geometry

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9Introductionto Geometry

9.1 Some Basic Definitions

9.2 Parallel andPerpendicular Lines

9.3 Polygons

9.4 Properties of Triangles

9.5 Perimeters and Areas ofPolygons

9.6 Circles

9.7 Surface Area andVolume

1

GEOMETRY COMES FROM THEGREEK WORDSGEO

(MEANING EARTH) AND METRON (MEANING MEASURE).

9.1 Some Basic Definitions

In this section, you will learn about• Points, lines, and planes• Angles• Adjacent and vertical angles• Complementary and supplementary angles

INTRODUCTION . In this chapter, we will study two-dimensional geometric figures suchas rectangles and circles. In daily life, it is often necessary to find the perimeter orarea of one of these figures. For example, to find the amount of fencing that is neededto enclose a circular garden, we must find the perimeter of a circle (called itscircum-ference). To find the amount of paint needed to paint a room, we must find the area ofits four rectangular walls.

We will also study three-dimensional figures such as cylinders and spheres. Tofind the amount of space enclosed within these figures, we must find their volumes.

Points, lines, and planesGeometry is based on three undefined words:point, line, and plane. Although wewill make no attempt to define these words formally, we can think of a point as ageometric figure that has position but no length, width, or depth. Points are alwayslabeled with capital letters. PointA is shown in Figure 9-1(a).

(a) (b) (c)

A

BB

AC

E

F

F I G U R E 9-1

2 Chapter 9 Introduction to Geometry

A line is infinitely long but has no width or depth. Figure 9-1(b) shows lineBC,passing through pointsB andC. A plane is a flat surface, like a table top, that haslength and width but no depth. In Figure 9-1(c), lineEF lies in the planeAB.

As Figure 9-1(b) illustrates, pointsB andC determine exactly one line, the lineBC. In Figure 9-1(c), the pointsE andF determine exactly one line, the lineEF. Ingeneral, any two points will determine exactly one line.

Other geometric figures can be created by using parts or combinations of points,lines, and planes.

Line segment The line segmentAB, denoted asAB, is the part of a line that consists of pointsAandB and all points in between (see Figure 9-2). PointsA andB are theendpointsof the segment.

Line segmentAB (AB)

Every line segment has amidpoint, which divides the segment into two parts ofequal length. In Figure 9-3,M is the midpoint of segmentAB, because the measure ofAM (denoted as m(AM)) is equal to the measure ofMB (denoted as m(MB)).

m(AM) 5 42 1

5 3

and

m(MB) 5 72 4

5 3

Since the measure of both segments is 3 units, m(AM) 5 m(MB).When two line segments have the same measure, we say that they arecongru-

ent. Since m(AM) 5 m(MB), we can write

AM 5; MB Read5; as “is congruent to.”

Another geometric figure is theray.

Ray A ray is the part of a line that begins at some point (say,A) and continues foreverin one direction. See Figure 9-4. PointA is theendpoint of the ray.

RayAB (AB→

) RayAB is denoted asAB→

. The endpointis always listed first.

Angles

Angle An angle is a figure formed by two rays with a common endpoint. The commonendpoint is called thevertex, and the rays are calledsides.

A

B

F I G U R E 9-2

MA B

1 2 3 4 5 6 7

3 units 3 units

F I G U R E 9-3

A

B

F I G U R E 9-4

9.1 Some Basic Definitions 3

The angle in Figure 9-5 can be denoted as

∠BAC, ∠CAB, ∠A, or ∠1 The symbol∠ means angle.

WARNING! When using three letters to name an angle, be sure the letter name ofthe vertex is the middle letter.

One unit of measurement of an angle is thedegree.It is 1360 of a full revolution.

We can use aprotractor to measure angles in degrees. See Figure 9-6.

If we read the protractor from left to right, we can see that the measure of∠GBF(denoted as m(∠GBF)) is 307.

When two angles have the same measure, we say that they are congruent. Sincem(∠ABC) 5 307 and m(∠GBF) 5 307, we can write

∠ABC5; ∠GBF

We classify angles according to their measure, as in Figure 9-7.

Classification of angles Acute angles:Angles whose measures are greater than 07 but less than 907.

Right angles:Angles whose measures are 907.

Obtuse angles:Angles whose measures are greater than 907 but less than 1807.

Straight angles:Angles whose measures are 1807.

A

B

C

1

Vertex ofthe angle

Sides ofthe angle

F I G U R E 9-5

Angle Measure in degrees

∠ABC 307∠ABD 607∠ABE 1107∠ABF 1507∠ABG 1807

9010080

80100

70110 60120 50130 40140

30150

2016010170

0180

11070120

60130

5014

040

150

30

160

20

170

1018

0 0

G AB

F

ED

C

F I G U R E 9-6

40°

Acute angle

(a)

90°

Right angle

(b)

130°

Obtuse angle

(c)

180°

Straight angle

(d)

F I G U R E 9-7


EXAMPLE 1 Classifying angles. Classify each angle in Figure9-8 as an acute angle, a right angle, an obtuse angle, or astraight angle.

Solution Since m(∠1), 907, it is an acute angle.

Since m(∠2). 907 but less than 1807,it is an obtuse angle.

Since m(∠BDE) 5 907, it is a right angle.

Since m(∠ABC) 5 1807, it is a straight angle.

Adjacent and vertical anglesTwo angles that have a common vertex and are side-by-side are calledadjacent angles.

EXAMPLE 2 Evaluating angles. Two angles withmeasures ofx7 and 357 are adjacent angles.

Use the information in Figure 9-9 to findx.

SolutionSince the sum of the measures of the angles is 807, we have

x1 355 80

x1 35!355 80!35 Subtract 35 from both sides.

x 5 45 352 355 0 and 802 355 45.

Thus,x5 45.

Self CheckIn the figure below, findx.

125°

180°

x°

Answer: 55

When two lines intersect, pairs of nonadjacent angles are calledvertical angles.In Figure 9-10(a),∠1 and∠3 are vertical angles, as are∠2 and∠4.

To illustrate that vertical angles always have the same measure, we refer to Fig-ure 9-10(b) with angles having measures ofx7, y7, and 307. Since the measure of anystraight angle is 1807, we have

301 x 5 180 and 301 y 5 180

x 5 150 y 5 150 Subtract 30 from both sides.

Sincex andy are both 150,x5 y.

Property of verticalangles

Vertical angles are congruent (have the same measure).

A

B

D

E

C

90°

12

F I G U R E 9-8

80°

35°

x°

F I G U R E 9-9

x°

y°30°4

(a) (b)

2

1

3

l1

l1

l2

l2

Note thatx andyare vertical angles.

F I G U R E 9-10


EXAMPLE 3 Evaluating angles. In Fig-ure 9-11, find a. m(∠1) andb. m(∠3).

Solutiona. The 507 angle and∠1 are vertical angles. Since ver-

tical angles are congruent, m(∠1)5 507.

b. SinceAD is a line, the sum of the measures of∠3,the 1007 angle, and the 507 angle is 1807. Ifm(∠3)5 x, we have

x1 1001 505 180

x1 1505 180 1001 505 150.

x 5 30 Subtract 150 from both sides.

Thus, m(∠3)5 307.

Self CheckIn Figure 9-11, find

a. m(∠2)

b. m(∠4)

Answer: 1007, 307

EXAMPLE 4 Evaluating angles.In Figure 9-12, findx.

SolutionSince the angles are vertical angles, they haveequal measures.

4x2 205 3x1 15

x2 205 15 To eliminate 3x from the right-hand side, subtract 3xfrom both sides.

x 5 35 To undo the subtraction of 20, add 20 to both sides.

Thus,x5 35.

Self CheckIn the figure below, findy.

(2y + 20)°(4y − 10)°

Answer: 15

Complementary and supplementary angles

Complementary andsupplementary angles

Two angles arecomplementary angleswhen the sum of their measures is 907.

Two angles aresupplementary angleswhen the sum of their measures is 1807.

EXAMPLE 5 Complementary and supplementary angles.

a. Angles of 607 and 307 are complementary angles,because the sum of their measures is 907. Eachangle is the complement of the other.

30°

60°

b. Angles of 1307 and 507 are supplementary, becausethe sum of their measures is 1807. Each angle is thesupplement of the other.

130°50°

100°

50°

1

2

4

3

A

D

F I G U R E 9-11

(3x + 15)°

(4x − 20)°

F I G U R E 9-12


c. The definition of supplementary angles requiresthat the sum oftwo angles be 1807. Three anglesof 407, 607, and 807 are not supplementary eventhough their sum is 1807.

40°80°

60°

EXAMPLE 6 Finding the complement andsupplement of an angle.

a. Find the complement of a 357 angle.

b. Find the supplement of a 1057 angle.

Solutiona. See Figure 9-13. Letx represent the complement of the

357 angle. Since the angles are complementary, we have

x1 355 90

x 5 55 Subtract 35 from both sides.

The complement of 357 is 557.

b. See Figure 9-14. Lety represent the supple-ment of the 1057 angle. Since the angles aresupplementary, we have

y1 1055 180

y 5 75 Subtract 105 from both sides.

The supplement of 1057 is 757.

Self Checka. Find the complement of a 507

angle.

b. Find the supplement of a 507angle.

Answer: 407, 1307

STUDY SET Section 9.1

VOCABULARY In Exercises 1–12, fill in the blanks to make a true statement.

1. A line segment has two endpoints. 2. Two points determine at most one line.

3. A midpoint divides a line segment into twoparts of equal length.

4. An angle is measured in degrees .

5. A protractor is used to measure angles. 6. An acute angle is less than 907.

7. A right angle measures 907. 8. An obtuse angle is greater than 907 but lessthan 1807.

9. The measure of a straight angle is 1807 . 10. Adjacent angles have the same vertex and areside-by-side .

11. The sum of two supplementary angles is 1807. 12. The sum of two complementary angles is907 .

35°x°

F I G U R E 9-13

105°y°

F I G U R E 9-14


CONCEPTS In Exercises 13–20, refer to Illustration 1 and tellwhether each statement is true. If a statement is false, explain why.

13. GF→

has pointG as its endpoint. true

14. AG has no endpoints.false

15. Line CD has three endpoints.false

16. PointD is the vertex of∠DGB. false

17. m(∠AGC) 5 m(∠BGD) true

18. ∠AGF 5; ∠BGE true

19. ∠FGB 5; ∠EGA true

20. ∠AGCand∠CGF are adjacent angles.true

In Exercises 21–28, refer to Illustration 1 and tell whether each angle is an acute angle, a right angle, an obtuse angle,or a straight angle.

21. ∠AGC acute 22. ∠EGA right 23. ∠FGD obtuse 24. ∠BGA straight

25. ∠BGE right 26. ∠AGD obtuse 27. ∠DGC straight 28. ∠DGB acute

In Exercises 29–32, refer to Illustration 2 and tell whethereach statement is true. If a statement is false, explain why.

29. ∠AGF and∠DGC are vertical angles.true

30. ∠FGE and∠BGAare vertical angles.false

31. m(∠AGB) 5 m(∠BGC). false

32. ∠AGC5; ∠DGF. true

In Exercises 33–38, refer to Illustration 3 and tell whethereach pair of angles are congruent.

33. ∠1 and∠2 yes

34. ∠FGB and∠CGE yes

35. ∠AGBand∠DGE yes

36. ∠CGD and∠CGB no

37. ∠AGF and∠FGE no

38. ∠AGBand∠BGD yes

In Exercises 39–46, refer to Illustration 3 and tell whether each statement is true.

39. ∠1 and∠CGD are adjacent angles.true 40. ∠2 and∠1 are adjacent angles.false

41. ∠FGA and∠AGCare supplementary.true 42. ∠AGBand∠BGCare complementary.false

43. ∠AGF and∠2 are complementary.true 44. ∠AGBand∠EGD are supplementary.true

45. ∠EGD and∠DGB are supplementary.true 46. ∠DGC and∠AGF are complementary.false

NOTATION In Exercises 47–50, fill in the blanks to make a true statement.

47. The symbol∠ means angle . 48. The symbolAB is read as “ line segment AB.”

49. The symbolAB→

is read as “ ray AB.” 50. The symbol 5; is read as “is congruentto.”

60°

30°60°

30°

90° 90°

D

B

F

C

A

E

G

I L L U S T R A T I O N 1

D

B

F C

A

E

G


90°

F

E

D

C

B

A

G

1

2



PRACTICE In Exercises 51–56, refer to Illustration 4 and find the length of each segment.

51. AC 3 52. BE 4

53. CE 3 54. BD 2

55. CD 1 56. DE 2

In Exercises 57–58, refer to Illustration 4 and find each midpoint.

57. Find the midpoint ofAD. B 58. Find the midpoint ofBE. D

In Exercises 59–62, use a protractor to measure each angle.

59.

407

60.

1057

61.

1357

62. 757

In Exercises 63–70, find x.

63.

45°

55°

x°

10 64.

45°

180°

x°

135

65.

50°

22.5°x°

27.5

66.130°

x°

40°

90

67.

(x + 30)°(2x)°

30

68. (6x − 5)°

(2x + 35)°

10

69. (4x + 15)°

(7x − 60)°

25

70.

