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Big Ideas Math Geometry Answers Chapter 8 Similarity

Studying & Practicing Math Geometry would be done in a fun learning process for a better understanding of the concepts. So, the best guide to prepare math in a fun learning way is our provided Big Ideas Math Geometry Answers Chapter 8 Similarity Guide. In this study guide, you will discover various exercise questions, chapter reviews, tests, chapter practices, cumulative assessment, etc. to learn all topics of chapter 8 similarity. These questions and answers are explained by the subject experts in a simple manner to make students learn so easily & score maximum marks in the exams.

Big Ideas Math Book Geometry Answer Key Chapter 8 Similarity

BIM Geometry Book Solutions are available for all chapters along with Chapter 8 Similarity on our website. So, make sure to check all the chapters of Big Ideas Math Book Geometry Answer Key and learn the subject thoroughly. Based on the common core 2019 curriculum, these Big Ideas Math Geometry Answers Chapter 8 Similarity are prepared. So, students can instantly take homework help from BIM Geometry Ch 8 Similarity Answers. Simply tap on the below direct links and refer to the solutions covered in the Big Ideas Math Book Geometry Answer Key Chapter 8 Similarity Guide.

• Similarity Maintaining Mathematical Proficiency – Page 415
• Similarity Mathematical Practices – Page 416
• 8.1 Similar Polygons – Page 417
• Lesson 8.1 Similar Polygons – Page (418-426)
• Exercise 8.1 Similar Polygons – Page (423-426)
• 8.2 Proving Triangle Similarity by AA – Page 427
• Lesson 8.2 Proving Triangle Similarity by AA – Page (428-432)
• Exercise 8.2 Proving Triangle Similarity by AA – Page (431-432)
• 8.1 & 8.2 Quiz – Page 434
• 8.3 Proving Triangle Similarity by SSS and SAS – Page 435
• Lesson 8.3 Proving Triangle Similarity by SSS and SAS – Page (436-444)
• Exercise 8.3 Proving Triangle Similarity by SSS and SAS – Page (441-444)
• 8.4 Proportionality Theorems – Page 445
• Lesson 8.4 Proportionality Theorems – Page (446-452)
• Exercise 8.4 Proportionality Theorems – Page (450-452)
• Similarity Chapter Review – Page (454-456)
• Similarity Chapter Test – Page 457
• Similarity Cumulative Assessment – Page (458-459)

Similarity Maintaining Mathematical Proficiency

Tell whether the ratios form a proportion.

Question 1. \(\frac{5}{3}, \frac{35}{21}\)

Answer: Yes, the ratios \(\frac{5}{3}, \frac{35}{21}\) form a proportion.

Explanation: A proportion means two ratios are equal. So, cross product of \(\frac{5}{3}, \frac{35}{21}\) is 21 x 5 = 105 = 35 x 3 Therefore, \(\frac{5}{3}, \frac{35}{21}\) form a proportion.

Question 2. \(\frac{9}{24}, \frac{24}{64}\)

Answer: Yes, the ratios \(\frac{9}{24}, \frac{24}{64}\) form a proportion.

Explanation: If the cross product of two ratios is equal, then it forms a proportion. So, 24 x 24 = 576 = 64 x 9 Therefore, the ratios \(\frac{9}{24}, \frac{24}{64}\) form a proportion.

Question 3. \(\frac{8}{56}, \frac{6}{28}\)

Answer: The ratios \(\frac{8}{56}, \frac{6}{28}\) do not form a proportion.

Explanation: If the cross product of two ratios is equal, then it forms a proportion. So, 56 x 6 = 336, 28 x 8 = 224 Therefore, the ratios \(\frac{8}{56}, \frac{6}{28}\) do not form a proportion.

Question 4. \(\frac{18}{4}, \frac{27}{9}\)

Answer: The ratios \(\frac{18}{4}, \frac{27}{9}\) do not form a proportion.

Explanation: If the cross product of two ratios is equal, then it forms a proportion. So, 9 x 18 = 162, 27 x 4 = 108 Therefore, the ratios \(\frac{18}{4}, \frac{27}{9}\) do not form a proportion.

Question 5. \(\frac{15}{21}, \frac{55}{77}\)

Answer: The ratios \(\frac{15}{21}, \frac{55}{77}\) form a proportion.

Explanation: If the cross product of two ratios is equal, then it forms a proportion. So, 15 x 77 = 1155, 55 x 21 = 1155 Therefore, the ratios \(\frac{15}{21}, \frac{55}{77}\) form a proportion.

Question 6. \(\frac{26}{8}, \frac{39}{12}\)

Answer: The ratios \(\frac{26}{8}, \frac{39}{12}\) form a proportion.

Explanation: If the cross product of two ratios is equal, then it forms a proportion. So, 26 x 12 = 312, 39 x 8 = 312 Therefore, the ratios \(\frac{26}{8}, \frac{39}{12}\) form a proportion.

Find the scale factor of the dilation.

Answer: k = \(\frac { 3 }{ 7 } \)

Explanation: The scale factor k = \(\frac { CP’ }{ CP } \) = \(\frac { 6 }{ 14 } \) = \(\frac { 3 }{ 7 } \)

Answer: k = \(\frac { 3 }{ 8 } \)

Explanation: The scale factor k = \(\frac { CP }{ CP’ } \) = \(\frac { 9 }{ 24 } \) = \(\frac { 3 }{ 8 } \)

Answer: k = \(\frac { 1 }{ 2 } \)

Explanation: The scale factor k = \(\frac { MK }{ M’K’ } \) = \(\frac { 14 }{ 28 } \) = \(\frac { 1 }{ 2 } \)

Question 10. ABSTRACT REASONING If ratio X and ratio Y form a proportion and ratio Y and ratio Z form a proportion, do ratio X and ratio Z form a proportion? Explain our reasoning.

Answer: Yes, ratio X and ratio Z form a proportion.

Explanation: If ratios are proportional means they are equal. So, ratio X and ratio Y form a proportion that means X = Y ratio Y and ratio Z form a proportion that means Y = Z From the above two equations, we can say that X = Z So, ratio X and ratio Z also form a proportion.

Similarity Mathematical Practices

Monitoring Progress

Answer: (a) Perimeter is 32 cm, area is 48 sq cm. (b) Perimeter is 48 cm, the area is 108 sq cm. (c) Perimeter is 64 cm, the area is 192 sq cm.

Explanation: The perimeter of trapezoid P = 2 + 5 + 6 + 3 = 16 cm Area of trapezoid A = \(\frac { (2 + 6)3 }{ 2 } \) = \(\frac { 3(8) }{ 2 } \) = \(\frac { 24 }{ 2 } \) = 12 sq cm (a) If scale factor k = 2, then Perimeter = kP = 2 x 16 = 32 Area = k²A = 2² x 12 = 4 x 12 = 48 (b) If scale factor k = 3, then, Perimeter = kP = 3 x 16 = 48 cm Area = k²A = 3² x 12 = 9 x 12 = 108 (c) If scale factor k = 4, then Perimeter = kP = 4 x 16 = 64 Area = k²A = 4² x 12 = 16 x 12 = 192

Answer: (a) Perimeter is 28 ft, area is 32 sq ft (b) Perimeter is 42 ft, area is 72 sq ft (c) Perimeter is 7 ft, area is 2 sq ft

Explanation: Perimeter of parallelogram P = 2(2 + 5) = 7 x 2 = 14 ft Area of the parallelogram = 2 x 4 = 8 sq ft (a) If scale factor k = 2, then Perimeter = kP = 2 x 14 = 28 Area = k²A = 2² x 8 = 4 x 8 = 32 (b) If scale factor k = 3, then Perimeter = kP = 3 x 14 = 42 Area = k²A = 3² x 8 = 72 (c) If scale factor k = \(\frac{1}{2}\), then Perimeter = kP = \(\frac{1}{2}\) x 14 = 7 Area = k²A = \(\frac{1}{2²}\) x 8 = \(\frac{1}{4}\) x 8 = 2

Question 3. A rectangular prism is 3 inches wide, 4 inches long, and 5 inches tall. Find the surface area and volume of the image of the prism when it is dilated by a scale factor of (a) 2, (b) 3, and (c) 4.

