Corollary 7.2. SOLUTION: Corollary 7.2 states, if three or more parallel lines cut off congruent segments on one transversal, then they cut off congru...

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ANSWER: 10 2. If XN = 6, XM = 2, and XY = 10, find NZ.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. XY =10, So, MY = 10 – 8 =2.

Use the Triangle Proportionality Theorem.

Use the Triangle Proportionality Theorem.

Substitute.

Solve for NZ.

Solve for MY.

Find XY.

ANSWER: 24 3. In

BC = 15, BE = 6, DC = 12, and AD = 8. Justify your answer.

Determine whether

ANSWER: 10 2. If XN = 6, XM = 2, and XY = 10, find NZ.

SOLUTION: If BC = 15, then EC = 15 – 6 = 9.

Use the Converse of the Triangle Proportionality Theorem. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. XY =10, So, MY = 10 – 8 =2. eSolutions Manual - Powered by Cognero

Use the Triangle Proportionality Theorem.

Therefore, ANSWER: yes;

. Page 1

Therefore,

.

ANSWER: ANSWER: 7-4 Parallel Lines and Proportional Parts 24 3. In

BC = 15, BE = 6, DC = 12, and AD = 8.

yes; 4. In

Justify your answer.

Determine whether

JK = 15, JM = 5, LK = 13, and PK = 9.

Determine whether

Justify your answer.

SOLUTION: JK = 15 and LK = 13. Therefore, MK = 15 – 5 = 10 and LP = 13 – 9 = 4.

SOLUTION: If BC = 15, then EC = 15 – 6 = 9.

Use the Converse of the Triangle Proportionality Theorem.

Use the Converse of the Triangle Proportionality Theorem.

If BC = 15, then EC = 15 – 6 = 9.

Use the Converse of the Triangle Proportionality Theorem.

Therefore,

.

ANSWER:

yes; 4. In

So,

JK = 15, JM = 5, LK = 13, and PK = 9.

Determine whether

Justify your answer.

and

are not parallel.

ANSWER: no; is a midsegment of

Find the value of

x. SOLUTION: JK = 15 and LK = 13. Therefore, MK = 15 – 5 = 10 and LP = 13 – 9 = 4.

Use the Converse of the Triangle Proportionality Theorem.

If BC = 15, then EC = 15 – 6 = 9.

5. SOLUTION:

By the Triangle Midsegment Theorem,

Use the Converse of the Triangle Proportionality Theorem.

Substitute.

ANSWER: 11

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So,

and

are not parallel.

Page 2

So,

and

are not parallel.

ANSWER: 7-4 Parallel Lines and Proportional Parts no; is a midsegment of

Find the value of

x.

ANSWER: 10 7. MAPS Refer to the map. 3rd Avenue and 5th Avenue are parallel. If the distance from 3rd Avenue to City Mall along State Street is 3201 feet, find the distance between 5th Avenue and City Mall along Union Street. Round to the nearest tenth.

5. SOLUTION: By the Triangle Midsegment Theorem, Substitute. SOLUTION: The distance between 5th Avenue and City Mall along State Street is 3201 − 1056 or 2145 feet. Let x be the distance between 5th Avenue and City Mall along Union Street. Use the Triangle Proportionality Theorem.

ANSWER: 11

The distance between 5th Avenue and City Mall along Union Street is 2360.3 ft. ANSWER: 2360.3 ft

6. SOLUTION:

ALGEBRA Find x and y. By the Triangle Midsegment Theorem, Substitute.

8. ANSWER: 10 7. MAPS Refer to the map. 3rd Avenue and 5th Avenue are parallel. If the distance from 3rd Avenue to City Mall along State Street is 3201 feet, find the distance between 5th Avenue and City Mall along Union Street. Round to the nearest tenth.

SOLUTION: We are given that

and

Solve for x.

Solve for y.

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SOLUTION: The distance between 5th Avenue and City Mall along State Street is 3201 − 1056 or 2145 feet.

Page 3

The distance between 5th Avenue and City Mall along Union Street is 2360.3 ft. ANSWER: x = 5; y = 8

ANSWER: 7-4 Parallel Lines and Proportional Parts 2360.3 ft ALGEBRA Find x and y.

9.

8.

SOLUTION: We are given that

SOLUTION: We are given that

and

.

Solve for y.

Solve for x.

By Corollary 7.2,

.

Solve for x.

Solve for y.

ANSWER: x = 5; y = 8

ANSWER: x = 20; y = 2 10. If AB = 6, BC = 4, and AE = 9, find ED.

9. SOLUTION: We are given that

.

Solve for y.

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By Corollary 7.2,

.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the Page sides4 into segments of proportional lengths.

ANSWER: 7-4 Parallel Lines and Proportional Parts x = 20; y = 2 10. If AB = 6, BC = 4, and AE = 9, find ED.

ANSWER: 6 11. If AB = 12, AC = 16, and ED = 5, find AE.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the Triangle Proportionality Theorem.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. AB =12. So, BC = 16 – 12 = 4.

Use the Triangle Proportionality Theorem.

Substitute.

Substitute.

Solve for ED.

Solve for AE. ANSWER: 6 11. If AB = 12, AC = 16, and ED = 5, find AE.

ANSWER: 15 12. If AC = 14, BC = 8, and AD = 21, find ED.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. AB =12. So, BC = 16 – 12 = 4.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

Use the Triangle Proportionality Theorem.

Here, BC = 8. So, AB = 14 – 8 = 6. Let x be the length of the segment AE. So, ED = 21 – x. Page 5

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Use the Triangle Proportionality Theorem.

So, AE = 9 and ED = 21 – 9 = 12. ANSWER: 7-4 Parallel Lines and Proportional Parts 15 12. If AC = 14, BC = 8, and AD = 21, find ED.

