Lesson 9.1 Using Similar Right Triangles

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Lesson 9.1 Using Similar Right Triangles. Students need scissors, rulers, and note cards. Today, we are going to… …use geometric mean to solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle. - PowerPoint PPT Presentation

Transcript of Lesson 9.1 Using Similar Right Triangles

Page 1: Lesson 9.1 Using Similar Right Triangles
Page 2: Lesson 9.1 Using Similar Right Triangles

Lesson 9.1Using Similar Right Triangles

Students need scissors, rulers, and note cards.

Today, we are going to……use geometric mean to solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle

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With a straight edge, draw one diagonal of the note card.

Draw an altitude from one vertex of the note card to the diagonal.

Cut the note card into three triangles by cutting along the segments.

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C

BAA D

B

DB

Chy

pote

nuse

hyp

hyp

short leg

short legshortlo

ng

leg

lon

g le

g

lon

g

Color code all 3 sides of all 3 triangles on the front and back.

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Arrange the small and

medium triangles on top of the

large triangle like this.?

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If the altitude is drawn to the hypotenuse of a right

triangle, then…

Theorems 9.1 – 9.3

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Theorem 9.1

…the three triangles formed are similar to each other.

Page 8: Lesson 9.1 Using Similar Right Triangles

BDBDCD

AD=

BD is a side of the medium and a side of the small

A

B

C D

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D AC

B

x

m n

xx=mn

BDBD=

CDAD

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Theorem 9.2

…the altitude is the geometric mean of the two

segments of the hypotenuse.

D AC

B

x

m n

xx=mn

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____ is the geometric mean of ____ and ____

When you do these problems, always tell yourself…

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D AC

B

x

8 3

xx=

83

1. Find x.

x ≈ 4.9

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D AC

B

4

8 x

44=

8x

2. Find x.

x = 2

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CBCD CB

CA=

A

B

C D

CB is a side of the large and

a side of the medium

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D AC

B

x

m

h

xx=mh

CBCB=

CDCA

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ABAD AB

AC=

A

B

C D

AB is a side of the large and

a side of the small

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D AC

B

x

nh

xx=nh

ABAB=

ADAC

Page 18: Lesson 9.1 Using Similar Right Triangles

Theorem 9.3

…the leg of the large triangle is the geometric mean of the

“adjacent leg” and the hypotenuse.

D AC

B

x

m n

xx=mh

y

h

yy=nh

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____ is the geometric mean of ____ and ____

When you do these problems, always tell yourself…

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D AC

B

x

9

14

xx=9

143. Find x.

x ≈ 11.2

Page 21: Lesson 9.1 Using Similar Right Triangles

D AC

B

x

410

xx=4

104. Find x.

x ≈ 6.3

Page 22: Lesson 9.1 Using Similar Right Triangles

5. Find x, y, zx y z

9 4

xx=9

13

x ≈ 10.8

yy=94

y = 6

zz=4

13

z ≈ 7.2

Page 23: Lesson 9.1 Using Similar Right Triangles

33=x56. Find h.

x = 1.8

h4

3

5

x1.8

3.2h

h=

3.2

1.8

h = 2.4

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Page 25: Lesson 9.1 Using Similar Right Triangles

Lesson 9.2 & 9.3 The Pythagorean Theorem

& Converse

Today, we are going to……prove the Pythagorean Theorem…use the Pythagorean Theorem and

its Converse to solve problems

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Theorem 9.4Pythagorean Theorem

hyp2 = leg2 + leg2 (c2 = a2 + b2 )

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2. Find x.

8

20

10 x

y

102 = y2 + 82

100 = y2 + 6436 = y2

14

x2 = 142 + 82

x2 = 196 + 64

x2 = 260

x ≈ 16.1

y = 6

Why can’t we use a geo mean proportion?

Page 28: Lesson 9.1 Using Similar Right Triangles

A Pythagorean Triple is a set of three positive

integers that satisfy the equation c 2 = a 2 + b 2.

The integers 3, 4, and 5 form a Pythagorean Triple because

52 = 32 + 42.

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Theorem 9.5Converse of the

Pythagorean Theorem

If c2 = a2 + b2, then the triangle is a right triangle.