(6x + 8)° (4x + 32)°

12

In Exercises 71–74, let x represent the unknown angle measure. Draw a diagram, write an appropriate equation, andsolve it for x.

71. Find the complement of a 307 angle. 607 72. Find the supplement of a 307 angle. 1507

73. Find the supplement of a 1057 angle. 757 74. Find the complement of a 757 angle. 157

In Exercises 75–78, refer to Illustration 5, in which m(∠1)$ 507. Find the measure of each angle or sum of angles.

75. ∠4 1307

76. ∠3 507

77. m(∠1)1 m(∠2)1 m(∠3) 2307

78. m(∠2)1 m(∠4) 2607

DB CA E

2 3 4 5 6 7 8 9


50° = 12

43



In Exercises 79–82, refer to Illustration 6, in which m(∠1)~m(∠3)~m(∠4)$ 1807, ∠3$; ∠4, and ∠4$; ∠5. Findthe measure of each angle.

79. ∠1 1007 80. ∠2 807

81. ∠3 407 82. ∠6 1407

APPLICATIONS

83. Cite five examples in real life where you see lines.

84. Cite two examples in real life where you see rightangles.

85. Cite two examples in real life where you see acuteangles.

86. Cite two examples in real life where you see obtuseangles.

87. MUSICAL INSTRUMENTS Suppose that you are abeginning band teacher describing the correct postureneeded to play various instruments. Use the diagramsin Illustration 7 to approximate the angle measure atwhich each instrument should be held in relation to thestudent’s body: a. flute 807 b. clarinet 307c. trumpet 657

88. PHRASES Explain what you think each of thesephrases means. How is geometry involved?a. The president did a complete 180-degree flip on the

subject of a tax cut.b. The rollerblader did a “360” as she jumped off the

ramp.

WRITING Write a paragraph using your own words.

89. Explain why an angle measuring 1057 cannot have acomplement.

90. Explain why an angle measuring 2107 cannot have asupplement.

REVIEW In Exercises 91–98, do the calculations.

91. Find 24. 16 92. Add:1

21

2

31

3

42312 or 1

1112

93. Subtract:3

42

1

82

1

3724 94. Multiply:

5

8?2

15?6

5110

95. Divide:12

174

4

346 96. 31 2 ? 4 11

97. 5 ? 31 4 ? 2 23 98. Find 30% of 60. 18

100°

21

6

5

43


a.

b.

c.



9.2 Parallel and Perpendicular Lines

In this section, you will learn about• Parallel and perpendicular lines• Transversals and angles• Properties of parallel lines

INTRODUCTION . In this section, we will considerparallel and perpendicular lines.Since parallel lines are always the same distance apart, the railroad tracks shown inFigure 9-15(a) illustrate one application of parallel lines.

Since perpendicular lines meet and form right angles, the monument and theground shown in Figure 9-15(b) illustrate one application of perpendicular lines.

Parallel and perpendicular linesIf two lines lie in the same plane, they are calledcoplanar. Two coplanar lines thatdo not intersect are calledparallel lines. See Figure 9-16(a).

Parallel lines Parallel lines are coplanar lines that do not intersect.

If lines l1 (l sub 1) andl2 (l sub 2) are parallel, we can writel1 ?? l2, where thesymbol ?? is read as “is parallel to.”

Perpendicular lines Perpendicular linesare lines that intersect and form right angles.

In Figure 9-16(b),l1 ⊥ l2, where the symbol⊥ is read as “is perpendicular to.”

(a) (b)

This symbol indicatesa right angle

F I G U R E 9-15

(a) (b)

l1 l1 l2l2

Parallel lines Perpendicular lines

F I G U R E 9-16

9.2 Parallel and Perpendicular Lines 11

Transversals and anglesA line that intersects two or more coplanar lines is called atransversal.For example,line l1 in Figure 9-17 is a transversal intersecting linesl2, l3, and l4.

When two lines are cut by a transversal, the following types of angles are formed.

Alternate interior angles:

∠4 and∠5∠3 and∠6

l3

l1

l2 1 2

3 4

5 6

7 8

Corresponding angles:

∠1 and∠5∠3 and∠7∠2 and∠6∠4 and∠8

l3

l1

l2 1 23 4

5 67 8

Interior angles:

∠3, ∠4, ∠5, and∠6

l3

l1

l2 1 2

3 4

5 6

7 8

EXAMPLE 1 Identifying angles. In Figure 9-18, identify a. all pairs of alternate interiorangles, b. all pairs of corresponding angles, andc. all interior angles.

Solution a. Pairs of alternate interior angles are

∠3 and∠5, ∠4 and∠6

b. Pairs of corresponding angles are

∠1 and∠5, ∠4 and∠8, ∠2 and∠6, ∠3 and∠7

c. Interior angles are

∠3, ∠4, ∠5, and∠6

Properties of parallel lines1. If two parallel lines are cut by a transversal, al-

ternate interior angles are congruent. (See Figure9-19.) If l1 ?? l2, then∠2 5; ∠4 and∠1 5; ∠3.

l1

l2

l3

l4

F I G U R E 9-17

2 1

3 4

65

7 8

F I G U R E 9-18

1 2

4 3

l3

l2

l1

F I G U R E 9-19


2. If two parallel lines are cut by a transversal, corresponding angles are congruent.(See Figure 9-20.) Ifl1 ?? l2, then∠1 5; ∠5, ∠3 5; ∠7, ∠2 5; ∠6, and∠4 5; ∠8.

3. If two parallel lines are cut by a transversal, interior angles on the same side of thetransversal are supplementary. (See Figure 9-21.) Ifl1 ?? l2, then∠1 is supplemen-tary to∠2 and∠4 is supplementary to∠3.

4. If a transversal is perpendicular to one of two parallel lines, it is also perpendicularto the other line. (See Figure 9-22.) Ifl1 ?? l2 and l3 ⊥ l1, then l3 ⊥ l2.

5. If two lines are parallel to a third line, they are parallel to each other. (See Figure9-23.) If l1 ?? l2 and l1 ?? l3, then l2 ?? l3.

EXAMPLE 2 Evaluating angles.See Figure 9-24. Ifl1 ?? l2

and m(∠3)5 1207, find the measures of theother angles.

Solutionm(∠1)5 607 ∠3 and∠1 are supplementary.

m(∠2)5 1207 Vertical angles are congruent:m(∠2)5 m(∠3).

m(∠4)5 607 Vertical angles are congruent:m(∠4)5 m(∠1).

m(∠5)5 607 If two parallel lines are cut by a transversal, alternate interior anglesare congruent: m(∠5)5 m(∠4).

m(∠6)5 1207 If two parallel lines are cut by a transversal, alternate interior anglesare congruent: m(∠6)5 m(∠3).

m(∠7)5 1207 Vertical angles are congruent: m(∠7)5 m(∠6).

m(∠8)5 607 Vertical angles are congruent: m(∠8)5 m(∠5).

Self CheckIf l1 ?? l2 and m(∠8)5 507, findthe measures of the other angles.(See Figure 9-24.)

Answer: m(∠5)5 507,m(∠7)5 1307, m(∠6)5 1307,m(∠3)5 1307, m(∠4)5 507,m(∠1)5 507, m(∠2)5 1307

EXAMPLE 3 Identifying congruent angles. See Figure9-25. If AB ?? DE, which pairs of angles are con-gruent?

Solution SinceAB ?? DE, corresponding angles are congru-ent. So we have

∠A 5; ∠1 and ∠B 5; ∠2

5 67 8

3 4

1 2

l3

l2

l1

F I G U R E 9-24

3 4

1 2

5 6

7 8

l3

l2

l1

F I G U R E 9-20

1 32 4

l3

l2

l1

F I G U R E 9-21

l3

l2

l1

F I G U R E 9-22

l2

l3

l1

F I G U R E 9-23

43

21

C

D E

A B

F I G U R E 9-25


EXAMPLE 4 Using algebra in ge-ometry. In Figure 9-26,l1 ?? l2. Find x.

SolutionThe angles involving x are correspondingangles. Sincel1 ?? l2, all pairs of correspondingangles are congruent.

9x2 155 6x1 30 The angle measures are equal.

3x2 155 30 Subtract 6x from both sides.

3x 5 45 To undo the subtraction of 15, add 15 to both sides.

x 5 15 To undo the multiplication by 3, divide both sides by 3.

Thus,x5 15.

Self CheckIn the figure below,l1 ?? l2. Find y.

(7y − 14)°(4y + 10)°

l2

l1

Answer: 8

EXAMPLE 5 Using algebra in geometry. In Figure 9-27,l1 ?? l2. Find x.

Solution Since the angles are interior angles on the sameside of the transversal, they are supplementary.

3x2 801 3x1 205 180 The sum of themeasures of twosupplementary angles is 1807.

6x2 605 180 Combine like terms.

6x 5 240 To undo the subtraction of 60, add 60 to both sides.

x 5 40 To undo the multiplication by 6, divide both sides by 6.

Thus,x5 40.



1. Two lines in the same plane are coplanar . 2. Parallel lines do not intersect.

3. If two lines intersect and form right angles, they areperpendicular .

4. A transversal intersects two or more coplanar lines.

5. In Illustration 1, ∠4 and ∠6 are alternate interiorangles.

6. In Illustration 1, ∠2 and ∠6 are correspondingangles.

CONCEPTS In Exercises 7–12, answer each question.

7. Which pairs of angles shown in Illustration 1 are alter-nate interior angles?∠4 and∠6, ∠3 and∠5

8. Which pairs of angles shown in Illustration 1 are corre-sponding angles?∠1 and∠5, ∠4 and∠8, ∠2 and∠6,∠3 and∠7

(9x − 15)°

(6x + 30)°l2

l1

F I G U R E 9-26

(3x + 20)°(3x − 80)°

l2

l1

F I G U R E 9-27

1 4

2 3

5 8

6 7



9. Which angles shown in Illustration 1 are interiorangles? ∠3, ∠4, ∠5, ∠6

10. In Illustration 2, l1 ?? l2. What can you conclude aboutl1 and l3? They are perpendicular.

11. In Illustration 3, l1 ?? l2 and l2 ?? l3. What can you con-clude aboutl1 and l3? They are parallel.

12. In Illustration 4,AB ?? DE. What pairs of angles are con-gruent? ∠A 5; ∠E, ∠D 5; ∠B, ∠1 5; ∠2


13. The symbol indicates a right angle . 14. The symbol?? is read as is parallel to .

15. The symbol⊥ is read asis perpendicular to. 16. The symboll1 is read as ell sub one .

PRACTICE In Exercises 17–20, find the measures of the missing angles.

17. In Illustration 5, l1 ?? l2 and m(∠4)5 1307. Find themeasures of the other angles.m(∠1)5 1307,m(∠2)5 507, m(∠3)5 507, m(∠5)5 1307, m(∠6)5 507,m(∠7)5 507, m(∠8)5 1307

18. In Illustration 6,l1 ?? l2 and m(∠2)5 407. Find the mea-sures of the other angles.m(∠1)5 1407,m(∠3)5 1407, m(∠4)5 407, m(∠5)5 407, m(∠6)5 1407,m(∠7)5 407, m(∠8)5 1407

19. In Illustration 7, l1 ?? AB. Find the measure of eachangle. m(∠A) 5 507, m(∠1)5 857, m(∠2)5 457,m(∠3)5 1357

20. In Illustration 8,AB ?? DE. Find m(∠B), m(∠E), andm(∠1). m(∠B) 5 607, m(∠E) 5 907, m(∠1)5 307

In Exercises 21–24, l1 ?? l2. Find x.

21.

(5x)°(6x − 10)°

l2

l1

10 22.

(4x − 8)°

(2x + 16)°

l1 l212

l3

l1 l2


l1

l2

l3


A

B

C

D

E1

2


56

7 8

34

1 2

l2

l1


3 7

84

1 2

56

l2

l1


l150° 1

AB

C2

345°


AC

1

30°

60°

E

D

B



23.

(2x + 10)°

(4x − 10)°

l2

l1

30 24.

(5x + 5)°

(2x + 80)°l2

l1

25


25. l1 ?? CA40

l1x°

C A

B

(3x + 20)°

26. AB ?? DE22

(5x − 40)°

(3x + 4)°

C

D E

A B

27. AB ?? DE12

A

C

(6x − 2)°

(9x − 38)°

E

DB

28. AC ?? BD7

(7x − 2)°(2x + 33)°

A B

DC

APPLICATIONS

29. BUILDING CONSTRUCTION List five exampleswhere you would see parallel lines in building construc-tion.

30. PLUMB LINES What is a plumb line? What geomet-ric principle does it illustrate?

31. BUILDING CONSTRUCTION List five exampleswhere you would see perpendicular lines in buildingconstruction.

32. HANGING WALLPAPER Explain why the conceptsof perpendicular and parallel are both important whenhanging wallpaper.

33. TOOLS See Illustration 9. What geometric conceptsdo the tools show?a. Scissors intersecting lines, vertical anglesb. Rake parallel and perpendicular lines

34. PARKING DESIGN Using terms from this chapter,write a paragraph describing the parking layout shownin Illustration 10.

