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

Transcript of 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...

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>