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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

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Chapter 5 3 Glencoe Geometry

5

Before you begin Chapter 5

• Read each statement.

• Decide whether you Agree (A) or Disagree (D) with the statement.

• Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure).

After you complete Chapter 5

• Reread each statement and complete the last column by entering an A or a D.

• Did any of your opinions about the statements change from the first column?

• For those statements that you mark with a D, use a piece of paper to write an example of why you disagree.

Anticipation GuideRelationships in Triangles

STEP 1 A, D, or NS

StatementSTEP 2A or D

1. Any point that is on the perpendicular bisector of a segment is equidistant from the endpoints of that segment.

A

2. The circumcenter of a triangle is equidistant from the midpoints of each side of the triangle. D

3. The altitudes of a triangle meet at the orthocenter. A

4. Three altitudes can be drawn for any one triangle. A

5. A median of a triangle is any segment that contains the midpoint of a side of the triangle. D

6. The measure of an exterior angle of a triangle is always greater than the measures of either of its corresponding remote interior angles.

A

7. The longest side in a triangle is opposite the smallest angle in that triangle. D

8. To write an indirect proof that two lines are perpendicular, begin by assuming the two lines are not perpendicular.

A

9. The length of the longest side of a triangle is always greater than the sum of the lengths of the other two sides.

D

10. In two triangles, if two pairs of sides are congruent, then the measure of the included angles determines which triangle has the longer third side.

A

Step 1

Step 2

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Perpendicular Bisector A perpendicular bisector is a line, segment, or ray that is perpendicular to the given segment and passes through its midpoint. Some theorems deal with perpendicular bisectors.

Perpendicular Bisector

Theorem

If a point is on the perpendicular bisector of a segment, then it is equidistant

from the endpoints of the segment.

Converse of Perpendicular

Bisector Theorem

If a point is equidistant from the endpoints of a segment, then it is on the

perpendicular bisector of the segment.

Circumcenter TheoremThe perpendicular bisectors of the sides of a triangle intersect at a point called

the circumcenter that is equidistant from the vertices of the triangle.

5-1 Study Guide and InterventionBisectors of Triangles

Find the measure of FM.

2.8

−− FK is the perpendicular bisector of

−−− GM .

FG = FM2.8 = FM

−−

BD is the perpendicular bisector of

−− AC . Find x.

3x + 8

5x - 6B

C

D

A

AD = DC 3x + 8 = 5x - 6 14 = 2x 7 = x

Example 1 Example 2

ExercisesFind each measure.

1. XW 2. BF

7.5 5

5

4.2

19 19

Point P is the circumcenter of �EMK. List any segment(s) congruent to each segment below.

3. −−−

MY 4. −−

KP

5. −−−

MN 6. −−−

ER

7.5 4.2

−−

YE

−−

NK −−RK

−−

MP , −−

EP

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Angle Bisectors Another special segment, ray, or line is an angle bisector, which divides an angle into two congruent angles.

Angle Bisector

Theorem

If a point is on the bisector of an angle, then it is equidistant from the sides

of the angle.

Converse of Angle

Bisector Theorem

If a point in the interior of an angle if equidistant from the sides of the angle, then

it is on the bisector of the angle.

Incenter TheoremThe angle bisectors of a triangle intersect at a point called the incenter that is

equidistant from the sides of the triangle.

5-1 Study Guide and Intervention (continued)

Bisectors of Triangles

⎯⎯ � MR is the angle bisector of ∠NMP. Find x if m∠1 = 5x + 8 and m∠2 = 8x - 16.

12

N R

PM

�� MR is the angle bisector of ∠NMP, so m∠1 = m∠2.5x + 8 = 8x - 16 24 = 3x 8 = x

ExercisesFind each measure.

1. ∠ABE 2. ∠YBA

43°

47° 8

8

3. MK 4. ∠EWL

3x - 82x + 1

(3x + 21)°(7x + 5)°

Point U is the incenter of �GHY. Find each measure below.

5. MU 6. ∠UGM

7. ∠PHU 8. HU

Example

28°

21°12

5

43 47

19 33

5

21

28

13

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5-1 Skills PracticeBisectors of Triangles

Find each measure.

1. FG 2. KL

5x - 1713

133x + 1

4.2

3. TU 4. ∠LYF

2x + 24

5x - 30

58°

5. IU 6. ∠MYW

19°19°

2x + 5

7x

(2x + 5)°(4x - 1)°

Point P is the circumcenter of �ABC. List any segment(s) congruent to each segment below.

7. −−−

BR

8. −−

CS

9. −−

BP

Point A is the incenter of �PQR. Find each measure below.

10. ∠ARU

11. AU

12. ∠QPK 40°

20°

(4x - 9)°(3x + 2)°

28 4.2

60 58

7 11

−−

AR

−−

AP , −−

CP

40

20

35

−−

AS

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Inscribed and Circumscribed CirclesThe three angle bisectors of a triangle intersect in a single point called the incenter. This point is the center of a circle that just touches the three sides of the triangle. Except for the three points where the circle touches the sides, the circle is inside the triangle. The circle is said to be inscribed in the triangle.

1. With a compass and a straightedge, construct the inscribed circle for �PQR by following the steps below.

Step 1 Construct the bisectors of ∠R and ∠Q. Label the point where the bisectors meet, A.

Step 2 Construct a perpendicular segment from A to −−−

RQ . Use the letter B to label the point where the perpendicular segment intersects

−−− RQ .

Step 3 Use a compass to draw the circle with center at A and radius

−− AB .

Construct the inscribed circle in each triangle.

2. 3.

The three perpendicular bisectors of the sides of a triangle also meet in a single point. This point is the center of the circumscribed circle, which passes through each vertex of the triangle. Except for the three points where the circle touches the triangle, the circle is outside the triangle.

4. Follow the steps below to construct the circumscribed circle for �FGH.

Step 1 Construct the perpendicular bisectors of −−−

FG and −−−

FH . Use the letter A to label the point where the perpendicular bisectors meet.

Step 2 Draw the circle that has center A and radius −−

AF .

Construct the circumscribed circle for each triangle.

5. 6.

F H

G

A

P

QR

A

B

5-1 Enrichment

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Medians A median is a line segment that connects a vertex of a triangle to the midpoint of the opposite side. The three medians of a triangle intersect at the centroid of the triangle. The centroid is located two thirds of the distance from a vertex to the midpoint of the side opposite the vertex on a median.

In �ABC, U is the centroid and

BU = 16. Find UK and BK.

BU = 2 − 3 BK

16 = 2 − 3 BK

24 = BK

BU + UK = BK 16 + UK = 24

UK = 8

ExercisesIn �ABC, AU = 16, BU = 12, and CF = 18. Find each measure.

1. UD 2. EU

3. CU 4. AD

5. UF 6. BE

In �CDE, U is the centroid, UK = 12, EM = 21, and UD = 9. Find each measure.

7. CU 8. MU

9. CK 10. JU

11. EU 12. JD

5-2 Study Guide and InterventionMedians and Altitudes of Triangles

Example

16

12

12

9

8

12

6

6

24

18

24

36

14

7

4.5

13.5

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Answers


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Altitudes An altitude of a triangle is a segment from a vertex to the line containing the opposite side meeting at a right angle. Every triangle has three altitudes which meet at a point called the orthocenter.