North side of street

South side of street

Planter





35. Why do you think that∠4 and∠6 shown in Illustra-tion 1 are called alternate interior angles?

36. Why do you think that∠4 and∠8 shown in Illustra-tion 1 are called corresponding angles?

37. Are pairs of alternate interior angles always congruent?Explain.

38. Are pairs of interior angles always supplementary? Ex-plain.

REVIEW

39. Find 60% of 120. 72 40. 80% of what number is 400?500

41. What percent of 500 is 225?45% 42. Simplify: 3.451 7.37? 2.98 25.4126

43. Is every whole number an integer?yes 44. Multiply: 21

5? 4

3

792635

9.3 PolygonsIn this section, you will learn about• Polygons• Triangles• Properties of isosceles triangles• The sum of the measures of the angles of a triangle• Quadrilaterals• Properties of rectangles• The sum of the measures of the angles of a polygon

INTRODUCTION . In this section, we will discuss figures calledpolygons.We see theseshapes every day. For example, the walls in most buildings are rectangular in shape.We also see rectangular shapes in doors, windows, and sheets of paper.

The gable ends of many houses are triangular in shape, as are the sides of theGreat Pyramid in Egypt. Triangular shapes are especially important because trianglesare rigid and contribute strength and stability to walls and towers.

The designs used in tile or linoleum floors often use the shapes of a pentagon ora hexagon. Stop signs are always in the shape of an octagon.

Polygons

Polygon A polygon is a closed geometric figure with at least three line segments for its sides.

The figures in Figure 9-28 arepolygons.They are classified according to the num-ber of sides they have. The points where the sides intersect are calledvertices.

Triangle(3 sides)

Quadrilateral(4 sides)

Pentagon(5 sides)

Hexagon(6 sides)

Octagon(8 sides)

F I G U R E 9-28

9.3 Polygons 17

EXAMPLE 1 Vertices of a polygon. Give the number of vertices ofa. a triangle and b. a hexagon.

Solutiona. From Figure 9-28, we see that a triangle has three angles and therefore three ver-

tices.

b. From Figure 9-28, we see that a hexagon has six angles and therefore six vertices.

Self CheckGive the number of vertices of

a. a quadrilateral

b. a pentagon

Answer: 4, 5

From the results of Example 1, we see that the number of vertices of a polygon isequal to the number of its sides.

TrianglesA triangle is a polygon with three sides. Figure 9-29 illustrates some common tri-angles. The slashes on the sides of a triangle indicate which sides are of equal length.

WARNING! Since equilateral triangles have at least two sides of equal length, theyare also isosceles. However, isosceles triangles are not necessarily equilateral.

Since every angle of an equilateral triangle has the same measure, an equilateraltriangle is alsoequiangular.

In an isosceles triangle, the angles opposite the sides of equal length are calledbase angles,the sides of equal length form thevertex angle,and the third side iscalled thebase.

The longest side of a right triangle is called thehypotenuse,and the other twosides are calledlegs.The hypotenuse of a right triangle is always opposite the 907angle

Properties of isosceles triangles

1. Base angles of an isosceles triangle are congruent.

2. If two angles in a triangle are congruent, the sides opposite the angles have thesame length, and the triangle is isosceles.

EXAMPLE 2 Determining whether a triangle is isosceles.Is the triangle in Figure 9-30an isosceles triangle?

Solution∠A and∠B are angles of the triangle. Since m(∠A) 5 m(∠B),we know that m(AC) 5 m(BC) and thatnABC(read as “triangleABC”) is isosceles.

Self CheckIn the figure below,l1 ?? AB. Is thetriangle an isosceles triangle?

l150°60°

A B

C

Answer: no

Equilateral triangle(all sides equal length)

Isosceles triangle(at least two sides of

equal length)

Scalene triangle(no sides equal length)

Right triangle(has a right angle)

90° Leg

LegBase angles

Vertex angle

Hypotenuse

F I G U R E 9-29

50°BA

C

50°

F I G U R E 9-30


The sum of the measures of the anglesof a triangleIf you draw several triangles and carefully measure each angle with a protractor, youwill find that the sum of the angle measures in each triangle is 1807.

Angles of a triangle The sum of the angle measures of any triangle is 1807.

EXAMPLE 3 Sum of the angles of a triangle.See Figure 9-31. Findx.

SolutionSince the sum of the angle measures of any triangle is 1807,we have

x1 401 905 180

x1 1305 180 401 905 130.

x 5 50 To undo the addition of 130, subtract 130 from both sides.

Thus,x5 50.

Self CheckIn the figure below, findy.

60°

30°

y°

Answer: 90

EXAMPLE 4 Vertex angle of an isosceles triangle.See Figure 9-32.If one base angle of an isosceles triangle measures 707, howlarge is the vertex angle?

Solution Since one of the base angles measures 707, so does the other.If we let x represent the measure of the vertex angle, we have

x1 701 705 180 The sum of the measures of the anglesof a triangle is 1807.

x1 1405 180 701 705 140.


The vertex angle measures 407.

QuadrilateralsA quadrilateral is a polygon with four sides. Some common quadrilaterals are shownin Figure 9-33.

Properties of rectangles1. All angles of a rectangle are right angles.

2. Opposite sides of a rectangle are parallel.

3. Opposite sides of a rectangle are of equal length.

40°

x°

F I G U R E 9-31

70°

F I G U R E 9-32

Parallelogram(Opposite sides

parallel)

Rectangle(Parallelogram with

four right angles)

Square(Rectangle with

sides of equal length)

Rhombus(Parallelogram with

sides of equal length)

Trapezoid(Exactly twosides parallel)

F I G U R E 9-33

9.3 Polygons 19

4. The diagonals of a rectangle are of equal length.

5. If the diagonals of a parallelogram are of equal length, the parallelogram is a rect-angle.

EXAMPLE 5 Squaring a foundation. A carpenter intends tobuild a shed with an 8-by-12-foot base. How can hemake sure that the rectangular foundation is “square”?

Solution See Figure 9-34. The carpenter can use a tape mea-sure to find the lengths of diagonalsAC andBD. Ifthese diagonals are of equal length, the figure will bea rectangle and have four right angles. Then the foun-dation will be “square.”

EXAMPLE 6 Properties of rectangles and tri-angles. In rectangleABCD (Figure

9-35), the length ofAC is 20 centimeters. Find each mea-sure: a. m(BD), b. m(∠1), and c. m(∠2).

Solutiona. Since the diagonals of a rectangle are of equal length,

m(BD) is also 20 centimeters.

b. We let m(∠1)5 x. Then, since the angles of a rectangle are right angles, we have

x1 305 90


Thus, m(∠1)5 607.

c. We let m(∠2)5 y. Then, since the sum of the angle measures of a triangle is 1807,we have

301 301 y 5 180

601 y 5 180 301 305 60.

y 5 120 To undo the addition of 60, subtract 60 from both sides.

Thus, m(∠2)5 1207.

Self CheckIn rectangleABCD, the length ofDC is 16 centimeters. Find eachmeasure:

a. m(AB)

b. m(∠3)

c. m(∠4)

Answer: 16 cm, 1207, 607

The parallel sides of a trapezoid are calledbases,the nonparallel sides are calledlegs,and the angles on either side of a base are calledbase angles.If the nonparallelsides are the same length, the trapezoid is anisosceles trapezoid.In an isosceles trap-ezoid, the base angles are congruent.

EXAMPLE 7 Cross section of a drainage ditch.A crosssection of a drainage ditch (Figure 9-36) is an isos-celes trapezoid withAB ?? CD. Find x andy.

Solution Since the figure is an isosceles trapezoid, its non-parallel sides have the same length. So m(AD) andm(BC) are equal, andx5 8.

Since the base angles of an isoscelestrapezoid are congruent, m(∠D) 5 m(∠C). Soy5 120.

30° 30°1

4

2

3

A B

D C

F I G U R E 9-35

12 ft

12 ft

8 ft 8 ft

A B

D C

F I G U R E 9-34

120°8 ft x ft

y°D C

A B

F I G U R E 9-36


The sum of the measures of the anglesof a polygonWe have seen that the sum of the angle measures of any triangle is 1807. Since apolygon withn sides can be divided inton2 2 triangles, the sum of the angle mea-sures of the polygon is (n2 2)1807.

Angles of a polygon The sumSof the measures of the angles of a polygon withn sides is given by theformula

S5 (n2 2)1807

EXAMPLE 8 Sum of the angles of a pentagon.Find the sum of theangle measures of a pentagon.

SolutionSince a pentagon has 5 sides, we substitute 5 forn in the formula and simplify.

S5 (n 2 2)1807

S5 (52 2)1807

5 3(1807)

5 5407

The sum of the angles of a pentagon is 5407.

Self CheckFind the sum of the angle mea-sures of a quadrilateral.

Answer: 3607



1. A polygon with four sides is called a quadrilateral . 2. A triangle is a polygon with three sides.

3. A hexagon is a polygon with six sides. 4. A polygon with five sides is called a pentagon .

5. An eight-sided polygon is an octagon . 6. The points where the sides of a polygon intersect arecalled vertices .

7. A triangle with three sides of equal length is called anequilateral triangle.

8. An isosceles triangle has two sides of equallength.

9. The longest side of a right triangle is thehypotenuse .

10. The base angles of an isoscelestriangle have the same measure.

11. A parallelogram with a right angle is a rectangle. 12. A rectangle with all sides of equal length is asquare .

13. A rhombus is a parallelogram with four sidesof equal length.

14. A trapezoid has two sides that are parallel andtwo sides that are not parallel.

15. The legs of an isosceles trapezoid have thesame length.

16. The perimeter of a polygon is the distancearound it.

9.3 Polygons 21

CONCEPTS In Exercises 17–24, give the number of sides of each polygon and classify it as a triangle, quadrilateral,pentagon, hexagon, or octagon. Then give the number of vertices.

17.

4, quadrilateral, 4

18.

6, hexagon, 6

19.

3, triangle, 3

20.

8, octagon, 8

21.

5, pentagon, 5

22.

4, quadrilateral, 4

23.

6, hexagon, 6

24.

3, triangle, 3

In Exercises 25–32, classify each triangle as an equilateral triangle, an isosceles triangle, a scalene triangle, or a righttriangle.

25.

scalene triangle

26.

55°

55°

isoscelestriangle

27. righttriangle

28.

equilateral triangle

29. 60° 60°

60°

equilateral triangle

30.

60°

30°

right triangle

31.

20 cm 20 cm

isosceles triangle

32.

60°70°

50°

scalenetriangle

In Exercises 33–40, classify each quadrilateral as a rectangle, a square, a rhombus, or a trapezoid.

33.

4 in. 4 in.

4 in.

4 in.

square 34.

trapezoid

35.

rhombus

36. 90°

rectangle

37.

90°

rectangle

38. 8cm

8cm

8cm8cm

rhombus

39.

trapezoid

40.

square


41. The symboln means triangle . 42. The symbol m(∠1) means the measure ofangle 1 .

PRACTICE In Exercises 43–48, the measures of two angles of nABC (shown in Illustration 1) are given. Find themeasure of the third angle.

43. m(∠A) 5 307 and m(∠B) 5 607.m(∠C) 5 907 .

44. m(∠A) 5 457 and m(∠C) 5 1057.m(∠B) 5 307 .

45. m(∠B) 5 1007 and m(∠A) 5 357.m(∠C) 5 457 .

46. m(∠B) 5 337 and m(∠C) 5 777.m(∠A) 5 707 .


47. m(∠A) 5 25.57 and m(∠B) 5 63.87.m(∠C) 5 90.77 .

48. m(∠B) 5 67.257 and m(∠C) 5 72.57.m(∠A) 5 40.257 .

In Exercises 49–52, refer to rectangle ABCD, shown in Illustration 2.

49. m(∠1)5 307 . 50. m(∠3)5 307 .

51. m(∠2)5 607 . 52. If m(AC) is 8 cm, then m(BD) 5 8 cm .

In Exercises 53–56, find the sum of the angle measures of each polygon.

53. A hexagon 7207 54. An octagon 1,0807

55. A decagon (10 sides)1,4407 56. A dodecagon (12 sides)1,8007

In Exercises 57–60, find the number of sides of the polygon with the given angle measure sum.