Answer: (a) Surface area is 376 sq in, volume is 480 cubic in (b) Surface area is 846 sq in, volume is 1620 cubic in (c) Surface area is 1504 sq in, volume is 3840 cubic in

Explanation: The surface area of the rectangular prism A = 2(3 x 4 + 4 x 5 + 5 x 3) = 2(12 + 20 + 15) = 2(47) = 94 in Volume of the rectangular prism V = 3 x 4 x 5 = 60 in³ (a) If the scale factor k = 2, then Surface Area = k²A = 2² x 94 = 4 x 94 = 376 sq in Volume = k³V = 2³ x 60 = 8 x 60 = 480 cubic in (b) If the scale factor k = 3, then Surface Area = k²A = 3² x 94 = 9 x 94 = 846 Volume = k³V = 3³ x 60 = 27 x 60 = 1620 (c) If the scale factor k = 4, then Surface Area = k²A = 4² x 94 = 16 x 94 = 1504 Volume = k³V = 4³ x 60 = 64 x 60 = 3840

8.1 Similar Polygons

Exploration 1

Comparing Triangles after a Dilation

Work with a partner: Use dynamic geometry software to draw any ∆ABC. Dilate ∆ABC to form a similar ∆A’B’C’ using an scale factor k and an center of dilation.

c. Repeat parts (a) and (b) for several other triangles, scale factors, and centers of dilation. Do you obtain similar results? Answer: From the newly transformed triangles in part (a) and (b), side lengths of the triangles are changed by the scale factor of k and angle measures remain the same for both of the triangles ΔABC.

Exploration 2

Work with a partner: Use dynamic geometry Software to draw any ∆ABC. Dilate ∆ABC to form a similar ∆A’B’C’ using any scale factor k and any center of dilation.

a. Compare the perimeters of ∆A’B’C and ∆ABC. What do you observe? Answer:

b. Compare the areas of ∆A’B’C’ and ∆ABC. What do you observe? Answer:

c. Repeat parts (a) and (b) for several other triangles, scale factors, and centers of dilation. Do you obtain similar results? LOOKING FOR STRUCTURE To be proficient in math, you need to look closely to discern a pattern or structure. Answer:

Communicate Your Answer

Question 3. How are similar polygons related?

Answer: if two polygons are similar means they have the same shape, corresponding angles are congruent and the ratios of lengths of their corresponding sides are equal. The common ratio is called the scale factor.

Question 4. A ∆RST is dilated by a scale factor of 3 to form ∆R’S’T’. The area of ∆RST is 1 square inch. What is the area of ∆R’S’T’?

Answer: Area of ∆R’S’T’ = 9 sq in

Explanation: Given that, Area of ∆RST = 1 sq inch Scale factor k = 3 Area of ∆R’S’T’ = k² x Area of ∆RST = 3² x 1 = 9 x 1 = 9

Lesson 8.1 Similar Polygons

Answer: The pairs of congruent angles are ∠K = ∠Q, ∠J = ∠P, ∠ L = ∠R The scale factor is \(\frac { 3 }{ 2 } \) The ratios of the corresponding side lengths in a statement of proportionality are \(\frac { PQ }{ JK } = \frac { PR }{ JL } = \frac { QR }{ LK } \)

Explanation: Given that, ∆JKL ~ ∆PQR The pairs of congruent angles are ∠K = ∠Q, ∠J = ∠P, ∠ L = ∠R To find the scale factor, \(\frac { PQ }{ JK } = \frac { 9 }{ 6 } \) = \(\frac { 3 }{ 2 } \), \(\frac { PR }{ JL } = \frac { 12 }{ 8 } \) = \(\frac { 3 }{ 2 } \), \(\frac { QR }{ LK } = \frac { 6 }{ 4 } \) = \(\frac { 3 }{ 2 } \) So, the scale factor is \(\frac { 3 }{ 2 } \)

Answer: x = 2

Explanation: The triangles are similar, so corresponding side lengths are proportional. \(\frac { RS }{ BC } \) = \(\frac { AB }{ QR } \) \(\frac { 4 }{ x } \) = \(\frac { 12 }{ 6 } \) \(\frac { 4 }{ x } \) = 2 4 = 2x x = 2

Answer: KM = 42

Explanation: Scale factor = \(\frac { JM }{ GH } \) = \(\frac { 48 }{ 40 } \) = \(\frac { 6 }{ 5 } \) Because the ratio of the lengths of the altitudes in similar triangles is equal to the scale factor, you can write the following proportion \(\frac { KM }{ HF } \) = \(\frac { 6 }{ 5 } \) \(\frac { KM }{ 35 } \) = \(\frac { 6 }{ 5 } \) KM = \(\frac { 6 }{ 5 } \) x 35 KM = 42

Answer: Perimeter of Gazebo A = 46 m

Explanation: Scale factor = \(\frac { AB }{ FG } \) = \(\frac { 10 }{ 15 } \) = \(\frac { 2 }{ 3 } \) So, \(\frac { AE }{ FK } \) = \(\frac { 2 }{ 3 } \) \(\frac { x }{ 18 } \) = \(\frac { 2 }{ 3 } \) x = 12 \(\frac { ED }{ KJ } \) = \(\frac { 2 }{ 3 } \) \(\frac { ED }{ 15 } \) = \(\frac { 2 }{ 3 } \) ED = 10 \(\frac { DC }{ JH } \) = \(\frac { 2 }{ 3 } \) \(\frac { DC }{ 12 } \) = \(\frac { 2 }{ 3 } \) DC = 8 \(\frac { BC }{ GH } \) = \(\frac { 2 }{ 3 } \) \(\frac { BC }{ 9 } \) = \(\frac { 2 }{ 3 } \) BC = 6 Therefore, perimeter = 6 + 8 + 10 + 12 + 10 = 46

Answer: Area of LMNP = 756 m 2

Explanation: As shapes are similar, their corresponding side lengths are proportional. Scale Factor k = \(\frac { NP }{ JK } \) = \(\frac { 21 }{ 7 } \) = 3 Area of LMNP = k² x Area of GHJK = 3² x 84 = 756 m 2

Answer: Both the hexagons are different. On the outer side of the hexagon, all the sides are equal. In the inside hexagon among the 6-sided 3 sides are different.

Answer: Both the hexagons are similar. Because all the sides of the outer hexagon are equal and all the sides of the inner hexagon are also equal.

Exercise 8.1 Similar Polygons

Vocabulary and Core Concept Check

Question 1. COMPLETE THE SENTENCE For two figures to be similar, the corresponding angles must be ____________ . and the corresponding side lengths must be _____________ .

What is the ratio of their areas?