The vertices of �ABC are A(1, 3), B(7, 7) and C(9, 3). Find the coordinates of the orthocenter of �ABC.Find the point where two of the three altitudes intersect.

Find the equation of the altitude from

A to −−−

BC .

If −−−

BC has a slope of −2, then the altitude

has a slope of 1 − 2 .

y - y1 = m(x – x1) Point-slope form

y - 3 = 1 − 2 (x – 1) m = 1 −

2 , (x

1, y

1) = A(1, 3)

y - 3 = 1 − 2 x – 1 −

2 Distributive Property

y = 1 − 2 x + 5 −

2 Simplify.

C to −−

AB .

If −−

AB has a slope of 2 − 3 , then the altitude has a

slope of - 3 − 2 .

y - y1 = m(x - x1) Point-slope form

y - 3 = - 3 − 2 (x - 9) m = - 3 −

2 , (x

1, y

1) = C(9, 3)

y - 3 = - 3 − 2 x + 27 −

2 Distributive Property

y = - 3 − 2 x + 33 −

2 Simplify.

Solve the system of equations and find where the altitudes meet.

y = 1 − 2 x + 5 −

2 y = - 3 −

2 x + 33 −

2

1 − 2 x + 5 −

2 = - 3 −

2 x + 33 −

2 Subtract 1 −

2 x from each side.

5 − 2 = −2x + 33 −

2 Subtract 33

− 2 from each side.

−14 = −2x Divide both sides by -2.

7 = xy = 1 −

2 x + 5 −

2 = 1 −

2 (7) + 5 −

2 = 7 −

2 + 5 −

2 = 6

The coordinates of the orthocenter of �ABC is (6, 7).

ExercisesCOORDINATE GEOMETRY Find the coordinates of the orthocenter of each triangle.

1. J(1, 0), H(6, 0), I(3, 6) 2. S(1, 0), T(4, 7), U(8, −3)


Medians and Altitudes of Triangles

Example

y

x

(9, 3)(1, 3)

(7, 7)

(3, 1) 5 −

2 , 3 −

2

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5-2

In �PQR, NQ = 6, RK = 3, and PK = 4. Find each length.

1. KM 2. KQ

2 4

3. LK 4. LR

1.5 4.5

5. NK 6. PM

2 6

In �STR, H is the centroid, EH = 6, DH = 4, and SM = 24. Find each length.

7. SH 8. HM

16 8

9. TH 10. HR

8 12

11. TD 12. ER

12 18

COORDINATE GEOMETRY Find the coordinates of the centroid of each triangle.

13. X(−3, 15) Y(1, 5), Z(5, 10) 14. S(2, 5), T(6, 5), R(10, 0)

(1, 10) (6, 3 1 − 3 )

COORDINATE GEOMETRY Find the coordinates of the orthocenter of each triangle.

15. L(8, 0), M(10, 8), N(14, 0) 16. D(−9, 9), E(−6, 6), F(0, 6) (10, 1) (-9, -3)

Skills PracticeMedians and Altitudes of Triangles

3

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In �ABC, CP = 30, EP = 18, and BF = 39. Find each length.

1. PD 2. FP

3. BP 4. CD

5. PA 6. EA

In �MIV, Z is the centroid, MZ = 6, YI = 18, and NZ = 12. Find each measure.

7. ZR 8. YZ

9. MR 10. ZV

11. NV 12. IZ

COORDINATE GEOMETRY Find the coordinates of the centroid of each triangle.

13. I(3, 1), J(6, 3), K(3, 5) 14. H(0, 1), U(4, 3), P(2, 5)

COORDINATE GEOMETRY Find the coordinates of the orthocenter of each triangle.

15. P(-1, 2), Q(5, 2), R(2, 1) 16. S(0, 0), T(3, 3), U(3, 6)

17. MOBILES Nabuko wants to construct a mobile out of flat triangles so that the surfaces of the triangles hang parallel to the floor when the mobile is suspended. How can Nabuko be certain that she hangs the triangles to achieve this effect?

5-2 PracticeMedians and Altitudes of Triangles

A

C

F

E

DP

B18 30

26 45

36 54

3 6

9 24

36 12

(4, 3) (2, 3)

(2, -1) (0, 9)

She needs to hang each triangle from its center of gravity or centroid, which is the point at which the three medians of the triangle intersect.

15 13

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1. BALANCING Johanna balanced a triangle flat on her finger tip. What point of the triangle must Johanna be touching?

2. REFLECTIONS Part of the working space in Paulette’s loft is partitioned in the shape of a nearly equilateral triangle with mirrors hanging on all three partitions. From which point could someone see the opposite corner behind his or her reflection in any of the three mirrors?

3. DISTANCES For what kind of triangle is there a point where the distance to each side is half the distance to each vertex? Explain.

4. MEDIANS Look at the right triangle below. What do you notice about the orthocenter and the vertices of the triangle?

5. PLAZAS An architect is designing a triangular plaza. For aesthetic purposes, the architect pays special attention to the location of the centroid C and the circumcenter O.

a. Give an example of a triangular plaza where C = O. If no such example exists, state that this is impossible.

b. Give an example of a triangular plaza where C is inside the plaza and O is outside the plaza. If no such example exists, state that this is impossible.

c. Give an example of a triangular plaza where C is outside the plaza and O is inside the plaza. If no such example exists, state that this is impossible.

5-2 Word Problem PracticeMedians and Altitudes of Triangles

centroid

orthocenter

equilateral: incenter = centroid = circumcenter

The orthocenter coincides with one of the vertices.

an equilateral triangle

an obtuse triangle

impossible

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Answers


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Constructing Centroids and OrthocentersThe three medians of a triangle intersect at a single point called the centroid. You can use a straightedge and compass to find the centroid of a triangle.

1. With a straightedge and compass, construct the centroid for �STU by following the steps below.

Step 1 Locate the midpoints of sides TU and SU. Label the midpoints A and B respectively.

Step 2 Draw the segments SA and TB. Use the letter H to label their point of intersection, which is the centroid of �STU.

Construct the centroid of each triangle.

2. 3.

The three altitudes of a triangle meet in a single point called the orthocenter of the triangle.

4. Follow the steps below to construct the orthocenter of �CDE using a straightedge and compass.

Step 1 Extend segments CD and DE past point D long enough to meet perpendiculars from E and C as shown.

Step 2 Construct the perpendicular from point C to the line DE and label the point of intersection X. Likewise, label the point of intersection of line CD with the perpendicular from E as point Z. In this case both X and Z lie outside �CDE.

Step 3 Label O the point where perpendiculars � �� CX and � �� EZ intersect. This is the orthocenter of �CDE.

Construct the orthocenter of each triangle. 5. 6.

5-2 Enrichment

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Angle Inequalities Properties of inequalities, including the Transitive, Addition, and Subtraction Properties of Inequality, can be used with measures of angles and segments. There is also a Comparison Property of Inequality.

For any real numbers a and b, either a < b, a = b, or a > b.