57. 9007 7-sided polygon 58. 1,2607 9-sided polygon

59. 2,1607 14-sided polygon 60. 3,6007 22-sided polygon

APPLICATIONS

61. Give three uses of triangles in everyday life. 62. Give three uses of rectangles in everyday life.

63. Give three uses of squares in everyday life. 64. Give a use of a trapezoid in everyday life.


65. Explain why a square is a rectangle. 66. Explain why a trapezoid is not a parallelogram.

REVIEW

67. Find 20% of 110. 22 68. Find 15% of 50. 7.5

69. Find 20% of6

11. 6

55 70. Find 30% of3

5. 0.18

71. What percent of 200 is 80?40% 72. What percent of 500 is 100?20%

73. 20% of what number is 500?2,500 74. 30% of what number is 21?70

75. Simplify: 0.854 2(0.25). 0.10625 76. Simplify: 3.251 124 0.4 ? 2. 63.25

9.4 Properties of TrianglesIn this section, you will learn about• Congruent triangles• Similar triangles• The Pythagorean theorem

INTRODUCTION . We can often use proportions and triangles to measure distances in-directly. For example, by using a proportion, Eratosthenes (275–195B.C.) was able to

A

C

B


60°1

23

A B

D C


9.4 Properties of Triangles 23

estimate the circumference of the earth to a remarkable degree of accuracy. On a sunnyday, we can use properties of similar triangles to calculate the height of a tree whilestaying safely on the ground. By using a theorem proved by the Greek mathematicianPythagoras (about 500B.C.), we can calculate the length of the third side of a righttriangle whenever we know the lengths of two sides.

Congruent trianglesTriangles that have the same size and the same shape are calledcongruent triangles.In Figure 9-37, trianglesABCandDEF are congruent:

nABC5; nDEF Read as “TriangleABC is congruent to triangleDEF.”

Corresponding angles and corresponding sides of congruent triangles are calledcor-responding parts. The notationnABC5; nDEF shows which vertices are corre-sponding parts.

Ä ÄÄ ÄÄ Ä

nABC5; nDEF

Corresponding parts of congruent triangles always have the same measure. In thecongruent triangles shown in Figure 9-37,

m(∠A) 5 m(∠D), m(∠B) 5 m(∠E), m(∠C) 5 m(∠F ),m(BC) 5 m(EF), m(AC) 5 m(DF), m(AB) 5 m(DE)

EXAMPLE 1 Corresponding parts of congruenttriangles. Name the corresponding partsof the congruent triangles in Figure 9-38.

Solution The corresponding angles are

∠A and∠E, ∠B and∠D, ∠C and∠F

Since corresponding sides are always opposite corresponding angles, the correspond-ing sides are

BC andDF, AC andEF, AB andED

We will discuss three ways of showing that two triangles are congruent.

SSS property If three sides of one triangle are congruent to three sides of a second triangle, thetriangles are congruent.

The triangles in Figure 9-39 are congru-ent because of the SSS property.

A

C

B D

F

E

F I G U R E 9-37

A

C F

B D E

F I G U R E 9-38

3 4

5

3 4

5

F I G U R E 9-39


SAS property If two sides and the angle between them in one triangle are congruent, respectively,to two sides and the angle between them in a second triangle, the triangles are con-gruent.

The triangles in Figure 9-40 are congru-ent because of the SAS property.

ASA property If two angles and the side between them in one triangle are congruent, respectively,to two angles and the side between them in a second triangle, the triangles are con-gruent.

The triangles in Figure 9-41 are congru-ent because of the ASA property.

EXAMPLE 2 Determining whether triangles are con-gruent. Explain why the triangles in Figure 9-42are congruent.

Solution Since vertical angles are congruent,

m(∠1)5 m(∠2)

From the figure, we see that

m(AC) 5 m(EC) and m(BC) 5 m(DC)

Since two sides and the angle between them in onetriangle are congruent, respectively, to two sides andthe angle between them in a second triangle, the tri-angles are congruent by the SAS property.

Similar trianglesIf two angles of one triangle are congruent to two angles of a second triangle, thetriangles will have the same shape. Triangles with the same shape are calledsimilartriangles. In Figure 9-43,nABC; nDEF (read the symbol; as “is similar to”).

WARNING! Note that congruent triangles are always similar, but similar trianglesare not always congruent.

3 4 3 490° 90°

F I G U R E 9-40

3 390°60° 60°

90°

F I G U R E 9-41

A

10 cm

12

10 cm

5 cm

5 cmE

B

C

D

F I G U R E 9-42

A30° 70°

B

C

D30° 70°

E

F

F I G U R E 9-43


Property of similartriangles

If two triangles are similar, all pairs of corresponding sides are in proportion.

In the similar triangles shown in Figure 9-43, the following proportions are true.

AB

DE5

BC

EF,

BC

EF5

CA

FD, and

CA

FD5

AB

DE

EXAMPLE 3 Finding the height of a tree. A tree casts a shadow 18 feet long at the sametime as a woman 5 feet tall casts a shadow that is 1.5 feet long. (See Figure 9-44.)Find the height of the tree.

Solution The figure shows the triangles determined by the tree and its shadow and the womanand her shadow. Since the triangles have the same shape, they are similar, and thelengths of their corresponding sides are in proportion. If we leth represent the heightof the tree, we can findh by solving the following proportion.

h

55

18

1.5Height of the tree

Height of the woman5

shadow of the tree

shadow of the woman.

1.5h 5 5(18) In a proportion, the product of the extremes is equal to the product ofthe means.

1.5h 5 90 Do the multiplication: 5(18)5 90.

h 5 60 To undo the multiplication by 1.5, divide both sides by 1.5 and simplify.

The tree is 60 feet tall.

The Pythagorean theoremIn the movieThe Wizard of Oz,the scarecrow was in search of a brain. To prove thathe had found one, he recited the Pythagorean theorem.

In a right triangle, the square of the hypotenuse is equal to the sum of squaresof the other two sides.

Pythagorean theorem If the length of the hypotenuse of a right triangle iscand the lengths of its legs area andb (as in Figure9-45), then

a2 1 b2 5 c2

18 ft1.5 ft

5 ft

h

F I G U R E 9-44

a

bc

Leg

LegHypotenuse

F I G U R E 9-45


EXAMPLE 4 Constructing a high-ropes adventure course.Abuilderof a high-ropes adventure course wants to secure the pole shown

in Figure 9-46 by attaching a cable from the ground anchor 20 feet from its base to apoint 15 feet up the pole. How long should the cable be?

SolutionWe can use the Pythagorean theorem witha5 20andb5 15.

c2 5 a2 1 b2

c2 5 202 1 152 Substitute 20 fora and 15for b.

c2 5 4001 225

c2 5 625=c2 5 =625 Since equal positive num-

bers have equal squareroots, take the positivesquare root of both sides.

c 5 25 =c2 5 c and=6255 25.

The cable will be 25 feet long.

Self CheckA 26-foot ladder rests against theside of a building. If the base ofthe ladder is 10 feet from the wall,how far up the side of the buildingwill the ladder reach?

Answer: 24 ft

Accent on Technology Finding the width of a television screen

The size of a television screen is the diago-nal measure of its rectangular screen. (SeeFigure 9-47.) To find the width of a 27-inch screen that is 17 inches high, we usethe Pythagorean theorem withc5 27 andb5 17.

c2 5 a2 1 b2

272 5 a2 1 172

272 2 172 5 a2

=272 2 172 5 =a2 Take the positivesquare root ofboth sides.

=272 2 172 5 a =a2 5 a.

Evaluate: =272 2 172

Keystrokes: ( 2 7 x2 2 1 7 x2 ) = 20.97617696

To the nearest inch, the width is 21 inches.

It is also true that

If the square of one side of a triangle is equal to the sum of the squares of theother two sides, the triangle is a right triangle.

EXAMPLE 5 Determining whether a triangle is a right triangle.Determine whether a triangle with sides of 5, 12, and 13 metersis a right triangle.

Self CheckDecide whether a triangle withsides of 9, 40, and 41 meters is aright triangle.

b = 15 ft

a = 20 ft

c ft

F I G U R E 9-46

17 in.

a in.

27 in.

F I G U R E 9-47


SolutionWe check to see whether the square of one side is equal to the sum of the squares ofthe other two sides.

132 5? 52 1 122 The longest side is the hypotenuse.

1695? 251 144

1695 169

Since 132 5 52 1 122, the triangle is a right triangle. Answer: yes



1. Congruent triangles are the same size and thesame shape.

2. All corresponding parts of congruent triangles havethe same measure.

3. If two triangles are similar , they have thesame shape.

4. The hypotenuse is the longest side of a right tri-angle.

CONCEPTS In Exercises 5–8, tell whether each statement is true. If a statement is false, tell why.

5. If three sides of one triangle are the same length as threesides of a second triangle, the triangles are congruent.true

6. If two sides of one triangle are the same length as twosides of a second triangle, the triangles are congruent.false

7. If two sides and an angle of one triangle are congruent,respectively, to two sides and an angle of a second tri-angle, the triangles are congruent.false

8. If two angles and the side between them in one triangleare congruent, respectively, to two angles and the sidebetween them in a second triangle, the triangles are con-gruent. true

9. Are the triangles shown in Illustration 1 congruent?yes

10. Are the triangles shown in Illustration 2 congruent?not necessarily

11. In a proportion, the product of the means isequal to the product of the extremes .

12. If two angles of one triangle are congruent totwo angles of a second triangle, the triangles are

similar .

4 cm

4 cm

8 cm8 cm

A

D

B

E

C

F


AC D

E

B

60°

60°



13. Are the triangles shownin Illustration 3 similar?yes

14. Are the triangles shown in Illustration 4 similar?yes

15. If x and y represent the lengths of two legs of a righttriangle andz represents the length of the hypotenuse,the Pythagorean theorem states thatz2 5 x2 1 y2. true

16. A triangle with sides of 3, 4, and 5 centimeters is a righttriangle. true


17. The symbol5; is read as “ is congruent to .” 18. The symbol; is read as “ is similar to .”

PRACTICE In Exercises 19–20, name the corresponding parts of the congruent triangles.

19. Refer to Illustration 5. (The slashes indicate pairs ofcongruent sides.)

AC corresponds toDF .

DE corresponds toAB .

BC corresponds toEF .

∠A corresponds to∠D.

∠E corresponds to∠B.

∠F corresponds to∠C.

20. Refer to Illustration 6.

AB corresponds toDE .

EC corresponds toBC.

AC corresponds toDC.

∠D corresponds to∠A.

∠B corresponds to∠E.

∠1 corresponds to∠2 .

In Exercises 21–28, determine whether each pair of triangles is congruent. If they are, tell why.

21.

3 cm

3 cm

5 cm

5 cm

6 cm

6 cm

yes, SSS 22.

5 cm 5 cm

6 cm

6 cm

yes, SAS 23.6 cm

6 cm

not necessarily

24.50°

50°

40°

40°

yes, ASA 25.

7 cm7 cm

4 cm4 cm yes, SSS 26.8 cm

8 cm40°

40°

not necessarily

27.

4 cm 4 cm

yes, SAS 28.

40° 40°6 cm 6 cm

yes, ASA

A B

E D

C

45°

45°


70° 70°40° 40°


C

DA

B

F

E


D

C

A

B

E

1

2

5 cm

5 cm

4 cm4 cm




29. 60°

60°

5 cm

5 cmx cm

6 cm

6 cm 30.

5 cm

7 cm

7 cm

x cm

5 cm

9 cm

9 cm

31.7 in.7 in.

7 in.7 in.

50°x°

507 32.

7 in.

7 in.

5 in.5 in.

50°

x°

507

In Exercises 33–34, tell whether the triangles are similar.

33.

60°

60°

40°

40°

yes 34. yes

In Exercises 35–40, refer to Illustration 7 and find the length of the unknown side.

35. a5 3 andb5 4. Findc. 5

36. a5 12 andb5 5. Findc. 13

37. a5 15 andc5 17. Findb. 8

38. b5 45 andc5 53. Finda. 28

39. a5 5 andc5 9. Findb. =56

40. a5 1 andb5 7. Findc. =50

In Exercises 41–44, the length of the three sides of a triangle are given. Determine whether the triangle is a right triangle.

41. 8, 15, 17 yes 42. 6, 8, 10 yes 43. 7, 24, 26 no 44. 9, 39, 40 no

APPLICATIONS In Exercises 45–54, use a calculator to help solve each problem. If an answer is not exact,give the answer to the nearest tenth.

45. HEIGHT OF A TREE The tree in Illustration 8 castsa shadow of 24 feet when a 6-foot man casts a shadowof 4 feet. Find the height of the tree.36 ft

46. HEIGHT OF A BUILDING Aman places a mirror onthe ground and sees the reflection of the top of a build-ing, as shown in Illustration 9. Find the height of thebuilding. 41.7 ft

24 ft

h

4 ft

6 ft


h

50 ft 6 ft

5 ft


a

bc



47. WIDTH OF A RIVER Use the dimensions in Illus-tration 10 to findw, the width of the river. 59.2 ft

48. FLIGHT PATH The airplane in Illustration 11 ascends200 feet as it flies a horizontal distance of 1,000 feet.How much altitude is gained as it flies a horizontal dis-tance of 1 mile? (Hint: 1 mile5 5,280 feet.) 1,056 ft

49. FLIGHT PATH An airplane descends 1,200 feet as itfiles a horizontal distance of 1.5 miles. How much al-titude is lost as it flies a horizontal distance of 5 miles?4,000 ft

50. GEOMETRY If segmentDE in Illustration 12 is par-allel to segmentAB, nABCwill be similar tonDEC.Find x. 4.2

51. ADJUSTING A LADDER A 20-foot ladder reachesa window 16 feet above the ground. How far from thewall is the base of the ladder?12 ft

52. LENGTH OF GUY WIRES A 30-foot tower is to befastened by three guy wires attached to the top of thetower and to the ground at positions 20 feet from itsbase. How much wire is needed?108.2 ft

53. BASEBALL A baseball diamond is a square witheach side 90 feet long. (See Illustration 13.) How far isit from home plate to second base?127.3 ft

54. TELEVISION What size is the television screenshown in Illustration 14? 31.4 in.


55. Explain the Pythagorean theorem. 56. Explain the procedure used to solve the equationc2 5 64.

REVIEW In Exercises 57–60, estimate the answer to each problem.