ANSWER: 12 13. If AD = 27, AB = 8, and AE = 12, find BC.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

Here, BC = 8. So, AB = 14 – 8 = 6. Let x be the length of the segment AE. So, ED = 21 – x.

Here, AE = 12. So, ED = 27 – 12 = 15.

Use the Triangle Proportionality Theorem.

Use the Triangle Proportionality Theorem.

Substitute.

Substitute in values and solve for BC.

Solve for x.

ANSWER: 10

So, AE = 9 and ED = 21 – 9 = 12.

Determine whether answer.

Justify your

ANSWER: 12 13. If AD = 27, AB = 8, and AE = 12, find BC.

14. ZX = 18, ZV = 6, WX = 24, and YX = 16 SOLUTION: ZV = 6 and YX = 16. Therefore, VX = 18 – 6 = 12 and WY = 24 – 16 = 8.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and eSolutions Manual - Powered by Cognero intersects the other two sides, then it divides the sides into segments of proportional lengths.

Use the Converse of the Triangle Proportionality Theorem.

Page 6

ANSWER: ANSWER: 7-4 Parallel Lines and Proportional Parts 10 Determine whether answer.

Justify your

yes;

16. ZV = 8, VX = 2, and YX =

WY

SOLUTION: Use the Converse of the Triangle Proportionality Theorem.

14. ZX = 18, ZV = 6, WX = 24, and YX = 16 SOLUTION: ZV = 6 and YX = 16. Therefore, VX = 18 – 6 = 12 and WY = 24 – 16 = 8.

Use the Converse of the Triangle Proportionality Theorem.

Because

,

and

are not parallel.

ANSWER:

no;

Since

17. WX = 31, YX = 21, and ZX = 4ZV , then

SOLUTION: YX = 21, so WY = 31 – 21 = 10 and since ZX = 4ZV, then VX = 3ZV. Use the Converse of the Triangle Proportionality Theorem.

.

ANSWER:

yes;

15. VX = 7.5, ZX = 24, WY = 27.5, and WX = 40

SOLUTION: VX = 7.5 and WY = 27.5. So, ZV = 24 – 7.5 = 16.5 and YX = 40 – 27.5 = 12.5.

Because

, we can say that

and

are not parallel.

Use the Converse of the Triangle Proportionality Theorem.

ANSWER: no;

are midsegments of Find the value of x.

Since ANSWER: yes;

18. SOLUTION:

eSolutions Powered Cognero 16. ZV =Manual 8, VX- = 2, andbyYX = WY

SOLUTION:

By the Triangle Midsegment Theorem, . 7 Page By the Alternate Interior Angles Theorem, x = 57. ANSWER:

are not parallel.

By the Alternate Interior Angles Theorem, .

ANSWER:

ANSWER: 60

7-4 Parallel Lines and Proportional Parts no; are midsegments of Find the value of x.

20. SOLUTION:

18. SOLUTION:

By the Triangle Midsegment Theorem,

By the Triangle Midsegment Theorem, . By the Alternate Interior Angles Theorem, x = 57.

Substitute.

ANSWER: 57 ANSWER: 50

19. SOLUTION: By the Triangle Midsegment Theorem,

21. .

SOLUTION: By the Triangle Midsegment Theorem,

By the Alternate Interior Angles Theorem, . ANSWER: 60

Substitute.

ANSWER: 1.35 22. CCSS MODELING In Charleston, South Carolina, Logan Street is parallel to both King Street and Smith Street between Beaufain Street and Queen Street. What is the distance from Smith to Logan along Beaufain? Round to the nearest foot.

20. SOLUTION: By the Triangle Midsegment Theorem, Substitute.

ANSWER: 50 eSolutions Manual - Powered by Cognero

SOLUTION: Let x be the distance from Smith to Logon along Beaufain. Use the Triangle Proportionality Theorem. Page 8

Solve for x.

So, the distance from Smith to Logan is 891 ft. ANSWER: 7-4 Parallel Lines and Proportional Parts 1.35 22. CCSS MODELING In Charleston, South Carolina, Logan Street is parallel to both King Street and Smith Street between Beaufain Street and Queen Street. What is the distance from Smith to Logan along Beaufain? Round to the nearest foot.

SOLUTION: Let x be the distance from Smith to Logon along Beaufain. Use the Triangle Proportionality Theorem.

ANSWER: about 891 ft 23. ART Tonisha drew the line of dancers shown below for her perspective project in art class. Each of the dancers is parallel. Find the lower distance between the first two dancers.

SOLUTION: Distance between second dancer and third dancer =

Solve for x.

So, the distance from Smith to Logan is 891 ft. ANSWER: about 891 ft 23. ART Tonisha drew the line of dancers shown below for her perspective project in art class. Each of the dancers is parallel. Find the lower distance between the first two dancers.

Let x be the lower distance between the first two dancers. Use the Triangle Proportionality Theorem.

SOLUTION: Distance between second dancer and third dancer =

So, the lower distance between the first two dancers is or 1.2 inches. ANSWER: 1.2 in. eSolutions Manual - Powered by Cognero

ALGEBRA Find x and y.

Page 9

So, the lower distance between the first two dancers is or 1.2 inches. ANSWER: x = 2; y = 5

7-4 Parallel Lines and Proportional Parts ANSWER: 1.2 in. ALGEBRA Find x and y.

25. 24.

SOLUTION: SOLUTION: We are given that

We are given that

and

and

Solve for x.

Solve for x.

Solve for y.

Solve for y.

ANSWER: x = 2; y = 5

25.

ANSWER: x = 18; y = 3

SOLUTION: We are given that

and

ALGEBRA Find x and y.

