Jacobs Geometry
Contents 9 sections

Chapter 10 — Answers

Selected answers for Chapter 10: Similarity, transcribed from the book’s appendix. They are meant for checking your work, not for copying — the book prints answers to selected exercises only.

Back to Chapter 10 · Teacher’s answers: Chapter 10 · Complete key: Answers to Exercises PDF · Printed answers: appendices PDF

Lesson 1 (pages 380–384)

1. 2.25.

4. The geometric mean.

5. 96=64\dfrac{9}{6}=\dfrac{6}{4}.

8. h≈45h\approx45 ft.

9. 27\dfrac{2}{7}.

15. Division.

19. Rectangles 2 and 4.

22. 30.

28. 36 in; 3615=2.4\dfrac{36}{15}=2.4.

31. 1910\dfrac{19}{10}.

36. The width of one stripe.

37. Approximately 12.3 ft.

39. Approximately 122 ft.

41. 98\dfrac{9}{8}.

42. 256243\dfrac{256}{243}.

48. The means are interchanged.

50. Division.

Lesson 2 (pages 387–391)

1. They are proportional.

2. They are equal.

3. ∠DAC = ∠FBC.

6. ADBF=DCFC\dfrac{AD}{BF}=\dfrac{DC}{FC}.

9. OE = 4 and OG = 8.

13. 24.2.

15. The center of the dilation.

18. 12\dfrac{1}{2}.

20. Corresponding sides of similar polygons are proportional.

24. A midsegment of a triangle is parallel to the third side.

28. Division.

30. Betweenness of Points Theorem.

34. Two triangles are similar if their corresponding angles are equal and their corresponding sides are proportional.

35. OC = 56 mm; r2=2.0r_2=2.0.

37. BF = 24 mm; BFAE=2421≈1.1\dfrac{BF}{AE}=\dfrac{24}{21}\approx1.1.

42. Approximately 840 mm.

44. Approximately 1 m².

47. 2\sqrt{2}.

49. w=0w=0.

Lesson 3 (pages 394–398)

2. If a line parallel to one side of a triangle intersects the other two sides in different points.

7. The area of the rectangle.

8. As lengths: AC = 3, AD = 2, and AE = 6.

13. ACCO=APPB\dfrac{AC}{CO}=\dfrac{AP}{PB}.

17. Substitution. (Both are equal to APPB\dfrac{AP}{PB}.)

18. x=2.4x=2.4.

22. Reasonable because BE∥CFBE\parallel CF in △OCF (the Side-Splitter Theorem).

26. It is a parallelogram because DF∥EBDF\parallel EB and DF = EB.

29. If a line parallel to one side of a triangle intersects the other two sides in different points, it divides the sides in the same ratio.

36. EO.

39. Point G is the same point as point A.

41. EF = FG; so EFFG=1\dfrac{EF}{FG}=1 by division.

43. In a plane, two lines perpendicular to a third line are parallel.

45. The opposite sides of a parallelogram are equal.

Lesson 4 (pages 401–405)

1. Parallel lines form equal corresponding angles.

3. AA.

6. Yes. Two triangles similar to a third triangle are similar to each other.

10. 30°, 30°, and 120°.

14. Corresponding sides of similar triangles are proportional.

17. They are not collinear.

19. That they are parallel.

21. △AHK ∼ △ABD and △AKL ∼ △ADE.

23. HKBD=KLDE\dfrac{HK}{BD}=\dfrac{KL}{DE}.

27. Addition.

28. Multiplication.

31. △ABD ∼ △CBA.

34. x=8x=8, y=10y=10, and z=10z=10.

39. AA (∠A = ∠EFA and ∠FHA = ∠A).

45. as=bb−s\dfrac{a}{s}=\dfrac{b}{b-s}.

Lesson 5 (pages 409–413)

1. CG and CF.

4. Corresponding altitudes.

7. CG, the distance of the fish from the camera.

10. 7.5 cm.

12. 2.5.

15. 0.4.

18. r2r_2 is the reciprocal of r1r_1.

21. Yes. Corresponding sides of similar triangles are proportional.

23. No. These areas cannot be equal if △AGB is larger.

26. B′(7, 2), C′(17, 8).

27. △A′B′C′ appears to be an enlargement of △ABC. (Also, the corresponding sides of the two triangles appear to be parallel.)

31. 555.4 ft.

34. Corresponding altitudes of similar triangles have the same ratio as that of the corresponding sides.

35. ASA.

36. AA.

40. Ruler Postulate.

41. Through a point not on a line, there is exactly one line parallel to the given line.

47. Parallel lines form equal corresponding angles.

49. AA.

53. Reflexive.

55. Corresponding sides of similar triangles are proportional.

58. SSS Congruence Theorem.

p. 757

Lesson 6 (pages 416–419)

1. 12\dfrac{1}{2}.