57.0.95? 3.89

2.997113 58. 21% of 42 8 59. 32% of 60 20 60.

4.9661 5.001

2.994313

A B

C

D Ex

5

12

10


25 ft 20 ft

74 ft w ft


1 mi

x ft 200 ft

1,000 ft


90 ft

90 ft


19 in.

25 in.

d in.



In Exercises 61–62, simplify each expression.

61. 21 4 ? 32 38 62. 32 (52 2)2 1 22 22

9.5 Perimeters and Areas of Polygons

In this section, you will learn how to find• Perimeters of polygons• Perimeters of figures that are combinations of polygons• Areas of polygons• Areas of figures that are combinations of polygons

INTRODUCTION . In this section, we will discuss how to find perimeters and areas ofpolygons. Finding perimeters is important when estimating the cost of fencing or es-timating the cost of woodwork in a house. Finding areas is important when calculat-ing the cost of carpeting, the cost of painting a house, or the cost of fertilizing a yard.

Perimeters of polygonsRecall that theperimeter of a polygon is the distance around it. Since a square hasfour sides of equal lengths, its perimeterP is s1 s1 s1 s, or 4s.

Perimeter of a square If a square has a side of lengths, its perimeterP is given bythe formula

P5 4s s s

s

s

EXAMPLE 1 Perimeter of a square. Find the perimeter of a squarewhose sides are 7.5 meters long.

SolutionSince the perimeter of a square is given by the formulaP5 4s, we substitute 7.5 fors and simplify.

P5 4s

P5 4(7.5)

P5 30

The perimeter is 30 meters.

Self CheckFind the perimeter of a squarewhose sides are 23.75 centimeterslong.

Answer: 95 cm

Since a rectangle has two lengthsl and two widthsw, its perimeterP isl 1 l 1 w1 w, or 2l 1 2w.

Perimeter of arectangle

If a rectangle has lengthl and widthw, its perimeterPis given by the formula

P5 2l 1 2ww

l


EXAMPLE 2 Perimeter of a rectangle.Find the perimeter of the rectanglein Figure 9-48.

SolutionSince the perimeter is given by the formulaP5 2l 1 2w,we substitute 10 forl and 6 forw and simplify.

P5 2l 1 2w

P5 2(10) 1 2(6)

5 201 12

5 32

The perimeter is 32 centimeters.

Self CheckFind the perimeter of the isoscelestrapezoid below.

10 cm

12 cm

8 cm 8 cm

Answer: 38 cm

EXAMPLE 3 Converting units. Find the perimeterof the rectangle in Figure 9-49, in meters.

SolutionSince 1 meter5 100 centimeters, we can convert 80 centime-ters to meters by multiplying 80 centimeters by the unit conver-sion factor 1 m

100 cm.

80 cm5 80 cm?1 m

100 cm1m

100 cm5 1.

580

100m The units of centimeters divide out.

5 0.8 m Do the division: 804 1005 0.8.

We can now substitute 3 forl and 0.8 forw to get

P5 2l 1 2w

P5 2(3) 1 2(0.8)

5 61 1.6

5 7.6

The perimeter is 7.6 meters.

Self CheckFind the perimeter of the trianglebelow, in inches.

2 ft

14 in. 12 in.

Answer: 50 in.

EXAMPLE 4 Finding the base of an isoscelestriangle. The perimeter of the isos-

celes triangle in Figure 9-50 is 50 meters. Find the length ofits base.

SolutionTwo sides are 12 meters long, and the perimeter is 50 meters.If x represents the length of the base, we have

121 121 x 5 50

241 x 5 50 121 125 24.


The length of the base is 26 meters.

Self CheckThe perimeter of an isosceles tri-angle is 60 meters. If one of itssides of equal length is 15 meterslong, how long is its base?

Answer: 30 m

10 cm

6 cm

F I G U R E 9-48

80 cm

3 m

F I G U R E 9-49

12 m 12 m

x m

F I G U R E 9-50

9.5 Perimeters and Areas of Polygons 33

Perimeters of figures that are combinationsof polygons

Accent on Technology Perimeter of a figure

See Figure 9-51. To find the perimeter, we need to know the values ofx and y.Since the figure is a combination of two rectangles, we can use a calculator to seethat

x5 20.252 10.17 and y5 12.52 4.75

5 10.08 5 7.75

The perimeterP of the figure is

P5 20.251 12.51 10.171 4.751 x 1 y

P5 20.251 12.51 10.171 4.751 10.081 7.75

Evaluate: 20.251 12.51 10.171 4.751 10.081 7.75

Keystrokes: 2 0 . 2 5 1 1 2 . 5 1 1 0 . 1 7 1

4 . 7 5 1 1 0 . 0 8 1 7 . 7 5 5

65.5

The perimeter is 65.5 centimeters.

Areas of polygonsRecall that thearea of a polygon is the measure of the amount of surface it encloses.Area is measured in square units, such as square inches or square centimeters. SeeFigure 9-52.

In everyday life, we commonly use areas. For example,

• To carpet a room, we buy square yards.

• A can of paint will cover a certain number of square feet.

• To measure real estate, we often use square miles.

• We buy house roofing by the “square.” One square is 100 square feet.

The rectangle shown in Figure 9-53 has a length of 10 centimeters and a width of3 centimeters. If we divide the rectangle into squares as shown in the figure, each

y cm

x cm

4.75 cm

10.17 cm

12.5 cm

20.25 cm

F I G U R E 9-51

One square inch(1 in.2)

One square centimeter(1 cm2)

1 in. 1 in. 1 cm 1 cm

1 cm

1 cm

1 in.

1 in.

F I G U R E 9-52


square represents an area of 1 square centimeter—a surface enclosed by a square mea-suring 1 centimeter on each side. Because there are 3 rows with 10 squares in eachrow, there are 30 squares. Since the rectangle encloses a surface area of 30 squares,its area is 30 square centimeters, often written as 30 cm2.

This example illustrates that to find the area of a rectangle, we multiply its lengthby its width.

WARNING! Do not confuse the concepts of perimeter and area. Perimeter is thedistance around a polygon. It is measured in linear units, such as centimeters, feet, ormiles. Area is a measure of the surface enclosed within a polygon. It is measured insquare units, such as square centimeters, square feet, or square miles.

In practice, we do not find areas by counting squares in a figure. Instead, we useformulas for finding areas of geometric figures.

Figure Name Formula for area

s s

s

s

Square A5 s2, wheres is the length of one side.

w w

l

l

Rectangle A5 lw, wherel is the length andw is thewidth.

h

b

Parallelogram A5 bh, whereb is the length of the baseandh is the height. (A height is alwaysperpendicular to the base.)

h

b

h

b

TriangleA5 1

2bh, whereb is the length of the baseandh is the height. The segment thatrepresents the height is called analtitude.

h

b1

b2

TrapezoidA5 1

2h(b1 1 b2), whereh is the height of thetrapezoid andb1 andb2 represent thelength of each base.

1 cm2

10 cm

3 cm

F I G U R E 9-53


EXAMPLE 5 Area of a square. Find the area ofthe square in Figure 9-54.

SolutionWe can see that the length of one side of the square is 15centimeters. We can find its area by using the formulaA5 s2 and substituting 15 fors.

A5 s2

A5 (15)2 Substitute 15 fors.

A5 15 ? 15 152 5 15 ? 15.

A5 225 15 ? 155 225.

The area of the square is 225 cm2.

Self CheckFind the area of the square shownbelow.

20 in.

20 in.

20 in. 20 in.

Answer: 400 in.2

EXAMPLE 6 Number of square feet in 1square yard. Find the number

of square feet in 1 square yard. (See Figure 9-55.)

SolutionSince 3 feet5 1 yard, each side of 1 square yard is 3feet long.

1 yd2 5 (1 yd)2

5 (3 ft)2 Substitute 3 feet for 1 yard.

5 9 ft2 (3 ft)2 5 (3 ft)(3 ft) 5 9 ft2.

There are 9 square feet in 1 square yard.

Self CheckFind the number of square centi-meters in 1 square meter.

Answer: 10,000 cm2

EXAMPLE 7 Area of a football field.Find the area of a rectangular

football field in square feet. Disregard the endzones. (See Figure 9-56.)

SolutionTo find the area in square yards, we substitute 100for l and 53.5 forw in the formulaA5 lw.

A5 lw

A5 (100)(53.5)

5 5,350

The area is 5,350 square yards. Since there are 9 square feet per square yard, we canconvert this number to square feet by multiplying 5,350 square yards by9 ft2

1 yd2.

5,350 yd2 5 5,350 yd2 ?9 ft2

1 yd2

5 5,350? 9 ft2 The units of square yards divide out.

5 48,150 ft2 5,350? 95 48,150.

The area of a football field is 48,150 ft2.

Self CheckFind the area of a rectangle withdimensions of 6 inches by 2 feet,in square inches.

Answer: 144 in.2

15 cm

15 cm

15 cm 15 cm

F I G U R E 9-54

3 ft

3 ft

1 yd

1 yd

F I G U R E 9-55

53.5 yd

100 yd

F I G U R E 9-56


EXAMPLE 8 Area of a parallelogram.Find the area of the parallelo-gram in Figure 9-57.

SolutionThe length of the base of the parallelogram is

5 feet1 25 feet5 30 feet

The height is 12 feet. To find the area, we substitute 30 forb and 12 forh in theformula for the area of a parallelogram and simplify.

A5 bh

A5 30 ? 12

5 360

The area of the parallelogram is 360 ft2.

Self CheckFind the area of the parallelogrambelow.

12 cm

10 cm8 cm

Answer: 96 cm2

EXAMPLE 9 Area of a triangle. Find the area ofthe triangle in Figure 9-58.

SolutionThe area of a triangle is found by substituting 8 forb and 5for h in the formula for the area of a triangle.

A 51

2bh

A 51

2(8)(5) Substitute forb andh.

5 4(5) Do the multiplication:12(8)5 4.

5 20

The area of the triangle is 20 cm2.

Self CheckFind the area of the trianglebelow.

15 cm

17 cm12 cm

Answer: 90 cm2

EXAMPLE 10 Area of a triangle. Find the area of thetriangle in Figure 9-59.

Solution In this case, the altitude falls outside the tri-angle.

A 51

2bh

A 51

2(9)(13) Substitute 9 forb

and 13 forh.

51

2S91DS131 D Write 9 as91 and13 as131 .

5117

2Multiply the fractions.

5 58.5 Do the division.

The area of the triangle is 58.5 cm2.

5 ft 25 ft

12 ft

F I G U R E 9-57

8 cm

5 cm

F I G U R E 9-58

9 cm

15 cm13 cm

F I G U R E 9-59


EXAMPLE 11 Area of a trapezoid. Find the area of the trapezoid in Fig-ure 9-60.

SolutionIn this example,b1 5 10 andb2 5 6. It is incorrect to say thath5 1, because theheight of 1 foot must be expressed as 12 inches to be consistent with the units of thebases.Thus, we substitute 10 forb1, 6 for b2, and 12 forh in the formula for finding thearea of a trapezoid and simplify.

A51

2h(b1 1 b2)

A51

2(12)(101 6)

51

2(12)(16)

5 6(16)

5 96

The area of the trapezoid is 96 in.2

Self CheckFind the area of the trapezoid.

12 m

6 m

6 m8 m

Answer: 54 m2

Areas of figures that are combinationsof polygons

EXAMPLE 12 Carpeting a room. A living room/dining room area has the floor plan shown inFigure 9-61. If carpet costs $29 per square yard, including pad and installation, howmuch will it cost to carpet the room? (Assume no waste.)

Solution First we must find the total area of the living room and the dining room:

Atotal5 Aliving room1 Adining room

SinceCF divides the space into two rectangles, the areas of the living room and thedining room are found by multiplying their respective lengths and widths.

6 in.

1 ft

10 in.

F I G U R E 9-60

9 yd

4 yd

4 yd

7 ydLivingroom

Diningroom

A

G

D

E

F

B

C

F I G U R E 9-61


Area of living room5 lw

5 7 ? 4

5 28

The area of the living room is 28 yd2.To find the area of the dining room, we find its length by subtracting 4 yards

from 9 yards to obtain 5 yards, and note that its width is 4 yards.

Area of dining room5 lw

5 5 ? 4

5 20

The area of the dining room is 20 yd2.The total area to be carpeted is the sum of these two areas.

Atotal5 Aliving room 1 Adining room

Atotal5 28 yd2 1 20 yd2

5 48 yd2

At $29 per square yard, the cost to carpet the room will be 48? $29, or $1,392.