Solve for x. 26. SOLUTION: eSolutions Manual - Powered by Cognero

It is given that

and

Page 10

ANSWER: x = 10; y = 3

ANSWER: 7-4 Parallel Lines and Proportional Parts x = 18; y = 3 ALGEBRA Find x and y.

26. 27.

SOLUTION: It is given that

SOLUTION:

and

We are given that

Solve for x.

x = 10

By Corollary 7.2,

Solve for y.

Solve for x.

.

Solve for y.

.

ANSWER: x = 10; y = 3

ANSWER: x = 48; y = 72 CCSS ARGUMENTS Write a paragraph proof. 28. Corollary 7.1

27. SOLUTION: We are given that

Solve for y.

.

SOLUTION: In Corollary 7.1, it is stated that, if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. A good approach to this proof it is apply the Triangle Proportionality theorem, one triangle at a time. Given:

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Page 11

Prove:

ANSWER: 7-4 Parallel x = 48; yLines = 72 and Proportional Parts CCSS ARGUMENTS Write a paragraph proof. 28. Corollary 7.1 SOLUTION: In Corollary 7.1, it is stated that, if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. A good approach to this proof it is apply the Triangle Proportionality theorem, one triangle at a time. Given: Prove:

proportional. In By the Triangle Proportionality Theorem, BC and EF are proportional. Therefore, 29. Corollary 7.2 SOLUTION: Corollary 7.2 states, if three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal. This proof can be approached by using Corollary 7.1 to establish that, since we have three parallel lines, then we know they cut off the transversals proportionally. If the ratio of one side of this proportion is equal to 1, since both parts are equal, then the other side of the proportion must also equal 1. Therefore, they are also equal, or congruent, parts.

Given: Prove:

Proof: In By the Triangle Proportionality Theorem, AB and DE are proportional . In By the Triangle Proportionality Theorem, BC and EF are proportional. Therefore,

Proof: From Corollary 7.1,

Since

AB = BC by definition of congruence.

ANSWER: Given:

Therefore,

Prove:

DE = EF. By definition of congruence,

= 1. By substitution, 1 =

Thus,

ANSWER: Given: Prove:

Proof: In By the Triangle Proportionality Theorem, AB and DE are proportional. In By the Triangle Proportionality Theorem, BC and EF are

Proof: From Corollary 7.1,

proportional. Therefore, eSolutions Manual - Powered by Cognero

29. Corollary 7.2

SOLUTION:

Since

AB = BC by definition of congruence. Therefore,

= 1. By substitution, 1 =

DE = EF. By definition of congruence,

Thus, Page 12

Thus,

by subtracting

one from each side. Proof: From Corollary 7.1, Since 7-4 Parallel Lines and Proportional Parts

ANSWER:

AB = BC by definition of congruence. Therefore,

= 1. By substitution, 1 =

Thus,

Given: Prove:

DE = EF. By definition of congruence, 30. Theorem 7.5 SOLUTION: Theorem 7.5 states, if a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. In order to prove that need to establish that

, we first , which can be

accomplished by showing that Then, by Segment Addition Postulate, we can state that CA = BA + CB and CE = DE + CD. Substitute these values in for CA and CE in the previous proportion and the simplify. Given:

Proof: , and because they are corresponding angles. By AA Similarity, From the definition of similar By the Segment Addition

polygons,

Postulate, CA = BA + CB and CE = DE + CD. By Rewriting as a

substitution,

From simplifying,

sum,

Prove:

Thus,

by subtracting

one from each side. CCSS ARGUMENTS Write a two-column proof. 31. Theorem 7.6

Proof: , and because they are corresponding angles. By AA Similarity, From the definition of similar By the Segment Addition

polygons,

Postulate, CA = BA + CB and CE = DE + CD. By Rewriting as a

substitution,

From simplifying,

sum, Thus, one from each side. ANSWER:

SOLUTION: Theorem 7.6 states, if a line intersects two sides of a triangle and separates the sides into proportional corresponding segments, then the line is parallel to the third side of the triangle. Thinking backwards, how can we prove that two lines are parallel to each other? We can prove that by proving that a pair of corresponding angles, formed by these parallel lines, are congruent to each other. Using SAS Similarity theorem, prove that . Then, you can use congruent corresponding angles as a result of similar triangles.

by subtracting Given: Prove:

Given: Prove: eSolutions Manual - Powered by Cognero

Proof: Statements (Reasons)

Page 13

Given: Lines and Proportional Parts 7-4 Parallel

(SAS Similarity) (Def. of polygons) (If corr. angles are , then the lines are

7. 8. 9. || .)

Prove: 32. Theorem 7.7 SOLUTION: Theorem 7.7 states that a midsegment of a triangle is parallel to one side of the triangle, and its length is half the length of that side. For this proof, use the given information that to prove that

Proof: Statements (Reasons) (Given)

1.

by AA Similarity. Then, since you know that D and E are both midpoints, then you can (Add. Prop.)

2.

(Subst.)

3.

prove eventually prove that

midpoint relationships and substitution. Then, using as a result of proving

4. AB = AD + DB, AC = AE + EC (Seg. Add. Post.) (Subst.)

5. 6. 7. 8. 9. || .)

, using

then you can substitute into

(Refl. Prop.) (SAS Similarity) (Def. of polygons) (If corr. angles are , then the lines are

ANSWER:

, into

and prove that

, using

algebra. Given: D is the midpoint of E is the midpoint of

Given: Prove: Prove: Proof: Statements (Reasons) 1. D is the midpoint of

Proof: Statements (Reasons)

(Given)

(Given)

1.

(Add. Prop.)

2.

(Subst.)

3.