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, find the scale factor. Then list all pairs of congruent angles and write the ratios of the corresponding side lengths in a statement of proportionality.

In Exercises 5-8, the polygons are similar. Find the value of x.

Answer: x = 20

Explanation: \(\frac { DF }{ GJ } \) = \(\frac { DE }{ GH } \) \(\frac { 16 }{ 12 } \) = \(\frac { x }{ 15 } \) x = \(\frac { 16 x 15 }{ 12 } \) = \(\frac { 240 }{ 12 } \) x = 20

Answer: x = 12

Explanation: \(\frac { PN }{ KJ } \) = \(\frac { MN }{ JH } \) \(\frac { x }{ 8 } \) = \(\frac { 9 }{ 6 } \) x = \(\frac { 9 x 8 }{ 6 } \) = \(\frac { 72 }{ 6 } \) x = 12

In Exercises 9 and 10, the black triangles are similar. Identify the type of segment shown in blue and find the value of the variab1e.

Explanation: \(\frac { y }{ 18 } \) = \(\frac { y – 1 }{ 16 } \) 18(y – 1) = 16y 18y – 18 = 16y 18y – 16y = 18 2y = 18 y = 9

In Exercises 11 and 12, RSTU ~ ABCD. Find the ratio of their perimeters.

In Exercises 13-16, two polygons are similar. The perimeter of one polygon and the ratio of the corresponding side lengths are given. Find the perimeter of the other polygon.

Question 14. perimeter of smaller polygon: 66 ft: ratio: \(\frac{3}{4}\)

Answer: The perimeter of larger polygon is 88 ft.

Explanation: \(\frac { smaller }{ larger } \) = \(\frac { 66 }{ x } \) = \(\frac{3}{4}\) 66 x 4 = 3x 3x = 264 x = \(\frac { 264 }{ 3 } \) = 88

Question 16. perimeter of larger polygon: 85 m; ratio: \(\frac{2}{5}\)

Answer: The perimeter of smaller polygon is 34 m.

Explanation: \(\frac { smaller }{ larger } \) = \(\frac { x }{ 85 } \) = \(\frac{2}{5}\) 85 x 2 = 5x 5x = 170 x = \(\frac { 170 }{ 5 } \) = 34

Question 18. MODELING WITH MATHEMATICS Your family has decided to put a rectangular patio in your backyard. similar to the shape of your backyard. Your backyard has a length of 45 feet and a width of 20 feet. The length of your new patio is 18 feet. Find the perimeters of your backyard and of the patio.

Answer: The perimeter of the backyard is 130 ft. Perimeter of patio is 52 ft

In Exercises 19-22, the polygons are similar. The area of one polygon is given. Find the area of the other polygon.

Answer: Area of the larger triangle is 90 cm²

Explanation: \(\frac { 10 }{ A } \) = (\(\frac { 4 }{ 12 } \))² \(\frac { 10 }{ A } \) = \(\frac { 1 }{ 9 } \) A = 10 x 9 A = 90

Answer: Area of smaller triangle = 6 sq cm

Explanation: \(\frac { A }{ 96 } \) = (\(\frac { 3 }{ 12 } \))² \(\frac { A }{ 96 } \) = \(\frac { 1 }{ 16 } \) 16A = 96 A = \(\frac { 96 }{ 16 } \) A = 6

Answer: Because the first ratio has a side of A over the side length of B, the square of the second ratio should have the area of B over the area of A. \(\frac { 24 }{ x } \) = (\(\frac { 6 }{ 18 } \))² \(\frac { 24 }{ x } \) = \(\frac { 1 }{ 9 } \) x = 24 x 9 x = 216

In Exercises 25 and 26, decide whether the red and blue polygons are similar.

Answer: Yes Because both shapes are apparent and their side lengths are proportional and their corresponding angles are congruent.

ANALYZING RELATIONSHIPS In Exercises 28 – 34, JKLM ~ EFGH.

Question 28. Find the scale factor of JKLM to EFGH.

Answer: scale factor = \(\frac { EF }{ JK } \) =\(\frac { 8 }{ 20 } \) k = \(\frac { 2 }{ 5 } \)

Question 30. Find the values of x, y, and z.

Answer: x = \(\frac { 55 }{ 2 } \) y = 12 z = 65°

Explanation: \(\frac { KL }{ GF } \) = \(\frac { x }{ 11 } \) = \(\frac { 5 }{ 2 } \) 2x = 55 x = \(\frac { 55 }{ 2 } \) \(\frac { MJ }{ HE } \) = \(\frac { 30 }{ y } \) = \(\frac { 5 }{ 2 } \) 5y = 60 y = 12

Question 32. Find the ratio of the perimeters of JKLM to EFGH.

Answer: The perimeter of JKLM : Perimeter of EFGH = 85 : 34

Question 34. Find the ratio of the areas of JKLM to EFGH.

Answer: Area of JKLM : Area of EFGH = 378.125 : 60.5 = 25 : 4

Answer: The tennis table and court are not similar

Explanation: If two figures are similar then their angles are congruent and sides are proportional. If the tennis court and table are similar, then \(\frac { length of table }{ length of court } \) = \(\frac { width of table }{ width of court } \) \(\frac { 9 }{ 78 } \) = \(\frac { 5 }{ 36 } \) 9 • 36 = 5 • 78 324 = 390 So, Table and court are not similar.

MATHEMATICAL CONNECTIONS In Exercises 37 and 38, the two polygons are similar. Find the values of x and y.

Answer: x = 7.5

Explanation: \(\frac { x }{ 5 } \) = \(\frac { 6 }{ 4 } \) x = \(\frac { 15 }{ 2 } \)

ATTENDING TO PRECISION In Exercises 39 – 42. the figures are similar. Find the missing corresponding side length.

Question 40. Figure A has a perimeter of 24 inches. Figure B has a perimeter of 36 inches and one of the side lengths is 12 inches.

Answer: The corresponding side length of figure A is 8 in

Explanation: \(\frac { Perimeter of A }{ Perimeter of B } \) = \(\frac { Side length of A }{ Side length of B } \) \(\frac { 24 }{ 36 } \) = \(\frac { x }{ 12 } \) \(\frac { 2 }{ 3 } \) = \(\frac { x }{ 12 } \) 12 • 2 = 3x x = 8

Question 42. Figure A has an area of 18 square feet. Figure B has an area of 98 square feet and one of the side lengths is 14 feet.

Answer: The corresponding side length of figure A is 6 ft.

Explanation: \(\frac { Area of A }{ Area of B } \) = (\(\frac { Side length of A }{ Side length of B } \))² \(\frac { 18 }{ 98 } \) = (\(\frac { x }{ 14 } \))² \(\frac { 9 }{ 49 } \) = \(\frac { x² }{ 196 } \) x² = 36 x = 6

CRITICAL THINKING In Exercises 43-48, tell whether the polygons are always, sometimes, or never similar.

Question 44. two isosceles trapezoids

Answer: Two isosceles trapezoids are sometimes similar.

Question 46. two squares

Answer: Two squares are always similar.

Question 48. a right triangle and an equilateral triangle

Answer: A right triangle and an equilateral triangle are never similar.

Question 52. PROVING A THEOREM Prove the Perimeters of Similar Polygons Theorem (Theorem 8.1) for similar rectangles. Include a diagram in your proof.