The Exterior Angle Inequality Theorem can be used to prove this inequality involving an exterior angle.

Exterior Angle

Inequality Theorem

The measure of an exterior angle of a triangle

is greater than the measure of either of its

corresponding remote interior angles.

A C D1

B

m∠1 > m∠A, m∠1 > m∠B

List all angles of �EFG whose measures are less than m∠1.

The measure of an exterior angle is greater than the measure of either remote interior angle. So m∠3 < m∠1 and m∠4 < m∠1.

ExercisesUse the Exterior Angle Inequality Theorem to list all of the angles that satisfy the stated condition.

1. measures are less than m∠1

2. measures are greater than m∠3









M J K

3

4 521

L

Exercises 1–2

H E F3

4

21

G

5-3 Study Guide and InterventionInequalities in One Triangle

Example

R O

Q

N

P3 456

Exercises 9–10

78

21

S

X T W V

3

4

5

67 2 1

U

Exercises 3–8

∠3, ∠4

∠1, ∠5

∠5, ∠6

∠7

∠4

∠1, ∠7, ∠TUV

∠2, ∠3

∠1, ∠3, ∠5, ∠6, ∠TUV

∠4, ∠5, ∠7, ∠NPR

∠1, ∠8, ∠OPN, ∠ROQ

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Angle-Side Relationships When the sides of triangles are not congruent, there is a relationship between the sides and angles of the triangles.

• If one side of a triangle is longer than another side, then the angle opposite the longer side has a greater measure than the angle opposite the shorter side.

• If one angle of a triangle has a greater measure than another angle, then the side opposite the greater angle is longer than the side opposite the lesser angle.

B C

A

List the angles in order from smallest to largest measure.

R T9 cm

6 cm 7 cm

S

∠T, ∠R, ∠S

List the sides in order from shortest to longest.

A B

C

20°

35°

125°

−−− CB ,

−− AB ,

−− AC

ExercisesList the angles and sides in order from smallest to largest.

1.

T S

R48 cm

23.7 cm

35 cm

2.

R T

S

60°

80°

40°

3.

A C

B

3.8 4.3

4.0

4. 14

11

5

5.

45

8

6.

20

12

7. 35°

120° 25°

8.

56° 58°

9.

60° 54°


Inequalities in One Triangle

If AC > AB, then m∠B > m∠C.

If m∠A > m∠C, then BC > AB.

Example 1 Example 2

∠T, ∠R, ∠S ∠T, ∠R, ∠S ∠C, ∠B, ∠A

−−

RS , −−

ST , −−

RT −−

RS , −−

ST , −−

RT −−

AB , −−

AC , −−

BC

∠S, ∠U, ∠T, ∠B, ∠C, ∠A, ∠Q, ∠P, ∠R,

−−

UT , −−

ST , −−

SU −−

AC , −−

BA , −−

CB −−

PR , −−

RQ , −−

QP

∠E, ∠C, ∠D, ∠X, ∠Z, ∠Y, ∠T, ∠S, ∠R,

−−

CD , −−

DE , −−

CE −−

YZ , −−

XY , −−

XZ −−

RS , −−

RT , −−

ST

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5-3

Use the Exterior Angle Inequality Theorem to list all of theangles that satisfy the stated condition.

1. measures less than m∠1

∠2, ∠3, ∠4, ∠5, ∠7, ∠8

2. measures less than m∠9

∠2, ∠4, ∠6, ∠7

3. measures greater than m∠5

∠1, ∠3

4. measures greater than m∠8

∠1, ∠3, ∠5

List the angles and sides of each triangle in order from smallest to largest.

5.

5

62

6. 24°

98°

∠Q, ∠R, ∠S, −−

RP , −−

PQ , −−

RQ ∠K, ∠M, ∠L, −−

ML , −−

KL , −−

KM

7.

15

916

8.

38

39

34

∠F, ∠H, ∠G, −−

HG , −−

FG , −−

FH ∠X, ∠Y, ∠Z, −−

YZ , −−

XZ , −−

XY

9. 10.

∠A, ∠B, ∠C, −−

BC , −−

AC , −−

AB ∠S, ∠U, ∠T, −−

UT , −−

ST , −−

SU

1

2 4

6

7

8 93 5

Skills PracticeInequalities in One Triangle

98°

43°

42°

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Use the figure at the right to determine which angle has the greatest measure.

1. ∠1, ∠3, ∠4 2. ∠4, ∠8, ∠9

3. ∠2, ∠3, ∠7 4. ∠7, ∠8, ∠10

Use the Exterior Angle Inequality Theorem to list all angles that satisfy the stated condition.





Use the figure at the right to determine the relationship between the measures of the given angles.

9. m∠QRW, m∠RWQ 10. m∠RTW, m∠TWR

11. m∠RST, m∠TRS 12. m∠WQR, m∠QRW

Use the figure at the right to determine the relationship between the lengths of the given sides.

13. −−−

DH , −−−

GH 14. −−−

DE , −−−

DG

15. −−−

EG , −−−

FG 16. −−−

DE , −−−

EG

17. SPORTS The figure shows the position of three trees on one part of a Frisbee™ course. At which tree position is the angle between the trees the greatest?

53 ft

40 ft

3

2

1

37.5 ft

120°32°

48° 113°

17°H

D E F

G

3447

45

44

22

14

35

Q

R

S

TW

12

4 6

7 89

35

12

4 678 9

10

3

5

5-3 PracticeInequalities in One Triangle

∠1 ∠4

∠7 ∠10

∠3, ∠4, ∠5, ∠7, ∠8

∠5, ∠7, ∠8

∠1, ∠3, ∠5, ∠9

∠6, ∠9

m∠QRW < m∠RWQ m∠RTW < m∠TWR

m∠RST > m∠TRS m∠WQR < m∠QRW

−−

DH > −−

GH −−

DE < −−

DG

−−

EG < −−

FG −−

DE > −−

EG

2

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1. DISTANCE Carl and Rose live on the same straight road. From their balconies they can see a flagpole in the distance. The angle that each person’s line of sight to the flagpole makes with the road is the same. How do their distances from the flagpole compare?

2. OBTUSE TRIANGLES Don notices that the side opposite the right angle in a right triangle is always the longest of the three sides. Is this also true of the side opposite the obtuse angle in an obtuse triangle? Explain.

3. STRING Jake built a triangular structure with three black sticks. He tied one end of a string to vertex Mand the other end to a point on the stick opposite M, pulling the string taut. Prove that the length of the string cannot exceed the longer of the two sides of the structure.

string

M

4. SQUARES Matthew has three different squares. He arranges the squares to form a triangle as shown. Based on the information, list the squares in order from the one with the smallest perimeter to the one with the largest perimeter.

54˚47˚

3

1 2

5. CITIES Stella is going to Texas to visit a friend. As she was looking at a map to see where she might want to go, she noticed the cities Austin, Dallas, and Abilene formed a triangle. She wanted to determine how the distances between the cities were related, so she used a protractor to measure two angles.

a. Based on the information in the figure, which of the two cities are nearest to each other?

b. Based on the information in the figure, which of the two cities are farthest apart from each other?