4. AB = AD + DB, AC = AE + EC (Seg. Add. Post.) (Subst.)

5. 6. 7. 8. 9. || .)

E is the midpoint of

(Refl. Prop.) (SAS Similarity) (Def. of polygons) (If corr. angles are , then the lines are

32. Theorem 7.7

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SOLUTION: Theorem 7.7 states that a midsegment of a triangle is

2. (Midpoint Thm.) 3. AD = DB, AE = EC (Def. of segs.) 4. AB = AD + DB, AC = AE + EC (Seg. Add. Post.) 5. AB = AD + AD, AC = AE + AE (Subst.) 6. AB = 2AD, AC = 2AE (Subst.) 7. 8. 9. 10. 11. 12. parallel.) 13. 14.

(Div. Prop.) (Trans. Prop.) (Refl. Prop.) (SAS Similarity) (Def. of polygons) (If corr. angles are , the lines are (Def. of

polygons)

(Substitution Prop.)

Page 14

9. (Refl. Prop.) 10. (SAS Similarity) 11. (Def. of polygons) 7-4 Parallel Lines and Proportional Parts 12. (If corr. angles are , the lines are parallel.) (Def. of

13.

polygons)

14.

(Substitution Prop.)

15. 2DE = BC (Mult. Prop.) 16.

(Division Prop.)

Refer to

(Substitution Prop.)

14.

15. 2DE = BC (Mult. Prop.) (Division Prop.)

16.

33. If ST = 8, TR = 4, and PT = 6, find QR. SOLUTION: Since , we know that and . Therefore, by AA Similarity, .

ANSWER: Given: D is the midpoint of E is the midpoint of

Use the definition of similar polygons to create a proportion:

Prove:

Proof: Statements (Reasons) 1. D is the midpoint of

We know that SR = 8 + 4 =12. Substitute values and solve for QR. E is the midpoint of

(Given) 2. (Midpoint Thm.) 3. AD = DB, AE = EC (Def. of segs.) 4. AB = AD + DB, AC = AE + EC (Seg. Add. Post.) 5. AB = AD + AD, AC = AE + AE (Subst.) 6. AB = 2AD, AC = 2AE (Subst.) 7. 8. 9. 10. 11. 12. parallel.) 13. 14.

(Div. Prop.) (Trans. Prop.) (Refl. Prop.) (SAS Similarity) (Def. of polygons) (If corr. angles are , the lines are (Def. of

polygons)

(Substitution Prop.)

15. 2DE = BC (Mult. Prop.) 16.

ANSWER: 9

(Division Prop.)

34. If SP = 4, PT = 6, and QR = 12, find SQ. SOLUTION: Since , we know that and . Therefore, by AA Similarity, .

Use the definition of similar polygons to set up a proportion:

Substitute and solve for SQ:

Refer to

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33. If ST = 8, TR = 4, and PT = 6, find QR.

Page 15

ANSWER: 8

ANSWER: 7-4 Parallel Lines and Proportional Parts 9 34. If SP = 4, PT = 6, and QR = 12, find SQ. SOLUTION: Since , we know that and . Therefore, by AA Similarity, .

ANSWER: 8 35. If CE = t – 2, EB = t + 1, CD = 2, and CA = 10, find t and CE.

Use the definition of similar polygons to set up a proportion:

Substitute and solve for SQ:

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

Use the Triangle Proportionality Theorem.

Since CA = 10 and CD = 2, then DA =10-2= 8.

ANSWER: 8 35. If CE = t – 2, EB = t + 1, CD = 2, and CA = 10, find t and CE.

Substitute and solve for t.

Find CE. SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

Use the Triangle Proportionality Theorem.

ANSWER: 3, 1 36. If WX = 7, WY = a, WV = 6, and VZ = a – 9, find WY.

Since CA = 10 and CD = 2, then DA =10-2= 8.

Substitute and solve for t.

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SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

Use the Triangle Proportionality Theorem.

Page 16

So, a= WY = 21. ANSWER: 7-4 Parallel Lines and Proportional Parts 3, 1 36. If WX = 7, WY = a, WV = 6, and VZ = a – 9, find WY.

ANSWER: 21 37. If QR = 2, XW = 12, QW = 15, and ST = 5, find RS and WV.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

Use the Triangle Proportionality Theorem.

Use the Triangle Proportionality Theorem.

Since WY = a and WX = 7, XY = a – 7.

Since QW = 15 and WX = 12, then QX = 3.

Substitute and solve for a.

Substitute and solve for RS.

So, a= WY = 21.

Additionally, we know that

.

Substitute and solve for WV.

ANSWER: 21 37. If QR = 2, XW = 12, QW = 15, and ST = 5, find RS and WV. ANSWER: 8, 7.5 SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

38. If LK = 4, MP = 3, PQ = 6, KJ = 2, RS = 6, and LP = 2, find ML, QR, QK, and JH.

Use the Triangle Proportionality Theorem.

Manual - Powered by Cognero eSolutions Since QW = 15 and WX = 12, then QX = 3.

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and Page 17 intersects the other two sides, then it divides the sides into segments of proportional lengths.

Finally, by Triangle Proportionality Theorem, 7-4 Parallel Lines and Proportional Parts SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

. Substitute and solve for JH.

Use the Triangle Proportionality Theorem.

Substitute and solve for ML.

ANSWER: 2, 3, 6, 4

.

Also, we know that

39. MATH HISTORY The sector compass was a tool perfected by Galileo in the sixteenth century for measurement. To draw a segment two-fifths the length of a given segment, align the ends of the arms with the given segment. Then draw a segment at the 40 mark. Write a justification that explains why the sector compass works for proportional measurement.

Substitute and solve for

Because know that

, by AA Similarity, we .

