meters and
the apothem is 56 meters. What type of regular polygon is this?
.
Say Cheese!: Area and Circumference of a Circle
Calculate the circumference and area of each circle. Use 3.14 to approximate π . Each square on the
grid represents a square that is one centimeter long and one centimeter wide.
10. What is the area of the annulus shown? Use 3.14 to approximate π .
.
Boundary Lines: Area of Parallelograms and Triangles
Define each term in your own words.
11. parallelogram
.
12. altitude of a parallelogram
.
ID: CH 3 Test Review
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13. height of a parallelogram
.
14. altitude of a triangle
.
15. height of a triangle
.
Boundary Lines:Area of Parallelograms and Triangles
Calculate the area of each parallelogram.
EXAMPLE:
A ==== 8(4) ==== 32 mi2
.
16.
.
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17.
.
Boundary Lines: Area of Parallelograms and Triangles
In each parallelogram, the base, height, or area is unknown. Calculate the
value of the unknown measure.
EXAMPLE:
A ==== bh
63 ==== 9h
7 ==== h
The height is 7 meters.
.
18.
ID: CH 3 Test Review
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Boundary Lines: Area of Parallelograms and Triangles
The base of each triangle is labeled. Draw a segment that represents the height of the triangle.
EXAMPLE:
19.
20.
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Boundary Lines: Area of Parallelograms and Triangles
Calculate the area of each triangle.
EXAMPLE:
A ====1
2(6)(8) ====
1
2(48) ==== 24 in.
2
.
21.
.
22.
ID: CH 3 Test Review
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Boundary Lines: Area of Parallelograms and Triangles
In each triangle, the base, height, or area is unknown. Calculate the value of the unknown measure.
EXAMPLE:
A ====1
2bh
30 ====1
2(15)h
60 ==== 15h
4 ==== h
The height of the triangle is 4 meters.
.
23.
.
24.
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The Keystone Effect: Area of a Trapezoid
Problem Set
Calculate the area of each trapezoid. Each square on the grid represents a square that is two inches
long and two inches wide.
EXAMPLE:
A =1
2(8 + 16)10 =
1
2(24)(10) = 120
120 square inches
.
25.
ID: CH 3 Test Review
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26.
The Keystone Effect: Area of a Trapezoid
Calculate the area of each trapezoid with the given dimensions, where h represents the height, b1
represents the length of a base, and b2 represents the length of the other base.
EXAMPLE:
h = 2,b 1 = 4,b 2 = 3
A =1
2(4 + 3)2 =
1
2(7)(2) = 7
The area is 7 square units.
.
27. h = 6, b 1 = 5, b 2 = 3
.
28. h = 5, b 1 = 9, b 2 = 1
.
29. h = 4, b 1 =2
3, b 2 =
4
3
ID: CH 3 Test Review
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The Keystone Effect: Area of a Trapezoid
In each trapezoid, one base, height, or area is unknown. Calculate the value of the unknown measure.
EXAMPLE:
A =1
2(2 + 4)3 =
1
2(6)(3) = 9
The trapezoid has an area of 9 square feet.
.
30.
.
31.
.
ID: CH 3 Test Review
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32.
.
33.
.
34.
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Signs, Signs, Every Place There Are Signs!: Area of Regular Polygons
Calculate the area of each regular polygon.
EXAMPLE:
A =1
2(8)(6.3)(6)
A = 151.2
The area is 151.2 square centimeters.
.
35.
.
36.
.
37.
ID: CH 3 Test Review
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Say Cheese!: Area and Circumference of a Circle
Write the term that best completes each statement.
concentric circles diameter radius circle
irrational number circumference annulus
38. The ________________ is the distance around a circle.
39. The distance that is equal to one half the diameter of a circle is the ________________.
40. The distance across a circle through the center is the ________________.
41. A decimal that never repeats or terminates is a(n) _________________.
42. ________________ are circles that share the same center.
43. The ________________ is the region bounded by two concentric circles.
Say Cheese!: Area and Circumference of a Circle
Problem Set: Calculate the diameter of each circle.
EXAMPLE:
d = 2r = 2(6) = 12cm
.
44.
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45.
.
Say Cheese!:Area and Circumference of a Circle
Calculate the radius of each circle.
r =d
2=
18
2= 9ft
.
46.
.
47.
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Say Cheese!: Area and Circumference of a Circle
Calculate the circumference of each circle given the radius r of the circle. Write your answers in terms
of π
EXAMPLE:
r = 8 cm
C = 2πr = 2(π)(8) = 16π cm
.