5-3

59˚

64˚

Abilene

Dallas

Austin

Word Problem PracticeInequalities in One Triangle

They are equal.

Yes. Since an obtuse triangle

only has 1 obtuse angle and 2 acute angles, the side opposite the obtuse angle is the longest

side.

Sample answer: The string

divides the triangle in two; one of these triangles is right or obtuse

because one side of the string

must make a right or obtuse angle with the stick. In this triangle, the side opposite the right or obtuse angle is longer

than the string and that side is also a side of the triangle.

2, 1, 3

Dallas and Abilene

Abilene and Austin

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Construction ProblemThe diagram below shows segment AB adjacent to a closed region. The problem requires that you construct another segment XY to the right of the closed region such that points A, B, X, and Y are collinear. You are not allowed to touch or cross the closed region with your compass or straightedge.

Follow these instructions to construct a segment XY so that it is collinear with segment AB.

1. Construct the perpendicular bisector of −−

AB . Label the midpoint as point C, and the line as m.

2. Mark two points P and Q on line m that lie well above the closed region. Construct the perpendicular bisector, n, of

−−− PQ . Label the intersection of lines m and n as point D.

3. Mark points R and S on line n that lie well to the right of the closed region. Construct the perpendicular bisector, k , of

−− RS . Label the intersection of lines n and k as point E.

4. Mark point X on line k so that X is below line n and so that −−

EX is congruent to −−−

DC .

5. Mark points T and V on line k and on opposite sides of X, so that −−

XT and −−

XV are congruent. Construct the perpendicular bisector, �, of

−− TV . Call the point where the

line � hits the boundary of the closed region point Y. −−

XY corresponds to the new road.

Q

P

mk

�

nD

R E

T

X

V

YBA

C

S

ExistingRoad

Closed Region(Lake)

5-3 Enrichment

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5-3

Cabri Junior can be used to investigate the relationships between angles and sides of a triangle. Step 1 Use Cabri Junior. to draw and label a triangle. • Select F2 Triangle to draw a triangle. • Move the cursor to where you want the first vertex. Press ENTER . • Repeat this procedure to determine the next two vertices of the triangle. • Select F5 Alph-num to label each vertex. • Move the cursor to a vertex, press ENTER , enter A, and press ENTER again. • Repeat this procedure to label vertex B and vertex C.Step 2 Draw an exterior angle of �ABC. • Select F2 Line to draw a line through

−−− BC .

• Select F2 Point, Point on to draw a point on � �� BC so that C is between B and the new point.

• Select F5 Alph-num to label the point D. Step 3 Find the measures of the three interior angles and the exterior angle, ∠ACD. • Select F5 Measure, Angle. • To find the measure of ∠ABC, select points A, B, and

C (with the vertex B as the second point selected). • Repeat to find the remaining angle measures.Step 4 Find the measure of each side of �ABC. • Select F5 Measure, D. & Length. • To find the length of

−− AB , select point A and then select point B.

• Repeat this procedure to find the lengths of −−−

BC and −−

CA .

ExercisesAnalyze your drawing. 1. What is the relationship between m∠ACD and m∠ABC? m∠ACD and m∠BAC?

Sample answer: m∠ACD > m∠ABC; m∠ACD > m∠BAC

2. Make a conjecture about the relationship between the measures of an exterior angle (∠ACD) and its two remote interior angles (∠ABC and ∠BAC).

The measure of an exterior angle is equal to the sum of the measure of

the two remote interior angles.

3. Change the dimensions of the triangle by moving point A. (Press CLEAR so the pointer becomes a black arrow. Move the pointer close to point A until the arrow becomes transparent and point A is blinking. Press ALPHA to change the arrow to a hand. Then move the point.) Is your conjecture still true? yes

4. Which side of the triangle is the longest? the shortest? See students’ work.

5. Which angle measure (not including the exterior angle) is the greatest? the least? See students’ work.

6. Make a conjecture about where the longest side is in relationship to the greatest angle and where the shortest side is in relationship to the least angle.

The longest side is opposite the greatest angle. The shortest side is opposite the least angle.

Graphing Calculator ActivityCabri Junior: Inequalities in One Triangle

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Answers


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lencoe/M

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panies, Inc.

NAME DATE PERIOD


5-3

The Geometer’s Sketchpad can be used to investigate the relationships between angles and sides of a triangle.

Step 1 Use The Geometer’s Sketchpad to draw a triangle and one exterior angle. • Construct a ray by selecting the

Ray tool from the toolbar. First, click where you want the first point. Then click a second point to draw the ray.

• Next, select the Segment tool from the toolbar. Use the endpoint of the ray as the first point for the segment and click on a second point to construct the segment.

• Construct another segment joining the second point of the previous segment to a point on the ray.

• Display the labels for each point. Use the Selection Arrow tool to select all four points. Display the labels by selecting Show Label from the Display menu.

Step 2 Find the measures of each angle. • To find the measure of ∠ABC, use the Selection Arrow tool to select points

A, B, and C (with the vertex B as the second point selected). Then, under the Measure menu, select Angle. Use this method to find the remaining angle measures, including the exterior angle, ∠BCD.

Step 3 Find the measures of each side of the triangle. • To find the measure of side AB, select A and then B. Next, under the Measure

menu, select Distance. Use this method to find the length of the other two sides.

ExercisesAnalyze your drawing. 1. What is the relationship between m∠BCD and m∠ABC? m∠BCD and m∠BAC?

Sample answer: m∠BCD > m∠ABC; m∠BCD > m∠BAC

2. Make a conjecture about the relationship between the measures of an exterior angle (∠BCD) and its two remote interior angles (∠ABC and ∠BAC).

The measure of an exterior angle is equal to the sum of the measure of

the two remote interior angles.

3. Change the dimensions of the triangle by selecting point A with the pointer tool and moving it. Is your conjecture still true? yes

4. Which side of the triangle is the longest? the shortest? See students’ work.

5. Which angle measure (not including the exterior angle) is the greatest? the least? See students’ work.

6. Make a conjecture about where the longest side is in relationship to the greatest angle and where the shortest side is in relationship to the least angle.

The longest side is opposite the greatest angle. The shortest side is opposite the least angle.

A

B

C D

m�ABC � 69.29˚m�BCA � 55.92˚m�BAC � 54.78˚m�BCD � 124.08˚AB � 2.20 cm

BC � 2.17 cm

AC � 2.49 cm

Geometer’s Sketchpad ActivityInequalities in One Triangle

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Indirect Algebraic Proof One way to prove that a statement is true is to temporarily assume that what you are trying to prove is false. By showing this assumption to be logically impossible, you prove your assumption false and the original conclusion true. This is known as an indirect proof.

Steps for Writing an Indirect Proof

1. Assume that the conclusion is false by assuming the oppposite is true.

2. Show that this assumption leads to a contradiction of the hypothesis or some other fact.

3. Point out that the assumption must be false, and therefore, the conclusion must be true.

Given: 3x + 5 > 8 Prove: x > 1

Step 1 Assume that x is not greater than 1. That is, x = 1 or x < 1.