EXAMPLE 13 Area of one side of a tent.Find the area of one side of the tent in Figure 9-62.

Solution Each side is a combination of a trapezoid and a triangle. Since the bases of each trap-ezoid are 30 feet and 20 feet and the height is 12 feet, we substitute 30 forb1, 20 forb2, and 12 forh into the formula for the area of a trapezoid.

Atrap.51

2h(b1 1 b2)

Atrap.51

2(12)(301 20)

5 6(50)

5 300

The area of the trapezoid is 300 ft2.Since the triangle has a base of 20 feet and a height of 8 feet, we substitute 20

for b and 8 forh in the formula for the area of a triangle.

Atriangle51

2bh

Atriangle51

2(20)(8)

5 80

The area of the triangle is 80 ft2.The total area of one side of the tent is

Atotal5 Atrap. 1 AtriangleAtotal5 300 ft2 1 80 ft2

5 380 ft2

The total area is 380 ft2.

30 ft

12 ft

8 ft

20 ft

F I G U R E 9-62

9.5 Perimeters and Areas of Polygos 39



1. The distance around a polygon is called theperimeter .

2. The perimeter of a polygon is measured inlinear units.

3. The measure of the surface enclosed by a polygon iscalled its area .

4. If each side of a square measures 1 foot, the area en-closed by the square is 1 square foot .

5. The area of a polygon is measured insquare units.

6. The segment that represents the height of a triangle iscalled an altitude .

CONCEPTS In Exercises 7–14, sketch and label each of the figures described.

7. Two different rectangles, each having a perimeter of40 in. length 15 in. and width 5 in.; length 16 in. andwidth 4 in. (answers may vary)

8. Two different rectangles, each having an area of 40 in2.length 10 in. and width 4 in.; length 8 in. and width 5 in.(answers may vary)

9. A square with an area of 25 m2. sides of length 5 m 10. A square with a perimeter of 20 m.sides of length 5 m

11. A parallelogram with an area of 15 yd2. base 5 ydand height 3 yd (answers may vary)

12. A triangle with an area of 20 ft2. base 10 ft and height4 ft (answers may vary)

13. A figure consisting of a combination of two rectangleswhose total area is 80 ft2. length 5 ft and width 4 ft;length 20 ft and width 3 ft (answers may vary)

14. A figure consisting of a combination of a rectangle anda square whose total area is 164 ft2. length 20 ft andwidth 5 ft; length of side of square 8 ft (answers may vary)


15. The formula for the perimeter of a square isP5 4s .

16. The formula for the perimeter of a rectangle isP5 2l 1 2w .

17. The symbol 1 in.2 means one square inch. 18. One square meter is expressed as 1 m2 .

19. The formula for the area of a square isA5 s2 .

20. The formula for the area of a rectangle isA5 lw .

21. The formulaA5 12bh gives the area of a

triangle .22. The formulaA5 1

2h(b1 1 b2) gives the area of atrapezoid .

PRACTICE In Exercises 23–28, find the perimeter of each figure.

23.

8 in.

8 in.

8 in.

8 in.

32 in.

24.

6 cm

12 cm

12 cm

6 cm

36 cm

25.

6 m

6 m

10 m

4 m

4 m

2 m

2 m

36 m

26.

5 in.

5 in.

4 in. 4 in.

5 in.

23 in.

27.

7 cm

10 cm

6 cm6 cm

8 cm

37 cm

28.

10 cm

6 cm6 cm

7 cm

2 cm 2 cm

33 cm


In Exercises 29–32, solve each problem.29. Find the perimeter of an isosceles triangle with a base

of length 21 centimeters and sides of length 32 centi-meters. 85 cm

30. The perimeter of an isosceles triangle is 80 meters. Ifthe length of one side is 22 meters, how long is thebase? 36 m

31. The perimeter of an equilateral triangle is 85 feet. Findthe length of each side.2813 ft

32. An isosceles triangle with sides of 49.3 inches has aperimeter of 121.7 inches. Find the length of the base.23.1 in.

In Exercises 33–46, find the area of the shaded part of each figure.33.

4 cm

4 cm

16 cm2 34.

5 in.

3 in.

15 in.2

35.

6 cm4 cm

15 cm

60 cm236.

7 m6 m

10 m

60 m2

37.

10 in.

5 in.

25 in.2

38.

9 cm

3 cm

13.5 cm2

39.

13 mm

9 mm

17 mm

169 mm2

40.

7 cm

10 cm

3 cm 3 cm

7 cm

91 cm2

41.

8 m

8 m

8 m4 m

80 m2

42.

20 ft

30 ft

2 ft

360 ft2

43.

10 yd 10 yd

5 yd

10 yd

75 yd2 44.

10 in.

6 in.

17 in.

119 in.2

45.

14 m

3 m

3 m

6 m

75 m2 46.

25 cm

10 cm

8 cm

15 cm

335 cm2


47. How many square inches are in 1 square foot?144 48. How many square inches are in 1 square yard?1,296

APPLICATIONS Use a calculator to help solve each problem.

49. FENCING A YARD A man wants to enclose a rect-angular yard with fencing that costs $2.44 a foot, in-cluding installation. Find the cost of enclosing the yardif its dimensions are 110 ft by 85 ft.$951.60

50. FRAMING A PICTURE Find the cost of framing arectangular picture with dimensions of 24 inches by 30inches if framing material costs $8.46 per foot, includ-ing matting. $76.14

51. PLANTING A SCREEN A woman wants to plant apine-tree screen around three sides of her backyard.(See Illustration 1.) If she plants the trees 3 feet apart,how many trees will she need?81

52. PLANTING MARIGOLDS Agardener wants to planta border of marigolds around the garden shown in Il-lustration 2 to keep out rabbits. How many plants willshe need if she allows 6 inches between plants?144

53. BUYING A FLOOR Which is more expensive: Aceramic-tile floor costing $3.75 per square foot or lino-leum costing $34.95 per square yard?linoleum

54. BUYING A FLOOR Which is cheaper: A hardwoodfloor costing $5.95 per square foot or a carpeted floorcosting $37.50 per square yard?carpeted

55. CARPETINGAROOM A rectangular room is 24 feetlong and 15 feet wide. At $30 per square yard, howmuch will it cost to carpet the room? (Assume nowaste.) $1,200

56. CARPETING A ROOM A rectangular living roommeasures 30 by 18 feet. At $32 per square yard, howmuch will it cost to carpet the room? (Assume nowaste.) $1,920

57. TILING A FLOOR A rectangular basement roommeasures 14 by 20 feet. Vinyl floor tiles that are 1 ft2

cost $1.29 each. How much will the tile cost to coverthe floor? (Disregard any waste.)$361.20

58. PAINTING A BARN The north wall of a barn is arectangle 23 feet high and 72 feet long. There are fivewindows in the wall, each 4 by 6 feet. If a gallon ofpaint will cover 300 ft2, how many gallons of paint mustthe painter buy to paint the wall?6 gal

59. MAKING A SAIL If nylon is $12 per square yard,howmuch would the fabric cost to make a triangular sailwith a base of 12 feet and a height of 24 feet?$192

60. PAINTING A GABLE The gable end of a warehouseis an isosceles triangle with a height of 4 yards and abase of 23 yards. It will require one coat of primer andone coat of finish to paint the triangle. Primer costs $17per gallon, and the finish paint costs $23 per gallon. Ifone gallon covers 300 square feet, how much will it costto paint the gable, excluding labor?$80

61. SODDING A LAWN A landscaper charges $1.17 persquare foot to sod a lawn. If the lawn is in the shape ofa trapezoid, as shown in Illustration 3, what will it costto sod the lawn? $1,658.48

62. COVERING A SWIMMING POOL A swimmingpool has the shape shown in Illustration 4. How manysquare meters of plastic sheeting will be needed to coverthe pool? How much will the sheeting cost if it is $2.95per square meter? (Assume no waste.)400 m2, $1,180

20 m

12 m

25 m


15 yd

20 yd

9 yd


70 ft

100 ft


20 ft

16 ft



63. CARPENTRY How many sheets of 4-foot-by-8-footsheetrock are needed to drywall the inside walls on thefirst floor of the barn shown in Illustration 5? (Assumethat the carpenters will cover each wall entirely and thencut out areas for the doors and windows.)51

64. CARPENTRY If it costs $90 per square foot to builda one-story home in northern Wisconsin, estimate thecost of building the house with the floor plan shown inIllustration 6. $94,320


65. Explain the difference between perimeter and area. 66. Why is it necessary that area be measured in squareunits?

REVIEW In Review Exercises 67–72, do the calculations. Write all improper fractions as mixed numbers.

67.3

41

2

31512 68.

7

82

2

3524 69. 3

3

41 2

1

36112

70. 75

82 2

5

641924 71. 7

1

24 5

2

51718 72. 5

3

4? 2

5

616724

9.6 Circles

In this section, you will learn about• Circles• Circumference of a circle• Area of a circle

INTRODUCTION . In this section, we will discuss circles, one of the most useful geo-metric figures. In fact, the discovery of fire and the circular wheel were two of themost important events in the history of the human race.

Circles

Circle A circle is the set of all points in a plane that lie a fixed distance from a pointcalled itscenter.

A segment drawn from the center of a circle to a point on the circle is called aradius. (The plural ofradius is radii.) From the definition, it follows that all radii ofthe same circle are the same length.

12 ft

20 ft

48 ft


12 ft30 ft

20 ft

14 ft


9.6 Circles 43

A chord of a circle is a line segment connecting two points on the circle. Adi-ameter is a chord that passes through the center of the circle. Since a diameterD of acircle is twice as long as a radiusr, we have

D 5 2r

Each of the previous definitions is illustrated in Figure 9-63, in whichO is thecenter of the circle.

Any part of a circle is called anarc. In Figure 9-64, the part of the circle frompoint A to pointB is ABXXXXXX, read as arcAB. CDXXXXXX is the part of the circle from pointC topoint D. An arc that is half of a circle is asemicircle.

Semicircle A semicircle is an arc of a circle whose endpoints are the endpoints of a diameter.

If point O is the center of the circle in Figure 9-64,AD is a diameter andAEDXXXXXX isa semicircle. The middle letterE is used to distinguish semicircleAEDXXXXXX from semi-circle ABCDXXXXXXXX.

An arc that is shorter than a semicircle is aminor arc. An arc that is longer thana semicircle is amajor arc. In Figure 9-64,

ABXXXXXX is a minor arc and ABCDEXXXXXXXX is a major arc.

Circumference of a circleSince early history, mathematicians have known that the ratio of the distance arounda circle (the circumference) divided by the length of its diameter is approximately 3.First Kings, Chapter 7 of the Bible describes a round bronze tank that was 15 feetfrom brim to brim and 45 feet in circumference, and45

15 5 3. Today, we have a bettervalue for this ratio, known asp (pi). If C is the circumference of a circle andD is thelength of its diameter, then

p 5C

D, wherep 5 3.141592653589. . .

227 and 3.14 are often used asestimates ofp.

If we multiply both sides ofp 5 CD by D, we have the following formula.

Circumferenceof a circle

The circumference of a circle is given by the formula

C5 pD whereC is the circumference andD is the length of the diameter.

Since a diameter of a circle is twice as long as a radiusr, we can substitute 2r forD in the formulaC5 pD to obtain another formula for the circumferenceC:

C5 2pr

A

O

E

B

C

D

Diameter COD

Chord AB

Radius OE

F I G U R E 9-63

O

C

B

AE

D

F I G U R E 9-64


EXAMPLE 1 Circumference of a circle. Findthe circumference of a circle with diam-

eter of 10 centimeters. (See Figure 9-65.)

SolutionWe substitute 10 forD in the formula for the circumferenceof a circle.

C5 pD

C5 p(10)

C' 31.41592653589

To the nearest tenth, the circumference is 31.4 centimeters.

Self CheckTo the nearest hundredth, find thecircumference of a circle with aradius of 12 meters.

Answer: 75.40 m

Accent on Technology Calculating revolutions of a tire

To calculate how many times a 15-inch tire rotates when a car makes a 25-miletrip, we first find the circumference of the tire.

C5 pD

C5 p(15) Substitute 15 forD, the diameter of the tire.

C' 47.1238898The circumference of the tire is 47.1238898 inches.

We then change 25 miles to inches.

25 miles ?5,280 feet

1 mile?12 inches

1 foot5 25(5,280)(12) inches

5 1,584,000 inchesFinally, we divide 1,584,000 inches by 47.1238898 inches to get

Total distance

Circumference of tire5

1,584,000

47.1238898

5 33,613.52398To do this work on a scientific calculator, we press these keys.

Evaluate:25 ? 5,280? 12

15 ? p

Keystrokes: ( 2 5 3 5 2 8 0 3 1 2 ) 4 p 4 1 5 5

33613.52398

The tire makes about 33,614 revolutions.

EXAMPLE 2 Perimeter of a figure. Find the perimeter of the figure shown in Figure 9-66.

Solution The figure is a combination of three sides of a rectangleand a semicircle. The perimeter of the rectangular part is

Prectangular part5 81 61 85 22

The perimeter of the semicircle is one-half of the circum-ference of a circle with a 6-meter diameter.

