1. find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6...

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1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a.√6 b.√36 = 6 c.√98 d.√900 = 30 The geometric mean of a and b is √ab

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Page 1: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

1. Find the geometric mean between:a) 2 and 3b) 4 and 9c) 7 and 14d) 15 and 60

§ 13.1

a. √6

b. √36 = 6

c. √98

d. √900 = 30

The geometric mean of a and b is √ab

Page 2: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

2. Complete each statement:a. If 4x = 7y, then x/y = ______ and y/4 = ______ .b. If 12m = 21, then 4m = ______ and m/7 = ______ .c. If 6x = 5 9, then x/5 = ______ and x/9 = ______ .d. If 15x/28y = 5a/4b, then bx = ______ and x/a = ______

a. Given 4x = 7y. Divide both sides by 4y; x/y = 7/4

Divide both sides by 28; x/7 = y/4

b. Given 12m = 21. Divide both sides by 3; 4m = 7;

Divide both sides by 84; 4/7 = 4

c. Given 6x = 5 9. Divide both sides by 30; x/5 = 9/6 = 3/2; Divide both sides by 54; x/9 =

45/54 = 5/6.d. Given 15x/28y = 5a/4b. Multiple both sides by 28by/15;

yielding bx = 7ay/3; Multiply both sides by 28y/15a;

x/a = 7y/3b

Page 3: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

3. Complete each statement:a. If 5/12 = 15/36, then (5 + 12)/12 = (15 + ?)/36 .b. If 7/9 = 28/36, then 7/2 = 28/(36 - ?) .c. If a/b = 6/5, then (a + b)/b = ______ and (a – b)/b =

______ .d. If (a + c)/c = 11/7, then a/c = ______ and c/a = ______ .

a. If 5/12 = 15/36, then (5 + 12)/12 = (15 + 36)/36b. If 7/9 = 28/36, then 7/2 = 28/(36 - 28) .c. If a/b = 6/5, then (a + b)/b = 11/5 and (a – b)/b = 1/5 .d. If (a + c)/c = 11/7, then a/c = 4/7 and c/a = 7/4 .

a c a b c d a b c dIf Remember then and

b d b d b d

Page 4: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

4. Make a table with several entries of two positive numbers and their arithmetic mean and geometric mean. Make a conjecture about a relationship between these two means. Can you prove your conjecture?

Geometric mean ≤ arithmetic mean

A B Arithmetic mean Geometric mean

1 1 1 1

2 18 10 6

3 7 5 4.5826

234 518 421 348.1551

.2 .6 .4 .3464

Page 5: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

4.

(a – b) 2 > 0

a 2 – 2ab + b 2 > 0

a 2 + 2ab + b 2 > 4ab

(a + b) 2 > 2√ab

(a + b) 2 /2 > √ab

Page 6: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

5. Research Pythagorean Triples. List five primitive triples.

Isn’t the internet wonderful!

Page 7: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

6. Find the length of the altitude of an equilateral triangle with side 20.

b = 10√3.

10 b 20

1 23

a b c

1 23

Page 8: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

7. Find the length of the altitude of an square with side 20.

c = 20√2.

a b c

1 1 2

20 20 c

1 1 2

Page 9: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

8. Find the longest interior dimension of a box measuring 2meters by 3 meters by 4 meters.

We will use Pythagoras twice. First to find the length of the green line and then to find the length of the magenta line.

24

3

G 2 = 2 2 + 3 2 = 13

M 2 = 4 2 + G 2 = 16 + 13 = 39 and then

M = 6.249

Page 10: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

10. How can you use similar triangles to find the height of the flag pole in front of the library?

h

l 1

l 2

x

2 1

x h

l l

Page 11: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

11. How can you use similar triangles to find the diameter of the earth?

Research Eratosthenes.

Page 12: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

12. Find the relationship between the areas of two similar triangles.

Area = ½ bh Area = ½ kbkh

If the sides have a ratio of k then the areas have a ratio of k 2.

b kb

hkh

Page 13: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

13. Prove that if a line parallel to one side of a triangle intersects the other two sides, then it cuts off a similar triangle.

Statement Reason

1. AB DE Given

3. CAB = CDE Corresponding angles

4. ∆ABC ~ ∆DEC AA

2. C = C Reflexive

A

C

B

D E

Given: AB DE.

Prove: ∆ABC ~ ∆DEC

Page 14: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

14. Prove SSS similarity.

Given:

Prove: ∆ABC ~ ∆DEF

Statement Reason

1. Given

3. AB/AE’ = AC/AF’ Substitution

4. A = A Reflexive

6. E’F’/BC/AE’/AB Similarity

2. AE’ = DE, AF’ = DF Construction

9. E’F’ = EF Transitive

5. ∆ABC ~ ∆AE’F’ SAS

7. E’F’ = BC· AE’/AB = BC· DE/AB Multiplication.

8. EF = BC· DE/AB Multiplication.

10. ∆AE’F’ ∆DEF SSS

D

FEB

A

C

F’E’

AB AC BC

DE DF EF

AB AC BC

DE DF EF

11. ∆ABC ~ ∆DEF (5) AND– (10)

Page 15: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

15. Prove that the altitude to the hypotenuse separates the triangle into two triangles which are similar to each other and to the original triangle.

Because of the right triangle and a common angle in each of the triangles it is easy to show the triangles similar by AA or AAA.

Page 16: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

16. Find the length of the altitude to the hypotenuse of a right triangle with legs of 15 and 20.

Use the Pythagorean Theorem to find the hypotenuse of 25. then

short side a h x

hypotenuse c b a

And you know a, b and c, so

15 hand h 12

25 20

Page 17: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

17. A method used by carpenters to divide a board into equal parts is to use the vertical studding of a building as parallel lines, and to place the board to be divided transversely across them. Why does this work?

Notice all of the similar triangles.

Page 18: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

18. In rectangle ABCD construct the diagonal AC. Construct the altitude from D to AC meeting AC at E. Prove that ∆CDE ~ ∆ABC.

Because of the right triangle and a common angle in each of the triangles it is easy to show the triangles similar by AA or AAA.

E

D C

A B

E

Page 19: 1. Find the geometric mean between: a) 2 and 3 b) 4 and 9 c) 7 and 14 d) 15 and 60 § 13.1 a. √6 b. √36 = 6 c. √98 d. √900 = 30 The geometric mean of a

19. In the figure below ∆ABC ~ ∆DAB. Prove that AB is the geometric mean between AD and BC.

Because ∆ABC ~ ∆DAB we have the following proportions -

A

H

D

C

B

H

A

D

C

B

AB BC AC

DA AB BD

And the first proportion gives us AB as the geometric mean between AD and BC>