Substitute and solve for QK.

SOLUTION: To prove that two corresponding sides of two triangles are the same ratio as another pair of corresponding sides, you need to first establish that the triangles are similar. Once this is completed, a proportion statement can be written, relating the proportional sides. Substitute in given values from the diagram to prove that .

Finally, by Triangle Proportionality Theorem, . Substitute and solve for JH.

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ANSWER:

Page 18

7-4 Parallel Lines and Proportional Parts Determine the value of x so that

40. AB = x + 5, BD = 12, AC = 3x + 1, and CF = 15

ANSWER:

SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

Use the Triangle Proportionality Theorem.

Substitute.

Determine the value of x so that

40. AB = x + 5, BD = 12, AC = 3x + 1, and CF = 15 SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

Use the Triangle Proportionality Theorem.

ANSWER: 3 41. AC = 15, BD = 3x – 2, CF = 3x + 2, and AB = 12 SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths.

Use the Triangle Proportionality Theorem.

Substitute.

Substitute.

ANSWER: 3 eSolutions Manual - Powered by Cognero

41. AC = 15, BD = 3x – 2, CF = 3x + 2, and AB = 12 SOLUTION:

ANSWER: 6

Page 19

The midpoint of The midpoint of

ANSWER: 7-4 Parallel Lines and Proportional Parts 3

= =

. .

41. AC = 15, BD = 3x – 2, CF = 3x + 2, and AB = 12 SOLUTION: Triangle Proportionality Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the sides into segments of proportional lengths. Use the distance formula.

Use the Triangle Proportionality Theorem.

Substitute.

The segment is parallel to are both

because the slopes

and the segment length is half of

Thus, the segment is the midsegment of

ANSWER: 6

ANSWER:

42. COORDINATE GEOMETRY has vertices A(–8, 7), B(0, 1), and C(7, 5). Draw Determine the coordinates of the midsegment of that is parallel to Justify your answer. SOLUTION:

The endpoints of the midsegment are (–4, 4) and (– 0.5, 6). Sample answer: The segment is parallel to because the slopes are both and the segment length is half of midsegment of

Use the midpoint formula determine the midpoints of The midpoint of The midpoint of

43. HOUSES Refer to the diagram of the gable. Each piece of siding is a uniform width. Find the lengths of and

to and

= =

Thus, the segment is the

. .

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SOLUTION: Page 20 All the triangles are isosceles. Segment EH is the midsegment of triangle ABC . Therefore, segment EH is the half of the length of AC, which is 35 ÷ 2 or

44. a segment separated into five congruent segments because the slopes are both and the segment length is Lines half ofand Proportional Thus, the segment is the 7-4 Parallel Parts midsegment of

SOLUTION: Step 1: Construct an angle with vertex A, as shown below:

43. HOUSES Refer to the diagram of the gable. Each piece of siding is a uniform width. Find the lengths of and

SOLUTION: All the triangles are isosceles. Segment EH is the midsegment of triangle ABC . Therefore, segment EH is the half of the length of AC, which is 35 ÷ 2 or 17.5 feet. Similarly, FG is the midsegment of triangle BEH, so FG = 17.5 ÷ 2 or 8.75 feet. To find DJ, use the vertical altitude which is 12 feet. Let the altitude from B to the segment AC meet the segment DJ at K. Find BC using the Pythagorean Theorem. 2

2

2

BC = BK + KC 2 2 2 BC = 12 + 17.5 BC = Since the width of each piece of siding is the same, BJ =

BC, which is about

Step 2: With your compass on vertex A, choose a radius and make an arc on the diagonal, as shown below:

Step 3: With your compass on the new point formed on the diagonal, keep the same radius and make another arc further down the diagonal side of the angle. Continue this process until you have five arcs, like below:

or 15.92 in.

Now, use the Triangle Proportionality Theorem.

ANSWER: 8.75 in., 17.5 in., 26.25 in.

Step 4: Using a straight edge, draw a segment that connects each new point back to the horizontal side of the angle, perpendicular to that side, as shown below:

CONSTRUCTIONS Construct each segment as directed. 44. a segment separated into five congruent segments SOLUTION: Step 1: Construct an angle with vertex A, as shown below: eSolutions Manual - Powered by Cognero

Step 5. Label the points formed on the horizontal side of the angle and erase any extra length beyond the last point. Page 21

7-4 Parallel Lines and Proportional Parts Step 5. Label the points formed on the horizontal side of the angle and erase any extra length beyond the last point.

Step 3: Continue this process until you have four arcs. When you connect the points on the diagonal back to the horizontal, make sure the connecting lines are all parallel to each other. ( Since you want segment lengths at a ratio of 1 to 3, this can be created by 4 equal smaller segments, where three can be pieced together to make one that is 3/4 the original length.)

ANSWER: Sample answer: Step 4: Label the first point B and the last point C. .

45. a segment separated into two segments in which their lengths have a ratio of 1 to 3 SOLUTION: Step 1: Make an angle, with vertex A, as shown below: ANSWER: Sample answer:

Step 2: With your compass on vertex A, make an arc that passes through the diagonal side of the angle. Connect this new point back to the horizontal side of the angle. Label B as the new point made on the horizontal side of the angle, as shown below.

46. a segment 3 inches long, separated into four congruent segments SOLUTION: Step 1: Copy a 3 inch segment. horizontally. Then, make an angle, with vertex A, as shown below:

StepManual 3: Continue thisbyprocess eSolutions - Powered Cognero until

you have four arcs. When you connect the points on the diagonal back to the horizontal, make sure the connecting lines

Page 22

congruent segments SOLUTION: Step 1: Copy a 3 inch segment. horizontally. Then, 7-4 Parallel Lines and Proportional Parts make an angle, with vertex A, as shown below:

Step 2: With your compass on vertex A, make an arc that passes through the diagonal side of the angle. Connect this new point back to the horizontal side of the angle. Label B as the new point made on the horizontal side of the angle, as shown below.