Psemicircle51

2pD

51

2p(6) Substitute 6 forD.

' 9.424777961 Use a calculator.

10 cm

F I G U R E 9-65

8 m 8 m

6 m

F I G U R E 9-66

9.6 Circles 45

The total perimeter is the sum of the two parts.

Ptotal' 221 9.424777961

' 31.424777961

To the nearest hundredth, the perimeter of the figure is 31.42 meters.

Area of a circleIf we divide the circle shown in Figure 9-67(a) into an even number of pie-shapedpieces and then rearrange them as shown in Figure 9-67(b), we have a figure thatlooks like a parallelogram. The figure has a base that is one-half the circumference ofthe circle, and its height is about the same length as a radius of the circle.

If we divide the circle into more and more pie-shaped pieces, the figure will lookmore and more like a parallelogram, and we can find its area by using the formula forthe area of a parallelogram.

A 5 bh The formula for the area of a parallelogram.

512Cr Substitute12 of the circumference forb, andr for the height.

51

2(2pr )r C5 2pr.

5 pr2 12 ? 25 1 andr ? r 5 r2.

Area of a circle Thearea of a circlewith radiusr is given by the formula

A5 pr2

EXAMPLE 3 Area of a circle. To the nearesttenth, find the area of the circle in Fig-ure 9-68.

SolutionSince the length of the diameter is 10 centimeters and thelength of a diameter is twice the length of a radius, the lengthof the radius is 5 centimeters. To find the area of the circle,we substitute 5 forr in the formula for the area of a circle.

A5 pr2

A5 p(5)2

5 25p

' 78.53981634 Use a calculator.

To the nearest tenth, the area is 78.5 cm2.

Self CheckTo the nearest tenth, find the areaof a circle with a diameter of 12feet.

Answer: 113.1 ft2

o

(a) (b)

h

b

F I G U R E 9-67

10 cm

F I G U R E 9-68


Accent on Technology Painting a helicopter pad

Orange paint is available in gallon containers at $19 each, and each gallon willcover 375 ft2. To calculate how much the paint will cost to cover a helicopter pad60 feet in diameter, we first calculate the area of the helicopter pad.

A5 pr2

A5 p(30)2 Substitute one-half of 60 forr.

5 900p 30 ? 305 900.

' 2,827.433388 Use a calculator.

The area of the pad is 2,827.433388 ft2. Since each gallon of paint will cover375 ft2, we can find the number of gallons of paint needed by dividing2,827.433388 by 375.

Number of gallons needed'2,827.433388

375

' 7.539822369

Because paint only comes in full gallons, the painter will need to purchase 8gallons. The cost of the paint will be 8($19), or $152. To do this work on acalculator, we press these keys.

Evaluate:(30)2 ? p

375

Keystrokes: 3 0 x2 3 p 5 4 3 7 5 5 7.539822369

Round to 8 gallons.

Keystrokes: 8 3 1 9 5 152

The cost of the paint will be $152.

EXAMPLE 4 Finding areas. Find the shaded area in Figure 9-69.

Solution The figure is a combination of a triangle and two semicircles. By the Pythagoreantheorem, the hypotenuseh of the right triangle is

h5 =62 1 82 5 =361 645 =1005 10

The area of the triangle is

Aright triangle51

2bh 5

1

2(6)(8) 5

1

2(48)5 24

The area enclosed by the smaller semicircle is

Asmaller semicircle51

2pr2 5

1

2p(4)2 5 8p

The area enclosed by the larger semicircle is

Alarger semicircle51

2pr2 5

1

2p(5)2 5 12.5p

The total area is

Atotal5 241 8p 1 12.5p ' 88.4026494 Use a calculator.

To the nearest hundredth, the area is 88.40 in.2

8 in.

6 in.

h in.

F I G U R E 9-69

9.6 Circles 47



1. A segment drawn from the center of a circle to a pointon the circle is called a radius .

2. A segment joining two points on a circle is called achord .

3. A diameter is a chord that passes through thecenter of a circle.

4. An arc that is one-half of a complete circle is asemicircle .

5. An arc that is shorter than a semicircle is calleda minor arc.

6. An arc that is longer than a semicircle is called amajor arc.

7. The distance around a circle is called itscircumference .

8. The surface enclosed by a circle is called itsarea .

CONCEPTS In Exercises 9–14, refer to Illustration 1.

9. Name each radius.OA, OC, andOB10. Name each diameter.AC

11. Name each chord.DA, DC, andAC12. Name each minor arc.ADXXXXXX, DCXXXXXX, CBXXXXXX, BAXXXXXX13. Name each semicircle.ABCXXXXXX andADCXXXXXX14. Name each major arc.ADCBXXXXXXXX, DCBAXXXXXXXX, CBADXXXXXXXX, BADCXXXXXXXX

15. If you know the radius of a circle, how can you find itsdiameter? Double the radius.

16. If you know the diameter of a circle, how can you findits radius? Divide the diameter by 2.


17. The symbolABXXXXXX is read as arcAB . 18. To the nearest hundredth, the value ofp is3.14 .

19. The formula for the circumference of a circle isC5 pD or C5 2pr .

20. The formulaA5 pr2 gives the area of acircle .

PRACTICE

21. To the nearest hundredth, find the circumference of acircle with a diameter of 12 inches.37.70 in.

22. To the nearest hundredth, find the circumference of acircle with a radius of 20 feet.125.66 ft

23. Find the diameter of a circle with a circumference of36p meters. 36 m

24. Find the radius of a circle with a circumference of 50pmeters. 25 m

In Exercises 25–28, find the perimeter of each figure to the nearest hundredth.

25. 8 ft

3 ft

25.42 ft

26.

12 cm

10 cm

50.85 cm

27.

6 m

8 m 8 m

31.42 m

28. 18 in.

18 in.

10 in.

67.42 in.

O

A

D

BC

1

2

3

4



In Exercises 29–30, find the area of each circle to the nearest tenth.

29.

3 in.

A5 28.3 in.2 30.

12 ft

A5 113.1 ft2

In Exercises 31–34, find the total area of each figure to the nearest tenth.

31.

6 in.

10 in.

88.3 in.2

32.

4 cm

8 cm

57.1 cm2

33.

12 cm

12 cm

128.5 cm2

34.

4 in.

4 in.

62.8 in.2

In Exercises 35–38, find the area of each shaded region to the nearest tenth.

35. 4 in.

10 in

27.4 in.2

36.

8 in.

8 in.

13.7 in.2

37.

h = 9 in.

13 in.

r = 4 in.

66.7 in.2

38.

8 ft 8 ft

100.5 ft2

APPLICATIONS In Exercises 39–46, give each answer to the nearest hundredth.

39. AREA OF ROUND LAKE Round Lake has a circu-lar shoreline 2 miles in diameter. Find the area of thelake. 3.14 mi2

40. TAKING A WALK Sam is planning to hike aroundRound Lake. in Exercise 39. How far will he walk?6.28 mi

41. JOGGING Joan wants to jog 10 miles on a circulartrack 1

4 mile in diameter. How many times must shecircle the track? 12.73 times

42. FIXING THE ROTUNDA The rotunda at a state capi-tol is a circular area 100 feet in diameter. The legisla-ture wishes to appropriate money to have the floor ofrotunda tiled. The lowest bid is $83 per square yard,including installation. How much must the legislaturespend? $72,431.16

43. BANDING THE EARTH A steel band is drawntightly about the earth’s equator. The band is then loos-ened by increasing its length by 10 feet, and the result-ing slack is distributed evenly along the band’s entirelength. How far above the earth’s surface is the band?(Hint: You don’t need to know the earth’s circumfer-ence.) 1.59 ft

44. CONCENTRIC CIRCLES Two circles are calledconcentric circles if they have the same center. Findthe area of the band between two concentric circles iftheir diameters are 10 centimeters and 6 centimeters.50.27 cm2

45. ARCHERY See Illustration 2. What percentage of thearea of the target is the bullseye?6.25%

46. LANDSCAPE DESIGN See Illustration 3. Howmuch of the lawn does not get watered by the sprin-klers at the center of each circle?193.14 ft2

4 ft

1 ft


30 ft

30 ft


9.6 Circles 49


47. Explain what is meant by the circumference of a circle.48. Explain what is meant by the area of a circle.

49. Explain the meaning ofp. 50. Distinguish between a major arc and a minor arc.

REVIEW In Exercises 51–56, solve each problem.

51. Change9

10to a percent. 90% 52. Change

7

8to a percent. 8712%

53. Find 30% of 1,600. 480 54. Find1

2% of 520. 2.6

55. COST OF A DRESS Maria bought a dress for 25%off the regular price of $98. How much did she pay?$73.50

56. COST OF A SHIRT Bill bought a shirt on sale for$17.50. Find its original cost if it was on sale at 30%off. $25

9.7 Surface Area and Volume

In this section, you will learn about• Volumes of solids• Surface areas of rectangular solids• Volumes and surface areas of spheres• Volumes of cylinders• Volumes of cones• Volumes of pyramids

INTRODUCTION . In this section, we will discuss a measure of capacity calledvolume.Volumes are measured in cubic units, such as cubic inches, cubic yards, or cubic cen-timeters. For example,

• We buy gravel or topsoil by the cubic yard.

• We measure the capacity of a refrigerator in cubic feet.

• We often measure amounts of medicine in cubic centimeters.

We will also discuss surface area. The ability to compute surface area is necessary tosolve problems such as calculating the amount of material necessary to make a card-board box or a plastic beach ball.

Volumes of solidsA rectangular solid and acubeare two common geometric solids. (See Figure 9-70.)

Height(h)

Length (l)

Width (w)

2 cm

2 cm

2 cm

A rectangular solid A cube

F I G U R E 9-70


The volume of a rectangular solid is a measure of the space it encloses. Twocommon units of volume are cubic inches (in.3) and cubic centimeters (cm3). (SeeFigure 9-71.)

If we divide the rectangular solid shown in Figure 9-72 into cubes, each cuberepresents a volume of 1 cm3. Because there are 2 levels with 12 cubes on each level,the volume of the rectangular solid is 24 cm3.

In practice, we do not find volumes by counting cubes. Instead, we use the fol-lowing formulas.

Figure Name Volume Figure Name Volume

s

s

s

Cube V5 s3

h

rCylinder V5 pr2h

wh

l

RectangularSolid

V5 lwh h

r

Cone V51

3pr2h

hPrism V5 Bh*

hPyramid V5

1

3Bh*

*B represents the area of the base.

(continued)

1 in.

1 in.

1 in.

1 cubic inch (1 in.3)

1 cm1 cm

1 cm

1 cubic centimeter (1 cm3)

F I G U R E 9-71

1 cm3

4 cm

3 cm

2 cm

F I G U R E 9-72

9.7 Surface Area and Volume 51

Figure Name Volume Figure Name Volume

r

Sphere V54

3pr3

WARNING! The height of a geometric solid is always measured along a line per-pendicular to its base. In each of the solids in Figure 9-73,h is the height.

EXAMPLE 1 Number of cubicinches in one cubic

foot. How many cubic inches are there in 1cubic foot? (See Figure 9-74.)

SolutionSince a cubic foot is a cube with each side mea-suring 1 foot, each side also measures 12 inches.Thus, the volume in cubic inches is

V 5 s3 The formula for the volume of a cube.

V 5 (12)3 Substitute 12 fors.

5 1,728

There are 1,728 cubic inches in 1 cubic foot.

Self CheckHow many cubic centimeters arein 1 cubic meter?

Answer: 1,000,000 cm3

EXAMPLE 2 Volume of an oil storage tank. An oil storage tank is inthe form of a rectangular solid with dimensions of 17 by 10 by 8

feet. (See Figure 9-75.) Find its volume.

SolutionTo find the volume, we substitute 17 forl, 10 for w, and 8 forh in the formulaV5 lwh and simplify.

V5 lwh

V5 (17)(10)(8)

5 1,360

The volume is 1,360 ft3.

Self CheckFind the volume of a rectangularsolid with dimensions of 8 by 12by 20 meters.

Answer: 1,920 m3

hhh hh

F I G U R E 9-73

1 ft 12 in.

12 in.12 in.

1 ft1 ft

F I G U R E 9-74

17 ft

8 ft

10 ft

F I G U R E 9-75


EXAMPLE 3 Volume of a triangularprism. Find the volume of

the triangular prism in Figure 9-76.

SolutionThe volume of the prism is the area of its base mul-tiplied by its height. Since there are 100 centime-ters in 1 meter, the height in centimeters is

0.5 m5 0.5(1 m)

5 0.5(100 cm) Substitute 100 centimetersfor 1 meter.

5 50 cm

Since the area of the triangular base is 24 square centimeters and the height of theprism is 50 centimeters, we have

V5 Bh

V5 24(50)

5 1,200

The volume of the prism is 1,200 cm3.

Self CheckFind the volume of the triangularprism below.

10 cm

4 cm5 cm

5 cm

6 cm

Answer: 120 cm3

Surface areas of rectangular solidsThe surface areaof a rectangular solid is the sum of the areas of its six faces. (SeeFigure 9-77.)