Step 2 Make a table for several possibilities for x = 1 or x < 1. When x = 1 or x < 1, then 3x + 5 is not greater than 8.

Step 3 This contradicts the given information that 3x + 5 > 8. The assumption that x is not greater than 1 must be false, which means that the statement “x > 1” must be true.

ExercisesState the assumption you would make to start an indirect proof of each statement.

1. If 2x > 14, then x > 7.

2. For all real numbers, if a + b > c, then a > c - b.

Complete the indirect proof.

Given: n is an integer and n2 is even.Prove: n is even.

3. Assume that

4. Then n can be expressed as 2a + 1 by

5. n2 = Substitution

6. = Multiply.

7. = Simplify.

8. = 2(2a2 + 2a) + 1

9. 2(2a2 + 2a)+ 1 is an odd number. This contradicts the given that n2 is even,

so the assumption must be

10. Therefore,

x ≤ 7

a ≤ c - b

n is not even. That is, assume n is odd.

the meaning of odd number.

(2a + 1)2

(2a + 1)(2a + 1)

Distributive Property

false.

n is even.

5-4 Study Guide and InterventionIndirect Proof

Example

x 3x + 5

1 8

0 5

-1 2

-2 -1

-3 -4

4a2 + 4a + 1

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Indirect Proof with Geometry To write an indirect proof in geometry, you assume that the conclusion is false. Then you show that the assumption leads to a contradiction. The contradiction shows that the conclusion cannot be false, so it must be true.

Given: m∠C = 100 Prove: ∠A is not a right angle.

Step 1 Assume that ∠A is a right angle.

Step 2 Show that this leads to a contradiction. If ∠A is a right angle, then m∠A = 90 and m∠C + m∠A = 100 + 90 = 190. Thus the sum of the measures of the angles of �ABC is greater than 180.

Step 3 The conclusion that the sum of the measures of the angles of �ABC is greater than 180 is a contradiction of a known property. The assumption that ∠A is a right angle must be false, which means that the statement “∠A is not a right angle” must be true.

ExercisesState the assumption you would make to start an indirect proof of each statement.

1. If m∠A = 90, then m∠B = 45.

2. If −−

AV is not congruent to −−

VE , then �AVE is not isosceles.

Complete the indirect proof.

Given: ∠1 � ∠2 and −−−

DG is not congruent to −−−

FG .Prove:

−−− DE is not congruent to

−− FE .

3. Assume that Assume the conclusion is false.

4. −−−

EG � −−−

EG

5. �EDG � �EFG

6.

7. This contradicts the given information, so the assumption must

be

8. Therefore,

12

D G

FE

A B

C

m∠B ≠ 45

�AVE is isosceles.

−−

DE � −−

FE .

Reflexive Property

SAS

−−

DG � −−

FG CPCTC

false.

−−

DE is not congruent to −−

FE .


Indirect Proof

Example

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5-4 Skills PracticeIndirect Proof

State the assumption you would make to start an indirect proof of each statement.

1. m∠ABC < m∠CBA

m∠ABC ≥ m∠CBA

2. �DEF � �RST

�DEF � �RST

3. Line a is perpendicular to line b.

Line a is not perpendicular to line b.

4. ∠5 is supplementary to ∠6.

∠5 is not supplementary to ∠6.

Write an indirect proof of each statement.

5. Given: x2 + 8 ≤ 12 Prove: x ≤ 2

Proof:

Step 1: Assume x > 2.

Step 2: If x > 2, then x2 > 4. But if x2 > 4, it follows that x2 + 8 > 12. This contradicts the given fact that x2 + 8 ≤ 12.

Step 3: Since the assumption of x > 2 leads to a contradiction, it must be false. Therefore, x ≤ 2 must be true.

6. Given: ∠D � ∠F Prove: DE ≠ EF

Proof:

Step 1: Assume DE = EF.

Step 2: If DE = EF, then −−

DE � −−

EF by the definition of congruent segments. But if

−− DE �

−− EF , then ∠D � ∠F by the Isosceles

Triangle Theorem. This contradicts the given information that ∠D � ∠F.

Step 3: Since the assumption that DE = EF leads to a contradiction, it must be false. Therefore, it must be true that DE ≠ EF.

D F

E

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Answers


P

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NAME DATE PERIOD


State the assumption you would make to start an indirect proof of each statement.

1. −−−

BD bisects ∠ABC.

2. RT = TS

Write an indirect proof of each statement.

3. Given: -4x + 2 < -10 Prove: x > 3

4. Given: m∠2 + m∠3 ≠ 180 Prove: a ∦ b

5. PHYSICS Sound travels through air at about 344 meters per second when the temperature is 20°C. If Enrique lives 2 kilometers from the fire station and it takes 5 seconds for the sound of the fire station siren to reach him, how can you prove indirectly that it is not 20°C when Enrique hears the siren?

12

3

a

b

−−

BD does not bisect ∠ABC.

RT ≠ TS

Proof:

Step 1 Assume x ≤ 3.

Step 2 If x ≤ 3, then -4x ≥ -12. But -4x ≥ -12 implies that -4x + 2 ≥ -10, which contradicts the given inequality.

Step 3 Since the assumption that x ≤ 3 leads to a contradiction, it must be true that x > 3.

Proof:

Step 1 Assume a || b.

Step 2 If a || b, then the consecutive interior angles ∠2 and ∠3 are supplementary. Thus m∠2 + m∠3 = 180. This contradicts

the given statement that m∠2 + m∠3 ≠ 180.

Step 3 Since the assumption leads to a contradiction, the statement

a || b must be false. Therefore, a ∦ b must be true.

Assume that it is 20°C when Enrique hears the siren, then show that at this temperature it will take more than 5 seconds for the sound of the siren to reach him. Since the assumption is false, you will have proved that it is not 20°C when Enrique hears the siren.

5-4 PracticeIndirect Proof

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1. CANOES Thirty-five students went on a canoeing expedition. They rented 17 canoes for the trip. Use an indirect proof to show that at least one canoe had more than two students in it.

2. AREA The area of the United States is about 6,000,000 square miles. The area of Hawaii is about 11,000 square miles. Use an indirect proof to show that at least one of the fifty states has an area greater than 120,000 square miles.

3. CONSECUTIVE NUMBERS David was trying to find a common factor other than 1 between various pairs of consecutive integers. Write an indirect proof to show David that two consecutive integers do not share a common factor other than 1.

4. WORDS The words accomplishment, counterexample, and extemporaneous all have 14 letters. Use an indirect proof to show that any word with 14 letters must use a repeated letter or have two letters that are consecutive in the alphabet.

Suppose the letters are distinct and nonconsecutive. Then the alphabet must have at least 14 + 13 or 27 letters, a contradiction.

5. LATTICE TRIANGLES A lattice point is a point whose coordinates are both integers. A lattice triangle is a triangle whose vertices are lattice points. It is a fact that a lattice triangle has an area of at least 0.5 square units.

y

xO

A

B

C

5

5

a. Suppose �ABC has a lattice point in its interior. Show that the lattice triangle can be partitioned into three smaller lattice triangles.

b. Prove indirectly that a lattice triangle with area 0.5 square units contains no lattice point. (Being on the boundary does not count as inside.)