Question 54. THOUGHT PROVOKING The postulates and theorems in this book represent Euclidean geometry. In spherical geometry. all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. A plane is the surface of the sphere. In spherical geometry, is it possible that two triangles are similar but not congruent? Explain your reasoning. Answer:

• In Euclidean geometry, a postulate exists starting that through a point, there exists only one parallel line.
• In spherical geometry, great circles are straight lines and all of them intersect all the others. So, they are all parallel to each other. Thus, there are no parallel lines in spherical geometry.
• In spherical geometry, a line is a great circle. If there are many lines that contain any point then they are perpendicular to another line.
• So, in spherical geometry, eight right angles are formed by two perpendicular lines.

Maintaining Mathematical proficiency

Find the value of x.

Answer: x = 66°

Explanation: x + 24 + 90 = 180 x + 114 = 180 x = 180 – 114 x = 66

Answer: x = 60°

Explanation: x + x + x = 180 3x = 180 x = 60

8.2 Proving Triangle Similarity by AA

Comparing Triangles

Work with a partner. Use dynamic geometry software.

c. Are the two triangles similar? Explain. CONSTRUCTING VIABLE ARGUMENTS To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments. Answer:

d. Repeat parts (a) – (c) to complete columns 2 and 3 of the table for the given angle measures. Answer:

e. Complete each remaining column of the table using your own choice of two pairs of equal corresponding angle measures. Can you construct two triangles in this way that are not similar? Answer:

f. Make a conjecture about any two triangles with two pairs of congruent corresponding angles. Answer:

Question 2. What can you conclude about two triangles when you know that two pairs of corresponding angles are congruent? Answer: If two pairs of corresponding angles in a pair of triangles are congruent, then the triangles are similar. Because if the two angle pairs are the same, then the third pair must also be equal when the three angle pairs are all equal, the three pairs of sides must be in proportion.

Question 3. Find RS in the figure at the left. Answer: In a triangle RST TS = RS We know that TS = 4 Then RS = 4.

Lesson 8.2 Proving Triangle Similarity by AA

Show that the triangles are similar. Write a similarity statement.

Answer: m∠CDF + 32 = 90 degrees m∠CDF = 58 degrees ∠CDF ≅∠DFE And ∠CFE ≅ DFE So, △CDF = △DEF.

Question 4. WHAT IF? A child who is 58 inches tall is standing next to the woman in Example 3. How long is the child’s shadow’? Answer: Given that, The child is 58 inches tall and is standing next to the woman. The child shadow is 60/58 = 40/x 2330 = 60x 2330/60 = x x = 38.8. Therefore Childs shadow is 38.8 cm.

Question 5. You are standing outside, and you measure the lengths 0f the shadows cast by both you and a tree. Write a proportion showing how you could find the height of the tree. Answer: The propagation shows to find the height of the tree is My height/tree height = my shadow/tree shadow.

Exercise 8.2 Proving Triangle Similarity by AA

Question 2. WRITING Can you assume that corresponding sides and corresponding angles of any two similar triangles are congruent? Explain. Answer: The corresponding angles of two similar triangles are always congruent but the corresponding angles of two analogous triangles are always harmonious but the corresponding sides of the two triangles don’t have to be harmonious. In an analogous triangle, the corresponding sides are commensurable which means that the rates of corresponding sides are equal. However, also the corresponding sides are harmonious, but in other cases, if these rates are 1.

In Exercises 3 – 6. determine whether the triangles are similar. If they are, write a similarity statement. Explain your reasoning.

In Exercises 7 – 10. show that the two triangles are similar.

In Exercises 11 – 18, use the diagram to copy and complete the statement.

Question 12. ∆DCF ~ _________ Answer: △DCF ~ △BCG

Question 14. m∠ECF = _________ Answer: m∠ECF = 37 degrees.

Question 16. CF = _________ Answer: CF = 16

Question 18. DE = _________ Answer: DE = 21

To estimate the height of the pole △CDE can be proved similar to △CAD as ∠C (common angle)

∠CED = ∠CBA (corresponding angles)

If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. △CDE ~ △CAB

Corresponding angles are congruent and corresponding sides are proportional for similar triangles. DE/AB = CE/CB is an equation 1 Measure my height (DE), shadows CE and CB to calculate the height of pole (AB) from equation 1.

REASONING In Exercises 23 – 26, is it possible for ∆JKL and ∆XYZ to be similar? Explain your reasoning.

Question 24. ∆JKL is a right triangle and m∠X + m∠Y= 150°. Answer: The sum of angles = 180 150 + 90 > 180 So, the answer is no.

Question 26. m∠J + m∠K = 85° and m∠Y + m∠Z = 80° Answer: The third angle in both triangles would be: For△JKL ​m∠L=180−85 m∠L=95 For△XYZ, ​m∠X=180 − 80 m∠X=100 So, there wouldn’t be any single similar angle. So, no the two triangles△JKL,△ XYZ could not be similar.

Question 30. THOUGHT PROVOKING Decide whether each is a valid method of showing that two quadrilaterals are similar. Justify your answer. a. AAA Answer: The AAA Similarity Theorem is for triangles only and therefore, cannot be applied in case of quadrilaterals. Hence the AAA system is not valid for showing two quadrilaterals to be analogous.

b. AAAA Answer: The AAAA method is not a valid method for showing two quadrilaterals to be similar as in case of the rectangle and a square, all the four and the rectangle are not similar because the corresponding sides are not proportional.

Maintaining Mathematical Practices

Determine whether there is enough information to prove that the triangles are congruent. Explain your reasoning.

8.1 & 8.2 Quiz

List all pairs of congruent angles. Then write the ratios of the corresponding side lengths in a statement of proportionality.

The polygons are similar. Find the value of x.

Determine whether the polygons are similar. If they are, write a similarity statement. Explain your reasoning. (Section 8.1 and Section 8.2)

Show that the two triangles are similar.

Question 11. The dimensions of an official hockey rink used by the National Hockey League (NHL) are 200 feet by 85 feet. The dimensions of an air hockey table are 96 inches by 408 inches. Assume corresponding angles are congruent. (Section 8.1) a. Determine whether the two surfaces are similar. Answer: Given, The dimensions of an official hockey rink used by the National Hockey League (NHL) are 200 feet by 85 feet. The dimensions of an air hockey table are 96 inches by 408 inches. 1 feet = 12 inches 200 feet = 200 × 12 = 2400 inches 85 feet = 85 × 12 = 1020 inches Dimensions of a hockey rink in inches are 2400 inches by 1020 inches If the measure of the corresponding sides that include the angles are proportional hockey rink and hockey table are similar 200/96 = 240/96 = 25 inches 85/40.8 = 1020/10.8 = 25 inches 2400/96 = 1020/40.8 = 25 inches Thus the two rinks are similar.

b. If the surfaces are similar, find the ratio of their perimeters and the ratio of their areas. If not, find the dimensions of an air hockey table that are similar to an NHL hockey rink. Answer: P1 = 2(l + w) 2(2400 + 1020) 2(3420) = 6840 inches P2 = 2(l + w) 2(96 + 40.8) 2(136.8) = 273.6 inches P1/P2 = 6840/273.6 = 25 inches A1 = lw A1 = 2400 × 1020 = 2448000 sq. inch A2 = lw A2 = 96 × 40.8 = 3916.8 sq. inch A1/A2 = 2448000/3916.8 = 625 sq. inch

8.3 Proving Triangle Similarity by SSS and SAS

Deciding Whether Triangles Are Similar

Work with a partner: Use dynamic geometry software.