ANSWER: Sample answer:

47. MULTIPLE REPRESENTATIONS In this problem, you will explore angle bisectors and proportions. a. GEOMETRIC Draw three triangles, one acute, one right, and one obtuse. Label one triangle ABC and draw angle bisector Label the second MNP with angle bisector and the third WXY with angle bisector b. TABULAR Complete the table at the right with the appropriate values. c. VERBAL Make a conjecture about the segments of a triangle created by an angle bisector.

Step 3: Continue this process until you have four arcs. When you connect the points on the diagonal back to the horizontal, make sure the connecting lines are all parallel to each other.

Step 4: Label the points as shown. AB = BC = CD = DE

SOLUTION: a. When drawing the triangles, pay close attention

to the directions and labeling instructions. Use a protractor, or construction tool, when making the angle bisectors, to ensure accurate measurement values for the table. Sample answer:

ANSWER: Sample answer: eSolutions Manual - Powered by Cognero

Page 23

protractor, or construction tool, when making the angle bisectors, to ensure accurate measurement values for the table. 7-4 Parallel Lines and Proportional Parts Sample answer:

b. Carefully measure the indicated lengths in centimeters.

c. Look for a pattern in the table, specifically comparing the lengths of the ratios of sides for each triangle. Sample answer: The proportion of the segments created by the angle bisector of a triangle is equal to the proportion of their respective consecutive sides.

Sample answer: The proportion of the segments created by the angle bisector of a triangle is equal to the proportion of their respective consecutive sides. ANSWER: a. Sample answer:

b.

c. Sample answer: The proportion of the segments created by the angle bisector of a triangle is equal to the proportion of their respective consecutive sides. 48. CCSS CRITIQUE Jacob and Sebastian are finding the value of x in Jacob says that MP is one half of JL, so x is 4.5. Sebastian says that JL is one half of MP, so x is 18. Is either of them correct? Explain.

ANSWER: a. Sample answer:

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SOLUTION: Jacob; sample answer: Since M is the midpoint Page of 24 and P is the midpoint of , then is the midsegment of . Therefore,

ANSWER: c. Sample answer: The proportion of the segments 7-4 Parallel Lines and Proportional Parts is equal to created by the angle bisector of a triangle the proportion of their respective consecutive sides. 48. CCSS CRITIQUE Jacob and Sebastian are finding the value of x in Jacob says that MP is one half of JL, so x is 4.5. Sebastian says that JL is one half of MP, so x is 18. Is either of them correct? Explain.

49. REASONING In If D is

AF = FB and AH = HC.

of the way from A to B and E is

of the

way from A to C, is DE sometimes, always, or never

SOLUTION: Jacob; sample answer: Since M is the midpoint of and P is the midpoint of , then is the midsegment of . Therefore,

is the midsegment, so

Jacob; sample answer:

of BC? Explain.

SOLUTION: Always; sample answer: Since FA=FB, then F is a midpoint of . Similarly, since AH=HC and H is the midpoint of .

Therefore, FH is a midsegment of and .

so

Because trapezoid, so

ANSWER: Jacob; sample answer:

49. REASONING In If D is

Let BC = x, then

, we know that FHCB is a

is the midsegment, so

AF = FB and AH = HC.

of the way from A to B and E is

of the

way from A to C, is DE sometimes, always, or never

of BC? Explain.

ANSWER: Always; sample answer: FH is a midsegment. Let BC = x, then SOLUTION: Always; sample answer: Since FA=FB, then F is a midpoint of . Similarly, since AH=HC and H is the midpoint of .

Therefore, FH is a midsegment of and .

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Let BC = x, then

so

FHCB is a trapezoid, so Therefore,

CHALLENGE Write a two-column proof. 50. Given: AB = 4, BC = 4, and CD = DE Prove:

Page 25

(Def. of polygons) (If corr. angles are , lines are ||.)

16. 17. 7-4 Parallel Lines and Proportional Parts CHALLENGE Write a two-column proof. 50. Given: AB = 4, BC = 4, and CD = DE Prove:

SOLUTION: An effective strategy for this proof is to think of a way to get , by SAS Similarity. We already know that , so we need to establish that

. You can show that 2BC =

AC and 2DC = EC, through the given information and substitution into Segment Addition Postulate statements. Once this is done, you can prove that by transitive property. Once the triangles are proven similar, then the lines can be proven parallel by choosing a pair of congruent corresponding angles from the similar triangles.

, Proof: Statements (Reasons) 1. AB = 4, BC = 4 (Given) 2. AB = BC (Subst.) 3. AB + BC = AC (Seg. Add. Post.) 4. BC + BC = AC (Subst.) 5. 2BC = AC (Subtraction property.) 6. AC = 2BC (Symm. Prop). 7.

(Div. Prop.)

ANSWER: Proof: Statements (Reasons) 1. AB = 4, BC = 4 (Given) 2. AB = BC (Sub.) 3. AB + BC = AC (Seg. Add. Post.) 4. BC + BC = AC (Subst.) 5. 2BC = AC (Sub.) 6. AC = 2BC (Symm. Prop). (Div. Prop.)

7.

8. ED = DC (Given) 9. ED + DC = EC (Seg. Add. Post.) 10. DC + DC = EC (Sub.) 11. 2DC = EC (Sub.) (Div. Prop.)

12.

(Trans. Prop.)

13. 14. 15. 16. 17.