Sample answer: Suppose all

canoes had ≤ 2 students, then the total would be less than or equal to 17 × 2 = 34,

a contradiction.

Sample answer: Suppose no state has area > 120,000 mi2.

Then the total area could not exceed 120,000 × 49 + 11,000 =

5,891,000, a contradiction.

Sample answer in diagram above.

5-4 Word Problem PracticeIndirect Proof

Sample answer: From Exercise 5a, the lattice triangle contains

3 smaller lattice triangles, each of which has area at least 0.5 square units. The original

would then have area at least 1.5 square units, a contradiction.

Sample answer: Assume x and y are integers with a common factor greater than 1. For consecutive integers one is even and the other is odd, so x = 2a and y = 2a + 1, for an integer a. Let n be the common factor greater that 1. Therefore

x − n =

2a −

n is an integer and

y −

n =

2a+1 −

n

is also an integer. But 2a+1

− n =

2a −

n

+ 1 − n and 1 −

n is not an integer unless

n = 1, a contradiction.

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More Counterexamples

Some statements in mathematics can be proven false by counterexamples. Consider the following statement.

For any numbers a and b, a - b = b - a.

You can prove that this statement is false in general if you can fi nd one example for which the statement is false.

Let a = 7 and b = 3. Substitute these values in the equation above.

7 - 3 � 3 - 7 4 ≠ -4

In general, for any numbers a and b, the statement a - b = b - a is false. You can make the equivalent verbal statement: subtraction is not a commutative operation.

In each of the following exercises a, b, and c are any numbers. Prove that the statement is false by counterexample.

1. a - (b - c) � (a - b) - c 2. a ÷ (b ÷ c) � (a ÷ b) ÷ c

3. a ÷ b � b ÷ a 4. a ÷ (b + c) � (a ÷ b) + (a ÷ c)

5. a + (bc) � (a + b)(a + c) 6. a2 + a2 � a4

7. Write the verbal equivalents for Exercises 1, 2, and 3.

8. For the Distributive Property, a(b + c) = ab + ac, it is said that multiplication distributes over addition. Exercises 4 and 5 prove that some operations do not distribute. Write a statement for each exercise that indicates this.

Sample answers are given.

6 - (4 - 2) � (6 - 4) - 2 6 ÷ (4 ÷ 2) � (6 ÷ 4) ÷ 2

6 - 2 � 2 - 2 6 − 2 � 1.5

− 2

4 ≠ 0 3 ≠ 0.75

1. Subtraction is not an associative operation.2. Division is not an associative operation.3. Division is not a commutative operation.

4. Division does not distribute over addition.5. Addition does not distribute over multiplication.

5-4 Enrichment

6 ÷ 4 � 4 ÷ 6 6 ÷ (4 + 2) � (6 ÷ 4) +(6 ÷ 2)

3 − 2 ≠

2 −

3 6 ÷ 6 � 1.5 + 3

1 ≠ 4.5

6 + (4 . 2) � (6 + 4) (6 + 2) 62 + 62 � 64

6 + 8 � (10) (8) 36 + 36 � 1296

14 � 80 72 ≠ 1296

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The Triangle Inequality If you take three straws of lengths 8 inches, 5 inches, and 1 inch and try to make a triangle with them, you will find that it is not possible. This illustrates the Triangle Inequality Theorem.

Triangle Inequality

Theorem

The sum of the lengths of any two sides of a

triangle must be greater than the length of the third side. BC

A

a

cb

The measures of two sides of a triangle are 5 and 8. Find a range for the length of the third side.

By the Triangle Inequality Theorem, all three of the following inequalities must be true. 5 + x > 8 8 + x > 5 5 + 8 > x

x > 3 x > -3 13 > xTherefore x must be between 3 and 13.

ExercisesIs it possible to form a triangle with the given side lengths? If not, explain why not.

1. 3, 4, 6 2. 6, 9, 15

3. 8, 8, 8 4. 2, 4, 5

5. 4, 8, 16 6. 1.5, 2.5, 3

Find the range for the measure of the third side of a triangle given the measures of two sides.

7. 1 cm and 6 cm 8. 12 yd and 18 yd

9. 1.5 ft and 5.5 ft 10. 82 m and 8 m

11. Suppose you have three different positive numbers arranged in order from least to greatest. What single comparison will let you see if the numbers can be the lengths of the sides of a triangle?

yes no; 6 + 9 = 15

yes yes

no; 4 + 8 < 16 yes

5 cm < n < 7 cm 6 yd < n < 30 yd

4 ft < n < 7 ft 74 m < n < 90 m

Find the sum of the two smaller numbers. If that sum is greater than the largest number, then the three numbers can be the lengths of the sides of a triangle.

5-5 Study Guide and InterventionThe Triangle Inequality

Example

a + b > cb + c > aa + c > b

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Proofs Using The Triangle Inequality Theorem You can use the Triangle Inequality Theorem as a reason in proofs.

Complete the following proof.

Given: �ABC � �DECProve: AB + DE > AD − BE

Proof:

Statements1. �ABC � �DEC2. AB + BC > AC DE + EC > CD3. AB > AC – BC DE > CD – EC 4. AB + DE > AC - BC + CD - EC 5. AB + DE > AC + CD - BC - EC 6. AB + DE > AC + CD - (BC + EC)7. AC + CD = AD BC + EC = BE8. AB + DE > AD - BE

Reasons1. Given

2. Triangle Inequality Theorem

3. Subtraction

4. Addition5. Commutative6. Distributive7. Segment Addition Postulate

8. Substitution

ExercisesPROOF Write a two column proof.

Given: −−

PL ‖ −−−

MT K is the midpoint of

−− PT .

Prove: PK + KM > PL

Proof:Statements

1. −− PL ‖ −−−

MT 2. ∠P � ∠T3. K is the midpoint of

−− PT .

4. PK = KT5. 6. �PKL � �TKM7. 8. 9. PK + KM > PL

Reasons

1. Given

2. Alternate Interior Angles Theorem

3. Given4. Definition of midpoint

5. Vertical Angles Theorem6. ASA

7. Triangle Inequality Theorem8. CPCTC9. Substitution


The Triangle Inequality

∠PKL � ∠MKT

PK + KL > PL

KL = KM

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5-5 Skills PracticeThe Triangle Inequality

Is it possible to form a triangle with the given side lengths? If not, explain why not.

1. 2 ft, 3 ft, 4 ft 2. 5 m, 7 m, 9 m

3. 4 mm, 8 mm, 11 mm 4. 13 in., 13 in., 26 in.

5. 9 cm, 10 cm, 20 cm 6. 15 km, 17 km, 19 km

7. 14 yd, 17 yd, 31 yd 8. 6 m, 7 m, 12 m


9. 5 ft, 9 ft 10. 7 in., 14 in.