b. Copy the table and complete column 1. Answer:

c. Are the triangles similar? Explain your reasoning. Answer:

d. Repeat parts (a) – (c) for columns 2 – 6 in the table. Answer:

e. How are the corresponding side lengths related in each pair of triangles that are similar? Is this true for each pair of triangles that are not similar? Answer:

f. Make a conjecture about the similarity of two triangles based on their corresponding side lengths. CONSTRUCTING VIABLE ARGUMENTS To be proficient in math, you need to analyze situations by breaking them into cases and recognize and use counterexamples. Answer:

g. Use your conjecture to write another set of side lengths of two similar triangles. Use the side lengths to complete column 7 of the table. Answer:

Work with a partner: Use dynamic geometry software. Construct any ∆ABC. a. Find AB, AC, and m∠A. Choose any positive rational number k and construct ∆DEF so that DE = k • AB, DF = k • AC, and m∠D = m∠A. Answer:

b. Is ∆DEF similar to ∆ABC? Explain your reasoning. Answer:

c. Repeat parts (a) and (b) several times by changing ∆ABC and k. Describe your results. Answer:

Question 3. What are two ways to use the corresponding sides of two triangles to determine that the triangles are similar? Answer: If the pairs of corresponding sides are in proportion and the included angle of each pair is equal. Then the two triangles they form are similar. Any time two sides of a triangle and their included angles are fixed. Then all three vertices of that triangle are fixed.

Lesson 8.3 Proving Triangle Similarity by SSS and SAS

Monitoring progress

Use the diagram.

Question 1. Which of the three triangles are similar? Write a similarity statement.

Answer: The ratios are equal. So, △LMN, △XYZ are similar. The ratios are not equal. So △LMN, △RST are not similar.

Explanation: Compare △LMN, △XYZ by finding the ratios of corresponding side lengths Shortest sides: \(\frac { LM }{ YZ } \) = \(\frac { 20 }{ 30 } \) = \(\frac { 2 }{ 3 } \) Longest sides: \(\frac { LN }{ XY } \) = \(\frac { 26 }{ 39 } \) = \(\frac { 2 }{ 3 } \) Remaining sides: \(\frac { MN }{ ZX } \) = \(\frac { 24 }{ 36 } \) = \(\frac { 2 }{ 3 } \) The ratios are equal. So, △LMN, △XYZ are similar. Compare △LMN, △RST by finding the ratios of corresponding side lengths Shortest sides: \(\frac { LM }{ RS } \) = \(\frac { 20 }{ 24 } \) = \(\frac { 5 }{ 6 } \) Longest sides: \(\frac { LN }{ ST } \) = \(\frac { 26 }{ 33 } \) Remaining sides: \(\frac { MN }{ RT } \) = \(\frac { 24 }{ 30 } \) = \(\frac { 4 }{ 5 } \) The ratios are not equal. So △LMN, △RST are not similar.

Question 2. The shortest side of a triangle similar to ∆RST is 12 units long. Find the other side 1enths of the triangle.

Answer: The other side lengths of the triangle are 15 units, 16.5 units.

Explanation: The shortest side of a triangle similar to ∆RST is 12 units Scale factor = \(\frac { 12 }{ 24 } \) = \(\frac { 1 }{ 2 } \) So, other sides are 33 x \(\frac { 12 }{ 2 } \) = 16.5, 30 x \(\frac { 12 }{ 2 } \) = 15.

Explain how to show that the indicated triangles are similar.

Answer: The shorter sides: \(\frac { 18 }{ 24 } \) = \(\frac { 3 }{ 4 } \) Longer sides: \(\frac { 21 }{ 28 } \) = \(\frac { 3 }{ 4 } \) The side lengths are proportional. So ∆SRT ~ ∆PNQ

Answer: ∆XZW and ∆YZX are not proportional.

Explanation: The shorter sides: \(\frac { 9 }{ 16 } \) Longer sides: \(\frac { 15 }{ 20 } \) = \(\frac { 3 }{ 4 } \) The side lengths are not proportional. So ∆XZW and ∆YZX are not proportional.

Exercise 8.3 Proving Triangle Similarity by SSS and SAS

Monitoring progress and Modeling with Mathematics

In Exercises 3 and 4, determine whether ∆JKL or ∆RST is similar to ∆ABC.

In Exercises 5 and 6, find the value of x that makes ∆DEF ~ ∆XYZ.

In Exercises 7 and 8, verify that ∆ABC ~ ∆DEF Find the scale factor of ∆ABC to ∆DEF

Question 8. ∆ABC: AB = 10, BC = 16, CA = 20 ∆DEF: DE = 25, EF = 40, FD =50 Answer: Compare △ABC and △DEF DE/AB = 25/10 = 5/2 EF/BC = 40/16 = 5/2 FD/CA = 50/20 = 5/2 All the ratios of corresponding length sides are equal. It is called a scale factor and its length is equal to 5/2. So, △ABC ~△DEF by the SSS similarity theorem.

In Exercises 9 and 10. determine whether the two triangles are similar. If they are similar, write a similarity statement and find the scale factor of triangle B to triangle A.

In Exercises 11 and 12, sketch the triangles using the given description. Then determine whether the two triangles can be similar.

Question 12. The side lengths of ∆ABC are 24, 8x, and 48, and the side lengths of ∆DEF are 15, 25, and 6x.

Answer: \(\frac { AB }{ DE } \) = \(\frac { AC }{ DF } \) = \(\frac { BC }{ EF } \) \(\frac { 24 }{ 15 } \) = \(\frac { 8x }{ 25 } \) x = 5

In Exercises 13 – 16. show that the triangles are similar and write a similarity statement. Explain your reasoning.

In Exercises 17 and 18, use ∆XYZ.

Question 18. The longest side of a triangle similar to ∆XYZ is 39 units long. Find the other side lengths of the triangle. Answer: The longest side of a triangle similar to ∆XYZ is 39 units long. 13/39 = 12/y 13y = 39 × 12 13y = 468 y = 468/13 y = 36 13/39 = 10/y 13y = 39(10) 13y = 390 y = 390/13 y = 30 The side lengths are 30, 36 and 39.

Question 20. MATHEMATICAL CONNECTIONS Find the value of n that makes ∆DEF ~ ∆XYZ when DE = 4, EF = 5, XY = 4(n + 1), YZ = 7n – 1, and ∠E ≅ ∠Y. Include a sketch.

Answer: \(\frac { DE }{ XY } \) = \(\frac { EF }{ YZ } \) \(\frac { 4 }{ 4(n + 1) } \) = \(\frac { 5 }{ 7n – 1 } \) cross multiply the fractions 4(7n – 1) = 20(n + 1) 28n – 4 = 20n + 20 28n – 20n = 20 + 4 8n = 24 n = \(\frac { 24 }{ 8 } \) n = 3

ATTENDING TO PRECISION In Exercises 21 – 26, use the diagram to copy and complete the statement.