48. r = 2 cm
.
49. r = 10 ft
.
50. r = 4.2 cm
.
Say Cheese!: Area and Circumference of a Circle
Calculate the area of each circle given the radius r of the circle. Write your answers in terms of π .
EXAMPLE:
r = 6 m
A = πr2
= π(62) = 36π m
2
.
51. r = 4 in.
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52. r = 12 cm
.
53. r = 9.8 yd
.
Say Cheese!: Area and Circumference of a Circle
Calculate the radius of each circle given the circumference C of the circle. Write your answers in terms
of π .
EXAMPLE:
C = 90π mm
C = 2πr
90π = 2πr
45 = r
r = 45 mm
.
54. C = 220π ft
.
55. C = 13π cm
.
56. C = 10.8π mi
ID: CH 3 Test Review
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Say Cheese!: Area and Circumference of a Circle
Calculate the radius of each circle given the area A of the circle. Write your answers in terms of π .
EXAMPLE:
A = 9π cm2
A = πr2
9π = r2
9 = r2
3 = r
r = 3 cm
.
57. A = 16π m2
.
58. A = 49π yd2
.
59. A =1
4π m
2
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Say Cheese!: Area and Circumference of a Circle
Use the given information to answer each question.
EXAMPLE:
If a circle has a circumference of 6π inches, what is its area?
C = 2πr
6π = 2πr
3 = r
A = πr2
A = π(32)
A = 9π
The area of the circle is 9π square inches.
.
60. If a circle has a circumference of 3π feet, what is its area?
.
61. If a circle has an area of 25π square feet, what is its circumference?
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Say Cheese!: Area and Circumference of a Circle
Calculate the area of each annulus shown. Use 3.14 to approximate π .
EXAMPLE:
Area of larger circle:
A = πr2
= π(122) = 144π ≈ 452.16 in.
2
Area of smaller circle:
A = πr2
= π(82) = 64π ≈ 200.96 in.
2
Area of annulus:
A ≈ 452.16 − 200.96 = 251.2 in.2
.
62.
.
63.
ID: CH 3 Test Review
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Installing Carpeting and Tile: Area and Perimeter of Composite Figures
Calculate the area of each figure. All measurements are in centimeters. Use 3.14 for π and round
decimal answers to the nearest hundredth.
EXAMPLE:
A = 14(2) + 7(3)
= 28 + 21
= 49 cm2
.
64.
.
65.
.
ID: CH 3 Test Review
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66.
.
Installing Carpeting and Tile: Area and Perimeter of Composite Figures
Calculate the area of the shaded portion of each figure. All measurements are in inches. Use 3.14 for π
and round decimal answers to the nearest hundredth.
EXAMPLE:
A ≈3
4(3.14)(3
2)
=3
4(3.14)(9)
= 21.20 in.2
.
67.
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68.
.
69.
.
70. All of the line segments in the figure are either vertical or horizontal. Determine the perimeter of the figure.
ID: CH 3 Test Review
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Determine the area of the region bounded by the line segments.
71.
72. All of the line segments in the figure shown are either vertical or horizontal. What is the perimeter of the
figure?
73. All of the line segments in the diagram of the bathroom floor shown are either vertical or horizontal. How
many one-inch square tiles would it take to tile the entire floor?
ID: A
1
CH 3 Test Review
Answer Section
1. ANS:
Area of left triangle =1
2(36)(12) = 216 square inches
Area of right triangle =1
2(36)(12) = 216 square inches
Totalarea of kite = 216 + 216 = 432 square inches
432 square inches ⋅
1 square foot
144 square inches= 3 square feet
You will need 432 square inches, or 3 square feet, of nylon fabric to make the kite.
PTS: 1 REF: Ch3.2 TOP: Assignment
2. ANS:
209 =1
2(15 + 23)h
209 = 19h
11 = h
The height of the trapezoid is 11 yards.
PTS: 1 REF: Ch3.3 TOP: Assignment
3. ANS:
Let x represent one base. Then x + 2 represents the other base.
150 =1
2(x + x + 2)(10)
150 = 5(2x + 2)
150 = 10x + 10
140 = 10x
14 = x
One base is 14 meters and the other base is 14 + 2 = 16 meters.