11. 8 m, 13 m 12. 10 mm, 12 mm

13. 12 yd, 15 yd 14. 15 km, 27 km

15. 17 cm, 28 cm, 16. 18 ft, 22 ft

17. Proof Complete the proof.

Given: �ABC and �CDE

Prove: AB + BC + CD + DE > AE

Proof:

Statements Reasons1. AB + BC > AC

CD + DE > CE 1. Triangle Inequality Theorem

2. AB + BC + CD + DE > AC + CE 2. Addition Property of Equality

3. AC + CE = AE 3. Seg. Addition Post

4. AB + BC + CD + DE > AE 4. Substitution

yes yes

yes no; 13 + 13 ≯ 26

yes

yes

no; 9 + 10 ≯ 20

no; 14 + 17 ≯ 31

4 ft < n < 14 ft

5 m < n < 21 m

3 yd < n < 27 yd

11 cm < n < 45 cm

7 in. < n < 21 in.

2 mm < n < 22 mm

12 km < n < 42 km

4 ft < n < 40 ft

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Is it possible to form a triangle with the given side lengths? If not explain why not.

1. 9, 12, 18 2. 8, 9, 17

3. 14, 14, 19 4. 23, 26, 50

5. 32, 41, 63 6. 2.7, 3.1, 4.3

7. 0.7, 1.4, 2.1 8. 12.3, 13.9, 25.2


9. 6 ft and 19 ft 10. 7 km and 29 km

11. 13 in. and 27 in. 12. 18 ft and 23 ft

13. 25 yd and 38 yd 14. 31 cm and 39 cm

15. 42 m and 6 m 16. 54 in. and 7 in.

17. Given: H is the centroid of �EDF

Prove: EY + FY > DE

Proof:Statements

1. H is the centroid of �EDF2.

−− EY is a median.

3. Y is the midpoint of −−

DF

4. DY = FY

5. EY + DY > DE6. EY + FY > DE

Reasons

1. Given2. Definition of centroid

3. Definition of median4. Definition of midpoint5. Triangle Inequality Theorem

6. Substitution

18. GARDENING Ha Poong has 4 lengths of wood from which he plans to make a border for a triangular-shaped herb garden. The lengths of the wood borders are 8 inches, 10 inches, 12 inches, and 18 inches. How many different triangular borders can Ha Poong make?

yes no; 8 + 9 = 17

yes no; 23 + 26 < 50

yes yes

no; 0.7 + 1.4 = 2.1 yes

13ft < n < 25ft 22 km < n < 36 km

14 in. < n < 40 in. 5 ft < n < 41 ft

13 yd < n < 63 yd 8 cm < n < 70 cm

36 m < n < 48 m 47 in. < n < 61 in.

3

5-5 PracticeThe Triangle Inequality

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Tanya’s home

Supermarket

Railroad

A B C

1. STICKS Jamila has 5 sticks of lengths 2, 4, 6, 8, and 10 inches. Using three sticks at a time as the sides of triangles, how many triangles can she make?

Use the figure at the right for Exercises 2 and 3.

2. PATHS To get to the nearest super market, Tanya must walk over a railroad track. There are two places where she can cross the track (points A and B). Which path is longer? Explain.

3. PATHS While out walking one day Tanya finds a third place to cross the railroad tracks. Show that the path through point C is longer than the path through point B.

4. CITIES The distance between New York City and Boston is 187 miles and the distance between New York City and Hartford is 97 miles. Hartford, Boston, and New York City form a triangle on a map. What must the distance between Boston and Hartford be greater than?

5. TRIANGLES The figure shows an equilateral triangle ABC and a point Poutside the triangle.

B´

A´

C´

C

P´

P

B

A

a. Draw the figure that is the result of turning the original figure 60° counterclockwise about A. Denote by P', the image of P under this turn.

b. Note that −−−P'B is congruent to

−−PC . It is

also true that −−−PP' is congruent to

−−PA .

Why?

c. Show that −−PA ,

−−PB , and

−−PC satisfy the

triangle inequalities.

3

By the Triangle Inequality

Theorem, the distance from Tanya’s home to point B and on to the supermarket is greater than

the straight distance from Tanya’s home to the Supermarket.

Sample answer: Let S be the Supermarket and T be Tanya’s

home. Because ∠SAB is 90, m∠SBA < 90, so m∠SBC > 90,

making SC > SB. Similarly, CT > BT. Therefore CT + CS > BT + BS.

90 mi

See figure.

Sample answer: �P'PB is a

triangle with side lengths equal to PA, PB, and PC.

5-5 Word Problem PracticeThe Triangle Inequality

Sample answer: PA is congruent to P'A and m∠PAP' is 60°, So by SAS, triangle PP'A is equilateral. Thus, PP' = PA

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Constructing Triangles

The measurements of the sides of a triangle are given. If a triangle having sides with these measurements is not possible, then write impossible. If a triangle is possible, draw it and measure each angle with a protractor.

1. AR = 5 cm m∠A = 2. PI = 8 cm m∠P =

RT = 3 cm m∠R = IN = 3 cm m∠I =

AT = 6 cm m∠T = PN = 2 cm m∠N =

3. ON = 10 cm m∠O = 4. TW = 6 cm m∠T =

NE = 5.3 cm m∠N = WO = 7 cm m∠W =

OE = 4.6 cm m∠E = TO = 2 cm m∠O =

5. BA = 3.l cm m∠B = 6. AR = 4 cm m∠A =

AT = 8 cm m∠A = RM = 5 cm m∠R =

BT = 5 cm m∠T = AM = 3 cm m∠M =

M

RAT

BA

W

T

O

A R

T

30

94

56

impossible

112

15

53

impossible

162 90

11 37

7 53

5-5 Enrichment

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Hinge Theorem The following theorem and its converse involve the relationship between the sides of two triangles and an angle in each triangle.

Hinge Theorem

If two sides of a triangle are congruent to two

sides of another triangle and the included

angle of the first is larger than the included

angle of the second, then the third side of

the first triangle is longer than

the third side of the second triangle. RT > AC

Converse of the

Hinge Theorem

If two sides of a triangle are congruent to

two sides of another triangle, and the

third side in the first is longer than the

third side in the second, then the included

angle in the first triangle is greater than

the included angle in the second triangle. m∠M > m∠R

ExercisesCompare the given measures.

1. MR and RP

N

R

P

M

21°

19°

2. AD and CD C

A

DB

22°

38°

MR > RP AD > CD

3. m∠C and m∠Z 4. m∠XYW and m∠WYZ

m∠C < m∠Z m∠XYW < m∠WYZ

Write an inequality for the range of values of x.5.

115°

120° 24

2440

(4x - 10) 6.

33°

60

60

36

30(3x - 3)°

5-6 Study Guide and InterventionInequalities in Two Triangles

S T80°

R

B C60°

A

3336

TR

SN

M P

Compare the measures

of −−

GF and −−

FE .

H

E

F

G

22°28°

Two sides of �HGF are congruent to two

sides of �HEF, and m∠GHF > m∠EHF. By the Hinge Theorem, GF > FE.

Compare the measures

of ∠ABD and ∠CBD.

13

16

C

D

A

B

Two sides of �ABD are congruent to

two sides of �CBD, and AD > CD. By the Converse of the Hinge Theorem, m∠ABD > m∠CBD.