Question 22. m∠NRQ = ___________ Answer: m∠NRQ = m∠NRP = 91° by the vertical congruence

Question 24. RQ = ___________

Answer: RQ = 4√3

Explanation: Using the pythogrean theorem NQ² = NR² + RQ² 8² = 4² + RQ² 64 = 16 + RQ² 64 – 16 = RQ² 48 = RQ² RQ = 4√3

Question 26. m∠NPR = ___________

Answer: m∠NPR = 28°

Explanation: m∠NPR + m∠NRP + m∠RNP = 180° m∠NPR + 91° + 61° = 180° m∠NPR + 152° = 180° m∠NPR = 180° – 152° m∠NPR = 28°

Explanation: Check the similarity of maroon and violet triangles. longest sides: \(\frac { 5 }{ 7 } \) shortest sides: \(\frac { 3 }{ 4 } \) remaining sides: \(\frac { 3 }{ 4 } \) The ratios are not equal. So those traingles are not similar. Check the similarity of blue and violet triangles. longest sides: \(\frac { 5 }{ 5.25 } \) = 1 shortest sides: \(\frac { 3 }{ 3 } \) = 1 remaining sides: \(\frac { 3 }{ 3 } \) = 1 The ratios are equal. So those traingles are similar.

Question 30. WRITING Are any two right triangles similar? Explain.

Answer: Yes, any two right triangles can be similar. If two right triangles are similar, then the ratio of their longest, smallest and remaining side lengths must be equal and their angles must be congruent.

Answer: By observing the triangle ABC, m∠BAC = 90° Join the midpoints of the sides of the triangle. m∠DEF = 90°

b. Write the ratio of the lengths of the second pair of corresponding legs. Answer: First find the length of AC using pythagorean theorem AC² + AB² = BC² AC² + 36 = 100 AC² = 64 AC = 8 Find the length of DF using pythagorean theorem DF² + DE² = EF² DF² + 18² = 30² DF² = 900 – 324 DF² = 576 DF= 24

c. Are these triangles similar? Does this suggest a Hypotenuse-Leg Similarity Theorem for right triangles? Explain. Answer: k = \(\frac { AC }{ DF } \) = \(\frac { 8 }{ 24 } \) = \(\frac { 1 }{ 3 } \) k = \(\frac { AB }{ DE } \) = \(\frac { 6 }{ 18 } \) = \(\frac { 1 }{ 3 } \) So, triangles are similar.

Question 38. THOUGHT PROVOKING Decide whether each is a valid method of showing that two quadrilaterals are similar. Justify your answer. a. SASA Answer: If an angle of one triangle is congruent to an angle of a second triangle and the lengths of the sides including these angles are proportional, then the triangles are similar.

b. SASAS Answer: If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

c. SSSS Answer: If the corresponding side lengths of two triangles are proportional, then the triangles are similar.

d. SASSS Answer: If two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then those two triangles are similar.

Question 41. PROVING A THEOREM Copy and complete the paragraph proot of the second part of the Slopes of Parallel Lines Theorem (Theorern 3. 13) from page 439. Given m l = m n , l and n are nonvertical. Prove l || n You are given that m l = m n . By the definition of slope. m l = \(\frac{B C}{A C}\) and m n = \(\frac{E F}{D F}\) By ____________, \(\frac{B C}{A C}=\frac{E F}{D F}\). Rewriting this proportion yields ___________,

Find the coordinates of point P along the directed line segment AB so that AP to PB is the given ratio.

Question 44. A(- 3, – 5), B(9, – 1); 1 to 3 Answer: In order to divide the segment in the ratio 1 to 3, think of dividing the segment into 1 + 3 or 4 congruent pieces. Point P is the point that is 1/4 of the way from point A to B. Slope of AB = (-1 + 5)/(9+3) = 4/12 = 1/3 = rise/run To find the coordinates of point P, add 1/4 of the run to the x-coordinate of A and 1/4 of the rise to the y-coordinate of A. run = 1/4 × 3 = 3/4 rise = 1/4 × 1 = 1/4

8.4 Proportionality Theorems

Discovering a Proportionality Relationship

b. Compare the ratios of AD to BD and AE to CE. Answer:

c. Move \(\overline{D E}\) to other locations Parallel to \(\overline{B C}\) with endpoints on \(\overline{A B}\) and \(\overline{A C}\), and repeat part (b). Answer:

d. Change ∆ABC and repeat parts (a) – (c) several times. Write a conjecture that summarizes your results. LOOKING FOR STRUCTURE To be proficient in math, you need to look closely to discern a pattern or structure. Answer:

Work with a partner. Use dynamic geometry software to draw any AABC.

a. Bisect ∆B and plot point D at the intersection of the angle bisector and \(\overline{A C}\). Answer:

b. Compare the ratios of AD to DC and BA to BC. Answer:

c. Change ∆ABC and repeat parts (a) and (b) several times. Write a conjecture that summarizes your results. Answer:

• The proportionality relationship that exists in a triangle intersected by an angle bisector is it divides the opposite side into segments whose lengths are proportional to the lengths of the other two sides.
• The proportionality relationships that exist in a triangle by a line parallel to one side of a triangle intersects the other two sides, then divides the two sides proportionally.

Lesson 8.4 Proportionality Theorems

Explanation: Triangle property thorem is \(\frac { XW }{ WV } \) = \(\frac { XY }{ YZ } \) \(\frac { 44 }{ 35 } \) = \(\frac { 36 }{ YZ } \) cross multiply the fractions 44 • YZ = 36 • 35 44 • YZ = 1260 YZ = \(\frac { 1260 }{ 44 } \) YZ = \(\frac { 315 }{ 11 } \)

Find the length of the given line segment.

Answer: \(\overline{B D}\) = 12

Explanation: All the angles are congruent. So, \(\overline{A B}\), \(\overline{C D}\), \(\overline{E F}\) are parallel. using the three parallel lines theorem \(\frac { BD }{ DF } \) = \(\frac { AC }{ CE } \) \(\frac { [latex]\overline{B D}\) }{ 30 } [/latex] = \(\frac { 16 }{ 40 } \) \(\overline{B D}\) = \(\frac { 16 }{ 40 } \) • 30 \(\overline{B D}\) = 12

Answer: \(\overline{J M}\) = \(\frac { 96 }{ 5 } \)

Explanation: All the angles are congruent. So, \(\overline{G H}\), \(\overline{J K}\), \(\overline{M N}\) are parallel. using the three parallel lines theorem \(\frac { HK }{ KN } \) = \(\frac { GJ }{ JM } \) \(\frac { 15 }{ 18 } \) = \(\frac { 16 }{ [latex]\overline{J M}\) } [/latex] Cross multiply 15 • \(\overline{J M}\) = 16 • 18 = 288 \(\overline{J M}\) = \(\frac { 288 }{ 15 } \) \(\overline{J M}\) = \(\frac { 96 }{ 5 } \)

Find the value of the variable.

Answer: x = 28

Explanation: \(\overline{T V}\) is the angle bisector So, \(\frac { ST }{ TU } \) = \(\frac { SV }{ VU } \) \(\frac { 14 }{ x } \) = \(\frac { 24 }{ 48 } \) cross multiply 24x = 14 • 48 = 672 x = \(\frac { 672 }{ 24 } \) x = 28

Answer: x = 4√2

Explanation: \(\overline{W Z}\) is the angle bisector So, \(\frac { YZ }{ ZX } \) = \(\frac { YW }{ WX } \) \(\frac { 4 }{ 4 } \) = \(\frac { 4√2 }{ x } \) cross multiply 4x = 4 • 4√2 = 16√2 x = 4√2

Exercise 8.4 Proportionality Theorems

Question 2. VOCABULARY In ∆ABC, point R lies on \(\overline{B C}\) and \(\vec{A}\)R bisects ∆CAB. Write the proportionality statement for the triangle that is based on the Triangle Angle Bisector Theorem (Theorem 8.9).