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The “hypotenuse” is too short for the opposite angle to be 90˚ c2 < a2 + b2

The “hypotenuse” is too long for the opposite angle to be 90˚

c2 > a2 + b2

The hypotenuse is the perfect length for the opposite angle to be 90˚

c2 = a2 + b2

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Theorem 9.6

If c2 < a2 + b2, then the triangle is an acute

triangle.

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Theorem 9.7

If c2 > a2 + b2, then the triangle is an obtuse

triangle.

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How do we know if 3 lengths can represent the side lengths

of a triangle?

L < M + S

Page 34: Lesson 9.1 Using Similar Right Triangles

What kind of triangle?

L2 = M2 + S2

L2 < M2 + S2

L2 > M2 + S2

Right

Acute

Obtuse

Can a triangle be formed?

L = M + S

L < M + S

L > M + S

No can be formed

Yes, can be formed

No can be formed

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Do the lengths represent the lengths of a triangle? Is it a right triangle, acute triangle, or obtuse triangle?

3. 10, 24, 26

4. 3, 5, 7

5. 5, 8, 9

262 = 102 + 242

72 > 32 + 52

92 < 52 + 82

right triangle

obtuse triangle

acute triangle

26 < 10 + 24?

7 < 5 + 3?

9 < 8 + 5?

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92 = 52 + 56 2

Do the lengths represent the lengths of a triangle? Is it a right triangle, acute triangle, or obtuse triangle?

6. 5, 56 , 9

7. 23, 44, 70

8. 12, 80, 87 872 > 122 + 802

right triangle

not a triangle

obtuse triangle

70 < 23 + 44?

87 < 12 + 80?

9 < 5 + 56 ?

Page 37: Lesson 9.1 Using Similar Right Triangles

Find the area of the triangle.

9.

6

9

81 = x2 + 36

92 = x2 + 62

45 = x2

6.7 = xA ≈ ½ (6)(6.7)

≈ 20.12 units2

Page 38: Lesson 9.1 Using Similar Right Triangles

Find the area of the triangle.

10.13

20400 = x2 + 169

202 = x2 + 132

231 = x2

15.2 = xA ≈ ½ (13)(15.2)

≈ 98.8 units2

Page 39: Lesson 9.1 Using Similar Right Triangles

Find the area of the triangle.

11.

49 = x2 + 4

72 = x2 + 22

45 = x2

6.7 = xA ≈ ½ (4)(6.7)

≈ 13.4 units2

7

2

x

Page 40: Lesson 9.1 Using Similar Right Triangles

How much ribbon is needed using Method 1?

Which method requires less ribbon?

The diagram shows the ribbon for Method 2. How much ribbon is needed to wrap the box?

(3+12+3+12) + (3+6+3+6) = 48 in.

?

?

30

18

35 in

302 + 182 =

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Project ideas…A 20-foot ladder leans against a wall so that the base of the ladder is 8 feet from the base of the building. How far up on the building will the ladder reach?

A 50-meter vertical tower is braced with a cable secured at the top of the tower and tied 30 meters from the base. How long is the cable?

The library is 5 miles north of the bank. Your house is 7 miles west of the bank. Find the distance from your house to the library.

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Project ideas…Playing baseball, the catcher must throw a ball to 2nd base so that the 2nd base player can tag the runner out. If there are 90 feet between home plate and 1st base and between 1st and 2nd bases, how far must the catcher throw the ball?

While flying a kite, you use 100 feet of string. You are standing 60 feet from the point on the ground directly below the kite. Find the height of the kite.

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Page 44: Lesson 9.1 Using Similar Right Triangles

Lesson 9.4Special Right Triangles

Today, we are going to……find the side lengths of special right triangles

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45˚

45˚

5

xy

1. Find x. Use the Pythagorean Theorem to find y. Leave y in simplest radical form.

x = 5y 5 2

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45˚

45˚

9

xy

2. Find x. Use the Pythagorean Theorem to find y. Leave y in simplest radical form.

x = 9y 9 2

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Do you notice a pattern?