Example 2Example 1

x > 12.5 x < 12

30C

A X

B30

5048 24

24Z Y

42

28

ZW

XY

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PROVE RELATIONSHIPS IN TWO TRIANGLES You can use the Hinge Theorem and its converse to prove relationships in two triangles.

Given: RX = XS ∠SXT = 97°Prove: ST > RT

Proof:

Statements Reasons 1. ∠SXT and ∠RXT are

supplementary 2. m ∠SXT + m∠RXT = 180°

3. m∠SXT = 97°

4. 97 + m∠RXT = 180 5. m∠RXT = 83 6. 97 > 83 7. m∠SXT > m∠RXT 8. RX = XS 9. TX = TX10. ST > RT

1. Defn of linear pair

2. Defn of supplementary 3. Given 4. Substitution 5. Subtraction 6. Inequality 7. Substitution 8. Given 9. Reflexive10. Hinge Theorem

ExercisesComplete the proof.

Given: rectangle AFBC ED = DCProve: AE > FBProof:Statements Reasons1. rectangle AFBC, ED = DC2. AD = AD3. m∠EDA > m∠ADC4. AE > AC

5. AC = FB

6. AE > FB

1. given2. reflexive3. exterior angle4. Hinge Theorem5. opp sides � in rectangle.6. Substitution


Inequalities Involving Two Triangles

Example

97°

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5-6

Compare the given measures.

1. m∠BXA and m∠DXA

m∠BXA < m∠DXA

2. BC and DC

BC > DC


3. m∠STR and m∠TRU 4. PQ and RQ 31

30

22 22

R S

U T

95°

7 785°P R

S

Q

m∠STR < m∠TRU PQ > RQ

5. In the figure, −−

BA , −−−

BD , −−−

BC , and −−−

BE are congruent and AC < DE. How does m∠1 compare with m∠3? Explain your thinking.

m∠1 < m∠3; From the given information and the SSS Inequality Theorem, it follows that in �ABC and �DBE we have m∠ABC < m∠DBE. Since m∠ABC = m∠1 + m∠2 and m∠DBE = m∠3 + m∠2, it follows that m∠1 + m∠2 < m∠3 + m∠2. Subtract m∠2 from each side of the last inequality to get m∠1 < m∠3.

6. PROOF Write a two-column proof. Given:

−− BA �

−−− DA

BC > DC Prove: m∠1 > m∠2

Proof:

Statements Reasons

1. BA � DA 1. Given 2. BC > DC 2. Given 3. AC � AC 3. Reflexive Property 4. m∠1 > m∠3 4. SSS Inequality

12

3

B

AD C

E

Skills PracticeInequalities Involving Two Triangles

6

98

3

3

B

A C

D

X

12

B

A

D

C

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1 2D F

E

G

20 21

R TS

J K

14 14

14

13

12C F

E

D

(x + 3)°(x - 3)°

10 10

R TS

Q

40°

30°

60°A KM

B


1. AB and BK 2. ST and SR

3. m∠CDF and m∠EDF 4. m∠R and m∠T

5. PROOF Write a two-column proof.

Given: G is the midpoint of −−−

DF .m∠1 > m∠2

Prove: ED > EF

Proof:

Statements Reasons

1. G is the midpoint of −−

DF . 1. Given

2. −−

DG � −−

FG 2. Definition of midpoint

3. −−

EG � −−

EG 3. Reflexive Property

4. m∠1 > m∠2 4. Given

5. ED > EF 5. Hinge Theorem

6. TOOLS Rebecca used a spring clamp to hold together a chair leg she repaired with wood glue. When she opened the clamp, she noticed that the angle between the handles of the clamp decreased as the distance between the handles of the clamp decreased. At the same time, the distance between the gripping ends of the clamp increased. When she released the handles, the distance between the gripping end of the clamp decreased and the distance between the handles increased. Is the clamp an example of the Hinge Theorem or its converse?

AB > BK ST > SR

m∠CDF < m∠EDF m∠R < m∠T

Hinge Theorem

5-6 PracticeInequalities in Two Triangles

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1. CLOCKS The minute hand of a grandfather clock is 3 feet long and the hour hand is 2 feet long. Is the distance between their ends greater at 3:00 or at 8:00?

2. FERRIS WHEEL A Ferris wheel has carriages located at the 10 vertices of a regular decagon.

12

3

4

56

7

8

9

10

Which carriages are farther away from carriage number 1 than carriage number 4?

3. WALKWAY Tyree wants to make two slightly different triangles for his walkway. He has three pieces of wood to construct the frame of his triangles. After Tyree makes the first concrete triangle, he adjusts two sides of the triangle so that the angle they create is smaller than the angle in the first triangle. Explain how this changes the triangle.

4. MOUNTAIN PEAKS Emily lives the same distance from three mountain peaks: High Point, Topper, and Cloud Nine. For a photography class, Emily must take a photograph from her house that shows two of the mountain peaks. Which two peaks would she have the best chance of capturing in one image?

Emily

CloudNine

HighPoint

Topper

12 miles

9 m

iles

10 miles

5. RUNNERS A photographer is taking pictures of three track stars, Amy, Noel, and Beth. The photographer stands on a track, which is shaped like a rectangle with semicircles on both ends.

118˚

36˚

146˚

Photographer

Amy

Noel

Beth

a. Based on the information in the figure, list the runners in order from nearest to farthest from the photographer.

b. Explain how to locate the point along the semicircular curve that the runners are on that is farthest away from the photographer.

8:00

5, 6, and 7

Sample answer: By the Hinge

Theorem, the third side opposite the angle that was made smaller is now shorter than the third side of

the first triangle.

Amy, Beth, Noel

5-6 Word Problem PracticeInequalities in Two Triangles

Topper and Cloud Nine

Extend the line through the photographer and the center

of the semicircle to where it intersects the semicircular track.

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Hinge Theorem

The Hinge Theorem that you studied in this section states that if two sides of a triangle are congruent to two sides of another triangle and the included angle in one triangle has a greater measure than the included angle in the other, then the third side of the first triangle is longer than the third side of the second triangle. In this activity, you will investigate whether the converse, inverse and contrapositive of the Hinge Theorem are also true.

X

Y

Z

Q

S

R

1 2

Hypothesis: XY = QR, YZ = RS, m∠1 > m∠2Conclusion: XZ > QS

1. What is the converse of the Hinge Theorem?

2. Can you find any counterexamples to prove that the converse is false?

3. What is the inverse of the Hinge Theorem?

4. Can you find any counterexamples to prove that the inverse is false?

5. What is the contrapositive of the Hinge Theorem?

6. Can you find any counterexamples to prove that the contrapositive is false?

If two sides of one triangle are congruent to two sides of another triangle, and the third side of the first is longer than the third side

of the second, then the included angle of the first is larger than the included angle of the second.

No, it appears to be true.

If two sides of a triangle are not congruent to two sides of another triangle or the included angle in one triangle does not have a greater

measure than the included angle in the other, then the third side of the first triangle is not longer than the third side of the second triangle.


If the third side of the first triangle is not longer than the third side of the second triangle, then the other two sides are not congruent or the

included angle does not have a greater measure.


5-6 Enrichment

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example 1 example 2 −− answers −−. · pdf file9. the length of the ......

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