Answer: According to the triangle angle bisector theorem \(\frac { CR }{ BR } \) = \(\frac { AC }{ AB } \)

In Exercises 3 and 4, find the length of \(\overline{A B}\) .

Answer: \(\frac { AE }{ ED } \) = \(\frac { AB }{ BC } \) \(\frac { 14 }{ 12 } \) = \(\frac { AB }{ 18 } \) AB = \(\frac { 14 }{ 12 } \) • 18 AB = 21 units.

In Exercises 5 – 8, determine whether \(\overline{K M}\) || \(\overline{J N}\).

Question 10. 2 in.; 2 to 3

Answer: Construct a 2 inch segment and divide the segment into 2 + 3 or 5 congruent pieces. Point P is the point that is \(\frac { 1 }{ 5 } \) of the way from point A to point B.

Question 12. 9 cm ; 2 to 5 Answer: Construct a 9 cm segment and divide the segment into 2 + 5 or 7 congruent pieces. Point p is the point that is \(\frac { 1 }{ 7 } \) of the way from point A to point B.

In Exercises 13 – 16, use the diagram to complete the proportion.

In Exercises 17 and 18, find the length of the indicated line segment.

In Exercises 19 – 22, find the value of the variable.

MATHEMATICAL CONNECTIONS In Exercises 25 and 26, find the value of x for which \(\overline{P Q}\) || \(\overline{R S}\).

Question 36. THOUGHT PROVOKING Write the converse of the Triangle Angle Bisector Theorem (Theorem 8.9). Is the converse true? Justify your answer. Answer: Stating the converse of the Triangle Angle Bisector Theorem. The converse of the Triangle Angle Bisector Theorem states that if a ray divides a side of a triangle into segments whose lengths are proportional to the lengths of the other two sides, it bisects the angle opposite to the first side. AB/AC = BD/DC Showing ΔABD and ΔCDE are similar By construction, CE ∣∣AB So ∠CED≅∠DAB And ∠ECD ≅ ∠DBA So ΔABD∼ΔCDE BD/CD = AB/CE AC = CE ∠CED≅∠DAB and ∠CAD≅∠DAB So, ∠CED≅∠DAB Hence AD is an angle bisector of ∠BAC

Maintaining Mathematical Proficiency

Use the triangle.

Question 42. Which side is the hypotenuse? Answer: The leg c is the hypotenuse.

Solve the equation.

Question 44. x 2 + 16 = 25 Answer: x² + 16 = 25 x² = 25 – 16 x² = 9 x = ∓ 3

Similarity Review

Find the scale factor. Then list all pairs of congruent angles and write the ratios of the corresponding side lengths in a statement of proportionality.

Question 3. Two similar triangles have a scale factor of 3 : 5. The altitude of the larger triangle is 24 inches. What is the altitude of the smaller triangle? Answer: Scale factor of smaller triangle to larger traingle is \(\frac { 3 }{ 5 } \) and larger traingle altitude is 24 inches Let x be the smaller triangle altitude \(\frac { altitude of smaller triangle }{ altitude of larger traingle } \) = scale factor \(\frac { x }{ 24 } \) = \(\frac { 3 }{ 5 } \) x = \(\frac { 3 }{ 5 } \) • 24 x = 14.4

Question 4. Two similar triangles have a pair of corresponding sides of length 12 meters and 8 meters. The larger triangle has a perimeter of 48 meters and an area of 180 square meters. Find the perimeter and area of the smaller triangle. Answer: Scale factor = \(\frac { 2 }{ 3 } \) perimeter of smaller triangle = 32 Area of smaller triangle = 80

Explanation: The scale factor of smaller to larger traingle = \(\frac { 8 }{ 12 } \) = \(\frac { 2 }{ 3 } \) \(\frac { perimeter of smaller triangle }{ perimeter of larger triangle } \) = scale factor\(\frac { perimeter of smaller triangle }{ 48 } \) = \(\frac { 2 }{ 3 } \) perimeter of smaller triangle = \(\frac { 2 }{ 3 } \) • 48 perimeter of smaller triangle = 32 \(\frac { Area of smaller triangle }{ Area of larger triangle } \) = (scale factor)² \(\frac { Area of smaller triangle }{ 180 } \) = ( \(\frac { 2 }{ 3 } \))² Area of smaller triangle = \(\frac { 4 }{ 9 } \) • 180 = 80

Question 7. A cellular telephone tower casts a shadow that is 72 feet long, while a nearby tree that is 27 feet tall casts a shadow that is 6 feet long. How tall is the tower? Answer: \(\frac { shadow of tree }{ shadow of tower } \) = \(\frac { height of tree }{ height of tower } \) \(\frac { 6 }{ 72 } \) = \(\frac { 27 }{ x } \) 6x = 1944 x = 324 ft The height of the tower is 324 ft.

Use the SSS Similarity Theorem (Theorem 8.4) or the SAS Similarity Theorem (Theorem 8.5) to show that the triangles are similar.

Answer: \(\frac { 24 }{ 6 } \) = 4 \(\frac { 32 }{ 2x } \) = 4 32 = 8x x = \(\frac { 32 }{ 8 } \) x = 4

Determine whether \(\overline{A B}\) || \(\overline{C D}\)

Answer: \(\frac { DB }{ BE } \) = \(\frac { 10 }{ 16 } \) = \(\frac { 5 }{ 8 } \) \(\frac { CA }{ AE } \) = \(\frac { 20 }{ 28 } \) = \(\frac { 5 }{ 7 } \) \(\frac { DB }{ BE } \) ≠ \(\frac { CA }{ AE } \) So, CD and AB are not parallel.

Answer: \(\frac { DB }{ BE } \) = \(\frac { 12 }{ 20 } \) = \(\frac { 3 }{ 5 } \) \(\frac { CA }{ AE } \) = \(\frac { 13.5 }{ 22.5 } \) = \(\frac { 3 }{ 5 } \) \(\frac { DB }{ BE } \) = \(\frac { CA }{ AE } \) So, AB and CD are parallel.

Answer: \(\frac { ZC }{ AZ } \) = \(\frac { 24 }{ 15 } \) = \(\frac { 8 }{ 5 } \) \(\frac { YB }{ AY } \) = \(\frac { ZC }{ AZ } \) \(\frac { YB }{ 7 } \) = \(\frac { 8 }{ 5 } \) YB = \(\frac { 8 }{ 5 } \) • 7 YB = \(\frac { 56 }{ 5 } \) The length of \(\overline{Y B}\) is \(\frac { 56 }{ 5 } \)

Find the length of \(\overline{A B}\).

Answer: \(\frac { AB }{ 7 } \) = \(\frac { 6 }{ 4 } \) \(\overline{A B}\) = \(\frac { 6 }{ 4 } \) • 7 \(\overline{A B}\) = \(\frac { 42 }{ 4 } \) \(\overline{A B}\) = \(\frac { 21 }{ 2 } \) The length of \(\overline{A B}\) is \(\frac { 21 }{ 2 } \).

Answer: \(\frac { DB }{ CD } \) = \(\frac { AB }{ AC } \) \(\frac { 4 }{ 10 } \) = \(\frac { AB }{ 18 } \) \(\frac { 2 }{ 5 } \) = \(\frac { AB }{ 18 } \) AB = \(\frac { 36 }{ 5 } \)

Similarity Test

Determine whether the triangles are similar. If they are, write a similarity statement. Explain your reasoning.