2. 14\dfrac{1}{4}.

10. 8 m.

11. 32 m.

12. 64 m².

19. 45\dfrac{4}{5}.

21. 1625\dfrac{16}{25}.

25. They must be proportional.

28. 440.

30. 163,360\dfrac{1}{63{,}360}.

32. Yes. △ADE ∼ △ABC by AA.

34. No.

37. Six.

40. 12\dfrac{1}{2}.

44. Because all circles have the same shape.

46. 90π≈28390\pi\approx283 ft and 120π≈377120\pi\approx377 ft.

48. 34=0.75\dfrac{3}{4}=0.75.

53. 921 ft.

Review (pages 421–424)

1. 12\dfrac{1}{2}.

4. 512\dfrac{5}{12}.

9. 2.5.

10. It is 6.25 times as great.

12. 6.4.

14. 12.8.

17. Concave pentagons.

18. ABHIG ≅ EKJGI.

20. GFEI is a parallelogram. Both pairs of its opposite sides are equal because GFEKJ ≅ EIGJK.

25. It increases.

30. 2.

32. They are multiplied by 3.

34. 10x=x5\dfrac{10}{x}=\dfrac{x}{5}. Approximately 7.1 cm.

36. 22≈0.71\dfrac{\sqrt{2}}{2}\approx0.71.

41. In a plane, two lines parallel to a third line are parallel to each other.

46. 30 ft.

48. 70 ft.

53. EF20=68\dfrac{EF}{20}=\dfrac{6}{8}; EF = 15.

55. 43\dfrac{4}{3}.

56. 80.

58. 169\dfrac{16}{9}.

Algebra Review 1 (page 425)

1. 4 and −7.

2. 1 and −15-\dfrac{1}{5}.

3. 4+114+\sqrt{11} and 4−114-\sqrt{11}.

4. 7 and −7.

5. −2+26-2+2\sqrt{6} and −2−26-2-2\sqrt{6}.

6. 13\dfrac{1}{3}.

7. 43\dfrac{4}{3} and −3.

8. 2 and 5.

9. −1 and −8.

10. 1+265\dfrac{1+\sqrt{26}}{5} and 1−265\dfrac{1-\sqrt{26}}{5}.

Algebra Review 2 (page 426)

1. r=c2πr=\dfrac{c}{2\pi}.

2. p1=p2v2v1p_1=\dfrac{p_2v_2}{v_1}.

3. v2=p1v1p2v_2=\dfrac{p_1v_1}{p_2}.

4. c=a2+b2c=\sqrt{a^2+b^2}.

5. a=c2−b2a=\sqrt{c^2-b^2}.

6. h=2Abh=\dfrac{2A}{b}.

7. h=v22gh=\dfrac{v^2}{2g}.

8. r=A4πr=\sqrt{\dfrac{A}{4\pi}} or r=12Aπr=\dfrac{1}{2}\sqrt{\dfrac{A}{\pi}}.

9. R=EIR=\dfrac{E}{I}.

10. h=3Vπr2h=\dfrac{3V}{\pi r^2}.

11. a=S(1−r)a=S(1-r) or a=S−Sra=S-Sr.

12. r=S−aSr=\dfrac{S-a}{S} or r=1−aSr=1-\dfrac{a}{S}.

13. w=8ML2w=\dfrac{8M}{L^2}.

14. w1=W2w2w_1=\dfrac{W^2}{w_2}.

15. r=RA−150r=\dfrac{R}{A-150}.

16. A=R+150rrA=\dfrac{R+150r}{r} or A=Rr+150A=\dfrac{R}{r}+150.

17. d=1Ed=\sqrt{\dfrac{1}{E}}.

18. l=gT24π2l=\dfrac{gT^2}{4\pi^2}.

19. A=In−Ixd2A=\dfrac{I_n-I_x}{d^2}.

20. d=In−IxAd=\sqrt{\dfrac{I_n-I_x}{A}}.

21. R=r1r2r1+r2R=\dfrac{r_1r_2}{r_1+r_2}.

22. r1=Rr2r2−Rr_1=\dfrac{Rr_2}{r_2-R}.

23. n=−vr+vuur−vun=\dfrac{-vr+vu}{ur-vu} or n=−v(r+u)u(r−v)n=\dfrac{-v(r+u)}{u(r-v)} or n=vr+vuvu−urn=\dfrac{vr+vu}{vu-ur} or n=v(r+u)u(v−r)n=\dfrac{v(r+u)}{u(v-r)}.

24. C=59(F−32)C=\dfrac{5}{9}(F-32).

25. v=cf02−fs2f02+fs2v=c\dfrac{f_0^2-f_s^2}{f_0^2+f_s^2}.