PTS: 1 REF: Ch3.3 TOP: Assignment
4. ANS:
A =1
2(18)(12.4)(5)
= 558 square feet
PTS: 1 REF: Ch3.4 TOP: Assignment
ID: A
2
5. ANS:
A =1
2(35)(53.9)(10)
= 9432.5 square centimeters
PTS: 1 REF: Ch3.4 TOP: Assignment
6. ANS:
A =1
2(24)(24.9)(7)
= 2091.6
The area of the regular heptagon is 2091.6 square inches.
PTS: 1 REF: Ch3.4 TOP: Assignment
7. ANS:
A =1
2(160)(24.1)
= 1928
The area of the stop sign is 1928 square inches.
PTS: 1 REF: Ch3.4 TOP: Assignment
8. ANS:
378 =1
2(™)(10.5)(9)
378 = 47.25™
8 = ™
The length of a side of the regular nonagon is 8 yards.
PTS: 1 REF: Ch3.4 TOP: Assignment
9. ANS:
10,080 =1
2(30)(56)(n)
10,080 = 840n
12 = n
The regular polygon has 12 sides. Therefore, the polygon is a regular 12-gon.
PTS: 1 REF: Ch3.4 TOP: Assignment
ID: A
3
10. ANS:
Area of larger circle:
A = πr2
= π(102) = 100π ≈ 314 m
2
Area of smaller circle:
A = πr2
= π(7.52) = 56.25π ≈ 176.625 m
2
Area of annulus:
A ≈ 314 − 176.625 = 137.375 m2
The area of the annulus is approximately 137.375 square meters.
PTS: 1 REF: Ch3.5 TOP: Assignment
11. ANS:
A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.
PTS: 1 REF: Ch3.2 TOP: Skills Practice
12. ANS:
An altitude of a parallelogram is a line segment drawn from a vertex, perpendicular to the line containing the
opposite side.
PTS: 1 REF: Ch3.2 TOP: Skills Practice
13. ANS:
A height of a parallelogram is the perpendicular distance from any point on one side to the line containing the
opposite side.
PTS: 1 REF: Ch3.2 TOP: Skills Practice
14. ANS:
An altitude of a triangle is a line segment drawn from a vertex perpendicular to the line containing the
opposite side.
PTS: 1 REF: Ch3.2 TOP: Skills Practice
15. ANS:
A height of a triangle is the perpendicular distance from a vertex to the line containing the base of the
triangle.
PTS: 1 REF: Ch3.2 TOP: Skills Practice
16. ANS:
A = 11(6) = 66 mi2
PTS: 1 REF: Ch3.2 TOP: Skills Practice
17. ANS:
A = 7(16) = 112 yd2
PTS: 1 REF: Ch3.2 TOP: Skills Practice
ID: A
4
18. ANS:
A = bh
96 = 12b
8 = b
The base is 8 feet.
PTS: 1 REF: Ch3.2 TOP: Skills Practice
19. ANS:
PTS: 1 REF: Ch3.2 TOP: Skills Practice
20. ANS:
PTS: 1 REF: Ch3.2 TOP: Skills Practice
21. ANS:
A =1
2(7)(7) =
1
2(49) = 24.5 ft
2
PTS: 1 REF: Ch3.2 TOP: Skills Practice
ID: A
5
22. ANS:
A =1
2(4.5)(4) =
1
2(18) = 9 m
2
PTS: 1 REF: Ch3.2 TOP: Skills Practice
23. ANS:
A =1
2bh
A =1
2(3)(2)
A =1
2(6)
A = 3
The area of the triangle is 3 square feet.
PTS: 1 REF: Ch3.2 TOP: Skills Practice
24. ANS:
A =1
2bh
6 =1
2b(3)
12 = 3b
4 = b
The base of the triangle is 4 yards.
PTS: 1 REF: Ch3.2 TOP: Skills Practice
25. ANS:
A =1
2(4 + 8)6 =
1
2(12)(6) = 36
36 square inches
PTS: 1 REF: Ch3.3 TOP: Skills Practice
26. ANS:
A =1
2(10 + 20)14 =
1
2(30)(14) = 210
210 square inches
PTS: 1 REF: Ch3.3 TOP: Skills Practice
ID: A
6
27. ANS:
A =1
2(5 + 3)6 =
1
2(8)(6) = 24
The area is 24 square units.
PTS: 1 REF: Ch3.3 TOP: Skills Practice
28. ANS:
A =1
2(9 + 1)5 =
1
2(10)(5) = 25
The area is 25 square units.