45˚

45˚

5

55

45˚

45˚

9

992 2

Page 48: Lesson 9.1 Using Similar Right Triangles

In a 45˚- 45˚- 90˚ Triangle,

hypotenuse = leg 2

Theorem 9.845˚- 45˚- 90˚

Triangle Theorem

45˚

45˚

x

xx 2

Page 49: Lesson 9.1 Using Similar Right Triangles

45˚

45˚

7

xy

3. Find x and y.

x = 7y = 7 2

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45˚

45˚

y

x3

4. Find x and y.

2x = 3

y = 3

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45˚

45˚

y

x10

5. Find x and y.

x = 5 2

x =

2

10

2

y = 5 2

Page 52: Lesson 9.1 Using Similar Right Triangles

6. Label the measures of all angles.Find x. Use the Pythagorean Theorem to find y in simplest radical form. 66

6

x

y

x = 3 60˚60˚

30˚30˚

y = 3 3

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7. Label the measures of all angles.Find x. Use the Pythagorean Theorem to find y in simplest radical form. 88

8

x

y

x = 4 60˚60˚

30˚30˚

y = 4 3

Page 54: Lesson 9.1 Using Similar Right Triangles

Do you notice a pattern?

30˚

60˚

3

63 3

60˚

30˚

8

4

34

Page 55: Lesson 9.1 Using Similar Right Triangles

hypotenuse = 2 short leg

long leg = short leg 3

Theorem 9.930˚-60˚-90˚

Triangle Theorem

60˚

30˚

2s

s

3s

In a 30˚-60˚-90˚ Triangle,

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8. Find x and y.

10

x

y

60˚

30˚x = 5

y = 5 3

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9. Find x and y.

x

10

y

60˚

30˚x = 20

y = 10 3

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10. Find x and y.

y

x

12

60˚

30˚

3

x = 12

y = 24

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11. Find x and y.

x

y

12

60˚

30˚

y 12

3

y = 4 3

x = 8 3

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12. In regular hexagon ABCDEF, find x and y.

36012

y =

y = 30

x = 60

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13. Find x and y.

x = 24

y = 12 3

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14. Find x and y.

x = 8

y = 8 2

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Page 64: Lesson 9.1 Using Similar Right Triangles

Lesson 9.5 & 9.6Trigonometric

RatiosToday, we are going to……find the sine, cosine, and tangent of an acute angle…use trigonometric ratios to solve problems

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Trigonometric Ratios

sinecosinetangent

Page 66: Lesson 9.1 Using Similar Right Triangles

A

B

C

hypotenuse

adjacent to A

opposite A

____ is the hypotenuse____ is opposite A____ is adjacent to AABBCAC

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A

B

C

hypotenuse

opposite B

adjacent to B

____ is the hypotenuse____ is opposite B____ is adjacent to BABACBC

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sin A =

leg opposite A

hypotenuse

Page 69: Lesson 9.1 Using Similar Right Triangles

cos A =

leg adjacent to A

hypotenuse

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tan A =

leg opposite A

leg adjacent to A

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SOHCAHTOA

Sine is Opp / Hyp

Cosine is Adj / Hyp

Tangent is Opp / Adj

Page 72: Lesson 9.1 Using Similar Right Triangles

1.

sin A = = 0.38465

13

12

513

A

Opp

Hyp

Adj

cos A =

tan A = = 0.4167 5 12

= 0.9231 12 13

B

Page 73: Lesson 9.1 Using Similar Right Triangles

2.

sin B = = 0.923112 13

12

513

A

Adj

Hyp

Opp

cos B =

tan B = = 2.400012 5

= 0.3846 5 13

B

Page 74: Lesson 9.1 Using Similar Right Triangles

Find the Sine, Cosine, and Tangent.

3. sin 32° =

5. tan 32° =

0.5299

0.5299

0.6249

4. cos 58°=

Page 75: Lesson 9.1 Using Similar Right Triangles

6. Find x and y to the nearest tenth.

OPP

ADJ

HYPsin 42˚ =

cos 42˚ = x121

12

x

y42°

x = 12 cos 42˚

x ≈ 8.9

y121

y = 12 sin 42˚ y ≈ 8.0

Page 76: Lesson 9.1 Using Similar Right Triangles

7. How tall is the tree?

64.34 ft

tan 65˚ = x301

x = 30 tan 65˚

65°30 ft

xOpp

Hyp

Adj

Page 77: Lesson 9.1 Using Similar Right Triangles

8. Find x to the nearest tenth.

42° 12

x

x = 13.3

tan 42˚ = 12 x1

12 = x tan 42˚

x =tan 42˚

12 ? 48˚

tan 48˚ = x121

x = 12 tan 48˚

x = 13.3

Opp

Hyp

AdjAdj

Hyp

Opp

Page 78: Lesson 9.1 Using Similar Right Triangles

Use Inverse Sine, Inverse Cosine, and Inverse Tangent.