SA5 Arectangle 11 Arectangle 21 Arectangle 31 Arectangle 41 Arectangle 51 Arectangle 6

5 lw 1 lh 1 lh 1 hw1 lw 1 hw

5 2lw 1 2lh 1 2hw Combine like terms.

Surface area of arectangular solid

The surface area of a rectangular solid is given by the formula

SA5 2lw 1 2lh 1 2hw

wherel is the length,w is the width, andh is the height.

EXAMPLE 4 Surface area of an oiltank. An oil storage

tank is in the form of a rectangular solid withdimensions of 17 by 10 by 8 feet. (See Figure9-78.) Find the surface area of the tank.

Self CheckFind the surface area of a rectan-gular solid with dimensions of 8by 12 by 20 meters.

0.5 m

10 cm

6 cm 8 cm

F I G U R E 9-76

17 ft

8 ft

10 ft

F I G U R E 9-78

wh

h

w

hh l

l

hl

1 4 5 6

2

3

F I G U R E 9-77


SolutionTo find the surface area, we substitute 17 forl,10 forw, and 8 forh in the formula for surfacearea and simplify.

SA5 2lw1 2lh 1 2hw

SA5 2(17)(10) 1 2(17)(8) 1 2(8)(10)

5 3401 2721 160

5 772

The surface area is 772 ft2. Answer: 992 m2

Volumes and surface areas of spheresA sphere is a hollow, round ball. (See Figure 9-79.) The points on a sphere all lie ata fixed distancer from a point called itscenter.A segment drawn from the center of asphere to a point on the sphere is called aradius.

Accent on Technology Filling a water tank

See Figure 9-80. To calculate how many cubic feet of water are needed to fill aspherical water tank with a radius of 15 feet, we substitute 15 forr in the formulafor the volume of a sphere and simplify.

V54

3pr3

V54

3p(15)3

54

3p(3,375)

5 4,500p

' 14,137.16694 Use a calculator.To do the arithmetic with a calculator, pressthese keys.

Evaluate:4

3p(15)3

Keystrokes: 1 5 yx 3 5 3 4 4 3 5 3 p 5

14137.16694

To the nearest tenth, 14,137.2 ft3 of water will be needed to fill the tank.

There is a formula to find the surface area of a sphere.

Surface areaof a sphere

The surface area of a sphere with radiusr is given by the formula

SA5 4pr2

EXAMPLE 5 Manufacturing beach balls. A beach ball is to have a diameter of 16 inches.(See Figure 9-81.) How many square inches of material will be needed to make theball? (Ignore any waste.)

Solution Since a radiusr of the ball is one-half the diameter,r 5 8 inches. We can now sub-stitute 8 forr in the formula for the surface area of a sphere.

r

F I G U R E 9-79

15 ft

F I G U R E 9-80


SA5 4pr2

SA5 4p(8)2

SA5 256p

' 804.2477193

A little more than 804 in.2 of material will be needed to makethe ball.

Volumes of cylindersA cylinder is a hollow figure like a piece of pipe. (See Figure 9-82.)

EXAMPLE 6 Find the volume of the cylinder in Figure 9-83.

Solution Since a radius is one-half of the diameter of the circularbase,r 5 3 cm. From the figure, we see that the heightof the cylinder is 10 cm. So we can substitute 3 forrand 10 forh in the formula for the volume of a cylinder.

V5 pr2h

V5 p(3)2(10)

5 90p

' 282.7433388

To the nearest hundredth, the volume of the cylinder is 282.74 cm3.

Accent on Technology Volume of a silo

The silo in Figure 9-84 is a cylinder 50 feet tall topped with ahemisphere(a half-sphere). To find the volume of the silo, we add the volume of the cylinder to thevolume of the dome.

Volumecylinder 1 volumedome5 (Areacylinder’s base)(heightcylinder) 112(volumesphere)

5 pr2h11

2S43pr3D5 pr2h 1

2pr3

31

2S43

pr3D 51

2?4

3pr3 5

4

6pr3 5

2pr3

3.

5 p(10)2(50) 12p(10)3

3Substitute 10 forr and 50 forh.

5 5,000p 12,000

3p

517,000

3p 5,000p 1

2,000

3p 5

15,000

3p 1

2,000

3p.

' 17,802.35837

To do the arithmetic with a scientific calculator, press these keys.

Evaluate: p(10)2(50)12p(10)3

3

Keystrokes: p 3 1 0 x2 3 5 0 5 1 ( 2 3 p 3

1 0 yx 3 4 3 ) 5 17802.35837

The volume of the silo is approximately 17,802 ft3.

F I G U R E 9-81

h

r

F I G U R E 9-82

10 ft

50 ft

F I G U R E 9-84

10 cm

6 cm

F I G U R E 9-83


EXAMPLE 7 Machining a block of metal.See Figure 9-85. Find the volume thatis left when the hole is drilled throughthe metal block.

Solution We must find the volume of the rect-angular solid and then subtract thevolume of the cylinder.

Vrect. solid5 lwh

Vrect. solid5 12(12)(18)

5 2,592

Vcylinder5 pr2h

Vcylinder5 p(4)2(18)

5 288p

' 904.7786842

Vdrilled block5 Vrect. solid2 Vcylinder' 2,5922 904.7786842

' 1,687.221316

To the nearest hundredth, the volume is 1,687.22 cm3.

Volumes of conesTwo conesare shown in Figure 9-86. Each cone has a heighth and a radiusr, whichis the radius of the circular base.

EXAMPLE 8 Volume of a cone. To the nearest tenth, find the volume of the cone in Figure9-87.

Solution Since the radius is one-half of the diameter,r 5 4 cm. We then substitute 4 forr and6 for h in the formula for the volume of a cone.

V51

3Bh

V51

3pr2h

V51

3p(4)2(6)

V5 32p

' 100.5309649

To the nearest tenth, the volume is 100.5 cubic centimeters.

12 cm

18 cm

8 cm12 cm

F I G U R E 9-85

hh

rr

F I G U R E 9-86

8 cm

6 cm

F I G U R E 9-87


Volumes of pyramidsTwo pyramids with a heighth are shown in Figure 9-88.

EXAMPLE 9 Volume of a pyramid. Find the volume of a pyramid witha square base with each side 6 meters long and a height of 9meters.

SolutionSince the base is a square with each side 6 meters long, the area of the base is 62 m2,or 36 m2. We can then substitute 36 for the area of the base and 9 for the height in theformula for the volume of a pyramid.

V51

3Bh

V51

3(36)(9)

5 108

The volume of the pyramid is 108 m3.

Self CheckFind the volume of the pyramidshown below.

10 cm

6 cm 8 cm

Answer: 80 cm3



1. The space contained within a geometric solid is calledits volume .

2. A rectangular solid is like a hollow shoe box.

3. A cube is a rectangular solid with all sidesof equal length.

4. The volume of a cube with each side 1inch long is 1 cubic inch.

5. The surface area of a rectangular solid is the sumof the areas of its faces.

6. The point that is equidistant from every point on asphere is its center .

7. A cylinder is a hollow figure like a drinkingstraw.

8. A hemisphere is one-half of a sphere.

9. A cone looks like a witch’s pointed hat. 10. A figure with a polygon for its base that rises to a pointis called a pyramid .

h

(a)

h

(b)

The base is a triangle. The base is a square.

F I G U R E 9-88


CONCEPTS In Exercises 11–16, write the formula used for finding the volume of each solid.

11. A rectangular solid V5 lwh 12. A prism V5 Bh 13. A sphere V5 43pr

3

14. A cylinder V5 Bh 15. A cone V5 13Bh 16. A pyramid V5 1

3Bh

17. Write the formula for finding the surface area of a rect-angular solid. SA5 2lw 1 2lh 1 2hw

18. Write the formula for finding the surface area of asphere. SA5 4pr2

19. How many cubic feet are in 1 cubic yard?27 ft3 20. How many cubic inches are in 1 cubic yard?46,656 in.3

21. How many cubic decimeters are in 1 cubic meter?1,000 dm3

22. How many cubic millimeters are in 1 cubic centime-ter? 1,000 mm3

In Exercises 23–24, tell what geometric concept (perimeter, circumference, area, volume, or surface area) should beapplied to find each of the following.

23. a. size of a room to be air conditionedvolumeb. amount of land in a national parkareac. amount of space in a refrigerator freezervolumed. amount of cardboard in a shoe boxsurface areae. distance around a checkerboardperimeterf. amount of material used to make a basketball

surface area

24. a. amount of cloth in a car coversurface areab. size of a trunk of a car volumec. amount of paper used for a postage stamparead. amount of storage in a cedar chestvolumee. amount of beach available for sunbathingareaf. distance the tip of a propeller travels

circumference

NOTATION Fill in the blanks to make a true statement.

25. The symbol in.3 is read as 1 cubic inch . 26. One cubic centimeter is represented as1 cm3 .

PRACTICE In Exercises 27–38, find the volume of each solid. If an answer is not exact, round to the nearest hun-dredth.

27. A rectangular solid with dimensions of 3 by 4 by 5 cen-timeters. 60 cm3

28. A rectangular solid with dimensions of 5 by 8 by 10meters. 400 m3

29. A prism whose base is a right triangle with legs 3 and4 meters long and whose height is 8 meters.48 m3

30. A prism whose base is a right triangle with legs 5 and12 feet long and whose height is 10 feet.300 ft3

31. A sphere with a radius of 9 inches.3,053.63 in.3 32. A sphere with a diameter of 10 feet.523.60 ft3

33. A cylinder with a height of 12 meters and a circularbase with a radius of 6 meters.1,357.17 m3

34. A cylinder with a height of 4 meters and a circular basewith diameter of 18 meters.1,017.88 m3

35. A cone with a height of 12 centimeters and a circularbase with diameter of 10 centimeters.314.16 cm3

36. A cone with a height of 3 inches and a circular basewith radius of 4 inches. 50.27 in.3

37. A pyramid with a square base 10 meters on each sideand a height of 12 meters.400 m3

38. A pyramid with a square base 6 inches on each sideand a height of 4 inches.48 in.3

In Exercises 39–42, find the surface area of each solid. If an answer is not exact, round to the nearest hundredth.

39. A rectangular solid with dimensions of 3 by 4 by 5 cen-timeters. 94 cm2

40. A cube with a side 5 centimeters long.150 cm2

41. A sphere with a radius of 10 inches.1,256.64 in.2 42. A sphere with a diameter of 12 meters.452.39 m2

In Exercises 43–46, find the volume of each figure. If an answer is not exact, round to the nearest hundredth.

43.

8 cm8 cm

8 cm

3 cm 576 cm3 44.

6 cm

16 cm 565.49 cm3


45.

8 in.

20 in.

10 in.

335.10 in.3 46.

6 in.

3 in. 4 in.

5 in.

8 in.

52 in.3

APPLICATIONS In Exercises 47–52, if an answer is not exact, round to the nearest hundredth.

47. VOLUME OF A SUGAR CUBE A sugar cube is12inch on each edge. How much volume does it occupy?0.13 in.3

48. VOLUME OF A CLASSROOM A classroom is 40feet long, 30 feet wide, and 9 feet high. Find the num-ber of cubic feet of air in the room.10,800 ft3

49. VOLUME OF AN OIL TANK A cylindrical oil tankhas a diameter of 6 feet and a length of 7 feet. Find thevolume of the tank. 197.92 ft3

50. VOLUME OF A DESSERT A restaurant serves pud-ding in a conical dish that has a diameter of 3 inches.

If the dish is 4 inches deep, how many cubic inches ofpudding are in each dish?9.42 in.3

51. HOT-AIR BALLOONS The lifting power of a spheri-cal balloon depends on its volume. How many cubicfeet of gas will a balloon hold if it is 40 feet in diam-eter? 33,510.32 ft3

52. VOLUME OFACEREAL BOX Abox of cereal mea-sures 3 by 8 by 10 inches. The manufacturer plans tomarket a smaller box that measures 21

2 by 7 by 8 inches.By how much will the volume be reduced?100 in.3


53. What is meant by thevolumeof a cube? 54. What is meant by thesurface areaof a cube?

REVIEW Do the operations.

55. 4(61 4)2 22 36 56. 25(52 2)2 1 3 242

57. 51 2(61 23) 33 58. 3(61 34) 2 24 245

59. BUYING PENCILS Carlos bought 6 pencils at $0.60each and a notebook for $1.25. He gave the clerk a $5bill. How much change did he receive?$0.15

60. BUYING CLOTHES Mary bought 3 pairs of socksat $3.29 each and a pair of shoes for $39.95. Can shebuy these clothes with three $20 bills?yes

61. BUYING GOLF EQUIPMENT George bought 3packages of golf balls for $1.99 each, a package of teesfor $0.49, and a golf glove for $6.95. How much didhe spend? $13.41

62. BUYING MUSIC Lisa bought 4 compact discs at$9.99 each, 3 tapes for $6.95 each, and a carrying casefor $10.25. How much did she spend?$71.06


geometry

Documents

right angle b

right angles

points b

straight angles

adjacent angles

obtuse angles

points e

angle warning