(Reflexive Prop.) (SAS Similarity) (Def. of polygons) (If corr. angles are , lines are ||.)

51. OPEN ENDED Draw three segments, a, b, and c, of all different lengths. Draw a fourth segment, d, such that SOLUTION: By Corollary 7.1, we know that if we draw three parallel lines intersected by two transversals, then they will cut the transversals proportionally or

, as seen in the diagram below.

8. ED = DC (Given) 9. ED + DC = EC (Seg. Add. Post.) 10. DC + DC = EC (Subst.) 11. 2DC = EC (Subst.) 12. 13. 14. 15. 16. 17.

(Div. Prop.) (Trans. Prop.) (Reflexive Prop.) (SAS Similarity) (Def. of polygons) (If corr. angles are , lines are ||.)

ANSWER: Proof: Statements (Reasons) 1. AB = 4, BC = 4 (Given) eSolutions Manual - Powered by Cognero 2. AB = BC (Sub.) 3. AB + BC = AC (Seg. Add. Post.) 4. BC + BC = AC (Subst.)

ANSWER:

By Corollary 7.1, 52. WRITING IN MATH Compare the Triangle

Page 26

14. (Reflexive Prop.) 15. (SAS Similarity) 16. (Def. of polygons) 7-4 Parallel Lines and Proportional 17. (If corr. angles are Parts , lines are ||.) 51. OPEN ENDED Draw three segments, a, b, and c, of all different lengths. Draw a fourth segment, d,

ANSWER: Both theorems deal with a parallel line inside the triangle. The Midsegment Theorem is a special case of the Converse of the Proportionality Theorem. 53. SHORT RESPONSE What is the value of x?

such that SOLUTION: By Corollary 7.1, we know that if we draw three parallel lines intersected by two transversals, then they will cut the transversals proportionally or

, as seen in the diagram below.

SOLUTION:

By Corollary 7.2,

.

Solve for x.

ANSWER: 8 54. If the vertices of triangle JKL are (0, 0), (0, 10) and (10, 10) then the area of triangle JKL is 2 A 20 units B 30 units2

ANSWER:

2

C 40 units D 50 units2 SOLUTION: By Corollary 7.1, 52. WRITING IN MATH Compare the Triangle Proportionality Theorem and the Triangle Midsegment Theorem. SOLUTION: Both theorems deal with a parallel line inside the triangle. The Midsegment Theorem is a special case of the Converse of the Proportionality Theorem. ANSWER: Both theorems deal with a parallel line inside the triangle. The Midsegment Theorem is a special case of the Converse of the Proportionality Theorem. 53. SHORT RESPONSE What is the value of x?

SOLUTION: eSolutions Manual - Powered by Cognero

By Corollary 7.2, Solve for x.

.

So, the correct choice is D. ANSWER: D 55. ALGEBRA A breakfast cereal contains wheat, rice, and oats in the ratio 2 : 4: 1. If the manufacturer makes a mixture using 110 pounds of wheat, how many pounds of rice will be used? F 120 lb G 220 lb H 240 lb J 440 lb SOLUTION: Since the ratio of rice to wheat is 4: 2, we can set up a proportion to find the amount of rice needed for 110 pounds of wheat. Page 27

So, the correct choice is D.

The correct answer is G, 220 lb.

ANSWER: 7-4 Parallel Lines and Proportional Parts D 55. ALGEBRA A breakfast cereal contains wheat, rice, and oats in the ratio 2 : 4: 1. If the manufacturer makes a mixture using 110 pounds of wheat, how many pounds of rice will be used? F 120 lb G 220 lb H 240 lb J 440 lb

ANSWER: G 56. SAT/ACT If the area of a circle is 16 square meters, what is its radius in meters? A B C

SOLUTION: Since the ratio of rice to wheat is 4: 2, we can set up a proportion to find the amount of rice needed for 110 pounds of wheat.

D 12π E 16π SOLUTION:

Since the area of a circle can be found with , we can substitute in 16 for the area (A) and solve for r.

The correct answer is G, 220 lb. ANSWER: G 56. SAT/ACT If the area of a circle is 16 square meters, what is its radius in meters? A B

Therefore, the answer is A.

C D 12π E 16π

ANSWER: A

SOLUTION:

Since the area of a circle can be found with , we can substitute in 16 for the area (A) and solve for r.

ALGEBRA Identify the similar triangles. Then find the measure(s) of the indicated segment(s). 57.

SOLUTION:

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by the Vertical Angles Theorem. Page 28 Since , by the Alternate Interior Angles Theorem.

ANSWER:

ANSWER: 7-4 Parallel Lines and Proportional Parts A

by AA Similarity; 6.25

ALGEBRA Identify the similar triangles. Then find the measure(s) of the indicated segment(s).

58.

57.

SOLUTION: , since right angles are

SOLUTION: by the Vertical Angles Theorem. Since , by the Alternate Interior Angles Theorem. Therefore, by AA Similarity, . To find AB or x, write a proportion using the definition of similar polygons.

congruent.

, since

.

Therefore, by SAS Similarity,

.

Write a proportion using the definition of similar polygons to find the value of x.

Substitute this value for x to find RT and RS.

ANSWER: by AA Similarity; 6.25

ANSWER:

58.

by SAS Similarity; 15, 20 59.

SOLUTION: , since right angles are congruent.

, since

Therefore, by SAS Similarity,

. .

Write a proportion using the definition of similar polygons to find the value of x.