Answer: Longer sides: \(\frac { 32 }{ 24 } \) = \(\frac { 4 }{ 3 } \) shorter sides: \(\frac { 18 }{ 14 } \) = \(\frac { 9 }{ 7 } \) remaining sides: \(\frac { 20 }{ 15 } \) = \(\frac { 4 }{ 3 } \) Those are not equal. So, triangles are not similar.

Answer: \(\frac { AC }{ KJ } \) = \(\frac { 6 }{ 8 } \) = \(\frac { 3 }{ 4 } \) \(\frac { BC }{ JL } \) = \(\frac { 8 }{ [latex]\frac { 32 }{ 3 } \) } [/latex] = \(\frac { 3 }{ 4 } \) \(\frac { AC }{ KJ } \) = \(\frac { BC }{ JL } \) ∠C = ∠J So, △ABC and △JLK are similar.

Answer: \(\frac { XY }{ XW } \) = \(\frac { PZ }{ PW } \)

Answer: \(\frac { 9 }{ w } \) = \(\frac { 15 }{ 5 } \) \(\frac { 9 }{ w } \) = 3 9 = 3w w = 3

Answer: \(\frac { 17.5 }{ 21 } \) = \(\frac { q }{ 33 } \) q = \(\frac { 17.5 }{ 21 } \) • 33 q = \(\frac { 55 }{ 2 } \)

Answer: \(\frac { 21 – p }{ p } \) = \(\frac { 12 }{ 24 } \) \(\frac { 21 – p }{ p } \) = \(\frac { 1 }{ 2 } \) cross multiply 2(21 – p) = p 42 – 2p = p 42 = p + 2p 42 = 3p p = 14

Question 7. Given ∆QRS ~ ∆MNP, list all pairs of congruent angles, Then write the ratios of the corresponding side lengths in a statement of proportionality.

Answer: The pairs of congruent anglres are m∠QRS, m∠RSQ, m∠SQR, m∠MNP, m∠NPM, m∠PMN. The ratios of side lengths are \(\frac { RQ }{ MN } \), \(\frac { QS }{ MP } \), \(\frac { RS }{ NP } \)

Question 8. Find the length of \(\overline{E F}\).

Answer: \(\frac { DE }{ EF } \) = \(\frac { CD }{ BC } \) \(\frac { 3.2 }{ EF } \) = \(\frac { 2.8 }{ 1.4 } \) EF = 1.6 The length of \(\overline{E F}\) is 1.6

Question 9. Find the length of \(\overline{F G}\).

Answer: \(\frac { EF }{ FG } \) = \(\frac { BC }{ AB } \) \(\frac { 1.6 }{ FG } \) = \(\frac { 1.4 }{ 4.2 } \) FG = 4.8 The length of \(\overline{F G}\) is 4.8

Question 10. Is quadrilateral FECB similar to quadrilateral GFBA? If so, what is the scale factor of the dilation that maps quadrilateral FECB to quadrilateral GFBA?

Answer: The scale factor of dilation from quadrilateral FECB to quadrilateral GFBA is \(\frac { GF }{ FE } \)

Question 11. You are visiting the Unisphere at Flushing Meadows Corona Park in New York. To estimate the height of the stainless steel model of Earth. you place a mirror on the ground and stand where you can see the top of the model in the mirror. Use the diagram to estimate the height of the model. Explain why this method works. Answer: We know that the corresponding sides of a similar triangle are proportional height of Unisphere/distance of Unisphere from mirror = height of person/distance of the person from mirror x/100 = 5.6/4 x = 5.6 × 100/4 x = 140 Thus the estimated height of the tree = 140 ft

Answer: we know that 1 yard = ft As per the similarity theorem \(\frac { AD }{ EH } \) = \(\frac { AB }{ EF } \) \(\frac { 2 }{ 800.3 } \) = \(\frac { 1.4 }{ EF } \) EF = 1200 • 1 • 4 EF = 1680 Perimete of the park P = 2(1680 + 2400) = 2(4080) = 8160 ft Area of the actual park = 1680 • 2400 = 4,032,000 sq ft Therefore, perimeter of the actual park = 8160 ft area of the actual park is 4,032,000 sq ft.

Similarity Cumulative Assessment

b. Are the quadrilaterals similar? Explain your reasoning. Answer: No. \(\frac { AD }{ QT } \) = \(\frac { 2 }{ 1.5 } \) \(\frac { CD }{ TS } \) = \(\frac { 2.8 }{ 1.4 } \) = 2 So, quadrilaterals are not similar.

SSS Congruence Theorem (Theorem 5.8) Answer: If all the three sides of one triangle are equivalent to the corresponding three sides of the second triangle, then the two triangles are said to be congruent by SSS rule.

HL Congruence Theorem (Theorem 5.9) Answer: A given set of triangles are congruent if the corresponding lengths of their hypotenuse and one leg are equal.

ASA Congruence Theorem (Theorem 5. 10) Answer: If any two angles and the side included between the angles of one triangle are equivalent to the corresponding two angles and side included between the angles of the second triangle, then the two triangles are said to be congruent by ASA rule

AAS Congruence Theorem (Theorem 5. 11) Answer: AAS stands for Angle-angle-side. When two angles and a non-included side of a triangle are equal to the corresponding angles and sides of another triangle, then the triangles are said to be congruent.

Question 4. The slope of line l is – \(\frac{3}{4}\). The slope of line n is \(\frac{4}{3}\) What must be true about lines l and n ? (A) Lines l and n are parallel. (B) Lines l and n arc perpendicular. (C) Lines l and n are skew. (D) Lines l and n are the same line. Answer: The slope of l = – \(\frac{3}{4}\) Slope of n = \(\frac{4}{3}\) lines slopes are reciprocal and opposite. So, they are perpendicular.

Question 6. The coordinates of the vertices of ∆DEF are D(- 8, 5), E(- 5, 8), and F(- 1, 4), The coordinates of the vertices of ∆JKL are J(16, – 10), K(10, – 16), and L(2, – 8), ∠D ≅ ∠J. Can you show that ∆DEF ∆JKL by using the AA Similarity Theorem (Theorem 8.3)? If so, do so by listing the congruent corresponding angles and writing a similarity transformation that maps ∆DEF to ∆JKL. If not, explain why not. Answer: AA similarity theorem states that ∠D = ∠J. So, ∆DEF and ∆JKL are similar.

Question 8. ‘Your friend makes the statement “Quadrilateral PQRS is similar to quadrilateral WXYZ.” Describe the relationships between corresponding angles and between corresponding sides that make this statement true.

Answer: When 2 figures are similar, then their corresponding angles are congruent and their corresponding lengths are proportional. hence if PQRS is similar to wxyZ, then the following statements are true. ∠P = ∠W, ∠Q = ∠X, ∠R = ∠Y and ∠S = ∠Z and \(\frac { PQ }{ WX } \) = \(\frac { QR }{ XY } \) = \(\frac { RS }{ YZ } \) = \(\frac { PS }{ WZ } \) = k Here is a constant of proportionality.

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Home > CC2 > Chapter 8 > Lesson 8.2.1

Lesson 8.1.1, lesson 8.1.2, lesson 8.2.1, lesson 8.2.2, lesson 8.3.1, lesson 8.3.2, lesson 8.3.3, lesson 8.3.4.

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Basic College Mathematics (9th Edition)

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