# 1. Find the geometric mean between: 2 and 3 4 and 9 7 and 14 15 and 60 - PowerPoint PPT Presentation

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§ 13.1. 1. Find the geometric mean between: 2 and 3 4 and 9 7 and 14 15 and 60. The geometric mean of a and b is √ab. √6 √36 = 6 √98 √900 = 30. 2. Complete each statement: If 4x = 7y, then x/y = ______ and y/4 = ______ . If 12m = 21, then 4m = ______ and m/7 = ______ .

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1. Find the geometric mean between: 2 and 3 4 and 9 7 and 14 15 and 60

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§ 13.1

• 1. Find the geometric mean between:

• 2 and 3

• 4 and 9

• 7 and 14

• 15 and 60

The geometric mean of a and b is √ab

√6

√36 = 6

√98

√900 = 30

• 2. Complete each statement:

• If 4x = 7y, then x/y = ______ and y/4 = ______ .

• If 12m = 21, then 4m = ______ and m/7 = ______ .

• If 6x = 5  9, then x/5 = ______ and x/9 = ______ .

• 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

• 3. Complete each statement:

• If 5/12 = 15/36, then (5 + 12)/12 = (15 + ?)/36 .

• If 7/9 = 28/36, then 7/2 = 28/(36 - ?) .

• If a/b = 6/5, then (a + b)/b = ______ and (a – b)/b = ______ .

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

a. If 5/12 = 15/36, then (5 + 12)/12 = (15 + 36)/36

• 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 .

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

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

5. Research Pythagorean Triples. List five primitive triples.

Isn’t the internet wonderful!

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

b = 10√3.

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

c = 20√2.

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.

3

2

4

G 2 = 2 2 + 3 2 = 13

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

M = 6.249

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

h

l 1

x

l 2

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

Research Eratosthenes.

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

kh

h

b

kb

Area = ½ bh Area = ½ kbkh

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

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.

C

D

E

Given: AB  DE.

Prove: ∆ABC ~ ∆DEC

B

A

A

D

B

E

C

F

14. Prove SSS similarity.

Given:

Prove: ∆ABC ~ ∆DEF

F’

E’

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.

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

And you know a, b and c, so

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.

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

C

D

E

E

A

B

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.

C

C

D

H

A

B

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

D

H

A

B

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

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