PTS: 1 REF: Ch3.3 TOP: Skills Practice
29. ANS:
A =1
2
2
3+
4
3
Ê
Ë
ÁÁÁÁÁÁ
ˆ
¯
˜̃˜̃˜̃4 =
1
2(2)(4) = 4
The area is 4 square units.
PTS: 1 REF: Ch3.3 TOP: Skills Practice
30. ANS:
1
2(b 1 + 3)2 = 8
b 1 + 3 = 8
b 1 = 5
The trapezoid has a base of 5 centimeters.
PTS: 1 REF: Ch3.3 TOP: Skills Practice
31. ANS:
1
2(b 1 + 8)5 = 25
b 1 + 8 = 10
b 1 = 2
The trapezoid has a base of 2 yards.
PTS: 1 REF: Ch3.3 TOP: Skills Practice
32. ANS:
1
2(5 + 15)h = 70
10h = 70
h = 7
The trapezoid has a height of 7 yards.
PTS: 1 REF: Ch3.3 TOP: Skills Practice
ID: A
7
33. ANS:
A =1
2(13 + 22)5 =
1
2(35)(5) = 87.5
The trapezoid has an area of 87.5 square feet.
PTS: 1 REF: Ch3.3 TOP: Skills Practice
34. ANS:
1
2(b 1 + 13)7 = 98
b 1 + 13 = 28
b 1 = 15
The trapezoid has a base of 15 meters.
PTS: 1 REF: Ch3.3 TOP: Skills Practice
35. ANS:
A =1
2(12)(8.3)(5)
= 249
The area is 249 square feet.
PTS: 1 REF: Ch3.4 TOP: Skills Practice
36. ANS:
A =1
2(10)(10.4)(7)
= 364
The area is 364 square yards.
PTS: 1 REF: Ch3.4 TOP: Skills Practice
37. ANS:
A =1
2(4)(7.5)(12)
= 180
The area is 180 square inches.
PTS: 1 REF: Ch3.4 TOP: Skills Practice
38. ANS: circumference
PTS: 1 REF: Ch3.5 TOP: Skills Practice
39. ANS: radius
PTS: 1 REF: Ch3.5 TOP: Skills Practice
40. ANS: diameter
PTS: 1 REF: Ch3.5 TOP: Skills Practice
ID: A
8
41. ANS: irrational number
PTS: 1 REF: Ch3.5 TOP: Skills Practice
42. ANS: concentric circles
PTS: 1 REF: Ch3.5 TOP: Skills Practice
43. ANS: annulus
PTS: 1 REF: Ch3.5 TOP: Skills Practice
44. ANS:
d = 2r = 2(22) = 44ft
PTS: 1 REF: Ch3.5 TOP: Skills Practice
45. ANS:
d = 2r = 2(9.75) = 19.5 in.
PTS: 1 REF: Ch3.5 TOP: Skills Practice
46. ANS:
r =d
2=
100
2= 50m
PTS: 1 REF: Ch3.5 TOP: Skills Practice
47. ANS:
r =d
2=
10.5
2= 5.25 yd
PTS: 1 REF: Ch3.5 TOP: Skills Practice
48. ANS:
C = 2πr = 2(π)(2) = 4π cm
PTS: 1 REF: Ch3.5 TOP: Skills Practice
49. ANS:
C = 2πr = 2(π)(10) = 20π ft
PTS: 1 REF: Ch3.5 TOP: Skills Practice
50. ANS:
C = 2πr = 2(π)(4.2) = 8.4π cm
PTS: 1 REF: Ch3.5 TOP: Skills Practice
51. ANS:
A = πr2
= π(42) = 16π in.
2
PTS: 1 REF: Ch3.5 TOP: Skills Practice
52. ANS:
A = πr2
= π(122) = 144π cm
2
PTS: 1 REF: Ch3.5 TOP: Skills Practice
ID: A
9
53. ANS:
A = πr2
= π(9.82) = 96.04π yd
2
PTS: 1 REF: Ch3.5 TOP: Skills Practice
54. ANS:
C = 2πr
220π = 2πr
110 = r
r = 110 ft
PTS: 1 REF: Ch3.5 TOP: Skills Practice
55. ANS:
C = 2πr
13π = 2πr
6.5 = r
r = 6.5 cm
PTS: 1 REF: Ch3.5 TOP: Skills Practice
56. ANS:
C = 2πr
10.8π = 2πr
5.4 = r
r = 5.4 mi
PTS: 1 REF: Ch3.5 TOP: Skills Practice
57. ANS:
A = πr2
16π = πr2
16 = r2
4 = r
r = 4 m
PTS: 1 REF: Ch3.5 TOP: Skills Practice
ID: A
10
58. ANS:
A = πr2
49π = πr2
49 = r2
7 = r
r = 7 yd
PTS: 1 REF: Ch3.5 TOP: Skills Practice
59. ANS:
A = πr2
1
4π = πr
2
1
4= r
2
1
2= r
r =1
2m
PTS: 1 REF: Ch3.5 TOP: Skills Practice
60. ANS:
C = 2πr
3π = 2πr
1.5 = r
A = πr2
A = π(1.52)
A = 2.25π
The area of the circle is 2.25π square feet.