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SOLUTION: by the Reflexive Property of Congruence. Since , by the Corresponding Angles Theorem. Therefore, by AA Similarity,

. Page 29

Write a proportion using the definition of similar

ANSWER:

ANSWER: 7-4 Parallel Lines and Proportional Parts by SAS Similarity; 15, 20

by AA Similarity; 7.5 60. SURVEYING Mr. Turner uses a carpenter’s square to find the distance across a stream. The carpenter’s square models right angle NOL. He puts the square on top of a pole that is high enough to sight along to point P across the river. Then he sights along to point M . If MK is 1.5 feet and OK is 4.5 feet, find the distance KP across the stream.

59.

SOLUTION: by the Reflexive Property of Congruence. Since , by the Corresponding Angles Theorem. Therefore, by AA Similarity,

.

SOLUTION: By AA Similarity,

Write a proportion using the definition of similar polygons to find the value of x.

.

Use the Pythagorean Theorem to find MO.

So, WT = 12.5. WT + TY = WY by the Segment Addition Postulate. Since WY = 20, you can solve for TY.

Write a proportion using corresponding sides of the two triangles:

ANSWER: by AA Similarity; 7.5 60. SURVEYING Mr. Turner uses a carpenter’s square to find the distance across a stream. The carpenter’s square models right angle NOL. He puts the square on top of a pole that is high enough to sight along to point P across the river. Then he sights along to point M . If MK is 1.5 feet and OK is 4.5 feet, find the distance KP across the stream.

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SOLUTION:

Therefore, the distance KP is about 13.5 feet. ANSWER: 13.5 ft COORDINATE GEOMETRY For each quadrilateral with the given vertices, verify that the quadrilateral is a trapezoid and determine whether the figure is an isosceles trapezoid. Page 30 61. Q(–12, 1), R(–9, 4), S(–4, 3), T(–11, –4) SOLUTION:

ANSWER: 13.5 ft

ANSWER: QRST is an isosceles trapezoid

COORDINATE 7-4 Parallel Lines andGEOMETRY Proportional For Partseach quadrilateral with the given vertices, verify that the quadrilateral is a trapezoid and determine whether the figure is an isosceles trapezoid. 61. Q(–12, 1), R(–9, 4), S(–4, 3), T(–11, –4)

since RS =

= QT.

62. A(–3, 3), B(–4, –1), C(5, –1), D(2, 3) SOLUTION:

SOLUTION:

Use the slope formula to find the slope of the sides of the quadrilateral.

Use the slope formula to find the slope of the sides of the quadrilateral.

The slopes of exactly one pair of opposite sides are equal. So, they are parallel. Therefore, the quadrilateral ABCD is a trapezoid.

The slopes of exactly one pair of opposite sides are equal. So, this quadrilateral has only one pair of parallel sides. Therefore, the quadrilateral QRST is a trapezoid.

Use the Distance Formula to find the lengths of the legs of the trapezoid.

Use the Distance Formula to find the lengths of the legs of the trapezoid.

The lengths of the legs are not equal. Therefore, ABCD is not an isosceles trapezoid.

The lengths of the legs are equal. Therefore, QRST is an isosceles trapezoid.

ANSWER:

ANSWER:

isosceles since AB = QRST is an isosceles trapezoid

since RS =

= QT.

3), B(–4, C(5, 62. A(–3, –1),by –1), eSolutions Manual - Powered Cognero SOLUTION:

D(2, 3)

ABCD is a trapezoid, but not and CD = 5.

Point S is the incenter of measure.

Find each

Page 31

SK= SQ = 6

ANSWER: ABCD is a trapezoid, but not 7-4 Parallel Lines and Proportional Parts isosceles since AB = and CD = 5. Point S is the incenter of measure.

Find each

ANSWER: 6 64. QJ SOLUTION: Since S is the incenter of , then it is formed by the angle bisectors of each vertex. is the angle bisector of therefore, by the Angle Bisector Theorem, SK= SQ.

63. SQ

Use Pythagorean Theorem in the right triangle JSK.

SOLUTION: Since S is the incenter of , then it is formed by the angle bisectors of each vertex. is the angle bisector of therefore, by the Angle Bisector Theorem, SK= SQ.

Use Pythagorean Theorem in the right triangle JSK.

Use Pythagorean Theorem in the right triangle JSQ to find QJ.

SK= SQ = 6 ANSWER: 6 64. QJ SOLUTION: Since S is the incenter of , then it is formed by the angle bisectors of each vertex. is the angle bisector of therefore, by the Angle Bisector Theorem, SK= SQ.

Use Pythagorean Theorem in the right triangle JSK.

Therefore QJ=8.

ANSWER: 8 65. m∠MPQ SOLUTION: Since S is the incenter of , then it is formed by the angle bisectors of each vertex. is the angle bisector of .

Therefore,

ANSWER: 56

Use Pythagorean Theorem in the right triangle JSQ to find QJ.

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66. m∠SJP SOLUTION: Page 32 Since S is the incenter of , then it is formed by the angle bisectors of each vertex. is the

ANSWER: 7-4 Parallel Lines and Proportional Parts 56

ANSWER: 37.5 Solve each proportion.

66. m∠SJP SOLUTION: Since S is the incenter of , then it is formed by the angle bisectors of each vertex. is the angle bisector of .

67. SOLUTION: Solve for x.

Therefore,

and similarly,

ANSWER:

68.

We know that sum of the measures of a triangle is 180.

SOLUTION: Solve for x.

We know that an angle bisector of

because . Therefore,

ANSWER: 6.7

is

69. SOLUTION: Solve for x.

ANSWER: 37.5 Solve each proportion. 67. SOLUTION: Solve for x.

ANSWER: 2.1 70.

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ANSWER:

SOLUTION: Solve for x.

Page 33

ANSWER: 7-4 Parallel Lines and Proportional Parts 2.1 70. SOLUTION: Solve for x.

ANSWER: 3.6 71. SOLUTION: Solve for x.

ANSWER: 8.7

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