PTS: 1 REF: Ch3.5 TOP: Skills Practice
61. ANS:
A = πr2
25π = πr2
25 = r2
5 = r
C = 2πr
C = 2π(5)
C = 10π
The circumference of the circle is 10π feet.
PTS: 1 REF: Ch3.5 TOP: Skills Practice
ID: A
11
62. ANS:
Area of larger circle:
A = πr2
= π(222) = 484π ≈ 1519.76 ft
2
Area of smaller circle:
A = πr2
= π(112) = 121π ≈ 379.94 ft
2
Area of annulus:
A ≈ 1519.76 − 379.94 = 1139.82 ft2
PTS: 1 REF: Ch3.5 TOP: Skills Practice
63. ANS:
Area of larger circle:
A = πr2
= π(342) = 1156π ≈ 3629.84 m
2
Area of smaller circle:
A = πr2
= π(27.22) = 739.84π ≈ 2323.10 m
2
Area of annulus:
A ≈ 3629.84 − 2323.10 = 1306.74 m2
PTS: 1 REF: Ch3.5 TOP: Skills Practice
64. ANS:
A =1
2(30 + 15)(15) +
1
2(20)(15)
= 337.5 + 150
= 487.5 cm2
PTS: 1 REF: Ch3.6 TOP: Skills Practice
65. ANS:
A = 21
2
Ê
Ë
ÁÁÁÁÁÁ
ˆ
¯
˜̃˜̃˜̃(1)(2) + 4(1)
= 2 + 4
= 6 cm2
PTS: 1 REF: Ch3.6 TOP: Skills Practice
66. ANS:
A =1
2(6)(4) +
1
2(5)(10)
= 12 + 25
= 37 cm2
PTS: 1 REF: Ch3.6 TOP: Skills Practice
ID: A
12
67. ANS:
A ≈ 8(5) −1
2(3.14)(2.5
2)
= 40 − 9.8125
= 30.19 in.2
PTS: 1 REF: Ch3.6 TOP: Skills Practice
68. ANS:
A =3
4
1
2(32)(27.7)
Ê
Ë
ÁÁÁÁÁÁ
ˆ
¯
˜̃˜̃˜̃
=3
4(443.2)
= 332.4 in.2
PTS: 1 REF: Ch3.6 TOP: Skills Practice
69. ANS:
A ≈ 20(20) − (3.14)(102)
= 400 − 314
= 86 in.2
PTS: 1 REF: Ch3.6 TOP: Skills Practice
70. ANS:
The perimeter is 12 + 12 + 8 + 8 = 40 centimeters.
PTS: 1 REF: Ch3.6 TOP: Mid Ch Test
71. ANS:
To determine the area, you can add the areas of the two rectangles. One rectangle is 11 units by 12 units and
the other rectangle is 10 units by 6 units.
Total area = 11(12) + 10(6) = 132 + 60 = 192
The area of the region bounded by the line segments is 192 square units.
PTS: 1 REF: Ch3.6 TOP: End Ch Test
72. ANS:
The sum of the shorter horizontal segments is 20 yards, and the sum of the shorter vertical segments is 18
yards. So, 20 + 20 + 18 + 18 = 76.
The perimeter is 76 yards.
PTS: 1 REF: Ch3.6 TOP: End Ch Test
ID: A
13
73. ANS:
First, calculate the sum of the areas of the two rectangles:7(6) + 5(9) = 42 + 45 = 89 square feet. Then,
multiply the number of square feet by 144, the number of square inches in one square foot:
89 × 144 = 12,816.
So, 12,816 one-inch square tiles are needed to tile the entire floor.
PTS: 1 REF: Ch3.6 TOP: End Ch Test