9. sin A =1625

10. cos A =4553

11. tan A = 0.4402

m A =

m A =

m A =

40°

32°

24°

Page 79: Lesson 9.1 Using Similar Right Triangles

Use Inverse Sine, Inverse Cosine, and Inverse Tangent.

12. sin-1(0.7660)= A

14. tan-1(11.4300) = A

m A =

m A =

m A =

13. cos-1 513

= A

50°

67°

85°

Page 80: Lesson 9.1 Using Similar Right Triangles

Calculator language:

cos-1 513

= A

cos A = 513

Human language:

Page 81: Lesson 9.1 Using Similar Right Triangles

15. Solve the right triangle.

30

16A

C

B

mA = 28°

mB = 62°

AB = 34

tan A = 1630

(AB)2 = 302 + 162

opp

adj

Page 82: Lesson 9.1 Using Similar Right Triangles

16. Solve the right triangle.

10

6A

C

B

mA = 37°

mB = 53°

AC = 8

sin A = 610

102 = (AC)2 + 62

opp

hyp

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angle of elevation

angle of depression

A

B

A

B

C

C

Page 84: Lesson 9.1 Using Similar Right Triangles

?

20°

tan 20°=x8

x = 8 tan 20°

x = 2.9 ft

Page 85: Lesson 9.1 Using Similar Right Triangles

sin 45°= 3026+x

sin 45° sin 45° 30 = (26+x) sin 45°

42.43 = 26+x

16.43 = x

For the most comfortable height, the handle should be 16.43 inches.

Page 86: Lesson 9.1 Using Similar Right Triangles
Page 87: Lesson 9.1 Using Similar Right Triangles
Page 88: Lesson 9.1 Using Similar Right Triangles

12 miles

44˚

x

sin 28° = 60 x

x 127.8 cos 22° =

300 x

x 323.56

x 16.68

tan 44° = x 12

x 11.59

12 miles

44˚

x

cos 44° = 12 x

Page 89: Lesson 9.1 Using Similar Right Triangles

x 116.6

tan 25° = x 250

x 31.2

tan 58° = 50 x

x 84.8

cos 32° = x 100

x 46.7

sin 40° = 30 x

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Lesson 9.7 Vectors

Today, we are going to……find the magnitude and the

direction of a vector …add two vectors

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The magnitude of a vector PQ is the distance from the

initial point to the terminal point and is written | PQ |.

?

In other words, it is “the length of the vector”We use absolute value

symbols because length cannot be negative.

Page 92: Lesson 9.1 Using Similar Right Triangles

To find the magnitude of a vector…

Step 1: Identify the vector

component form X, Y

Step 2: Find the magnitude by

simplifying (X)2 + (Y)2

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The component form of AB is

__________-5, -3

For example, to make AB, we go left 5 units and down 3 units.

To find the magnitude of AB,

we simplify (-5)2 + (-3)2

and get 34 5.8

Page 94: Lesson 9.1 Using Similar Right Triangles

+4

+5?

A

B

| AB | ≈ 6.4

1. Find the magnitude of the vector.

4,5

(4)2 + (5)2

First, write the component form of

the vector.

Now, find the magnitude using the

formula.

Page 95: Lesson 9.1 Using Similar Right Triangles

-7

+5

| AB | ≈ 8.6

A

B2. Find the magnitude of the vector.

-7,5(-7)2 + (5)2

magnitude formula?component form?

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The direction of a vector is determined by the angle it

makes with a horizontal line.

45˚

45˚ southeast

Page 97: Lesson 9.1 Using Similar Right Triangles

30˚ southwest30˚

Identify the direction of the vector.

3.

4.

50˚ northwest

50˚

Page 98: Lesson 9.1 Using Similar Right Triangles

+5?

5. Find the direction of the vector

51° northeast+4

tan A = 54

m A = 51˚

opposite

adjacent

hypo

tenu

se

mark the hypotenuse

mark the opposite leg mark the adjacent leg

sin, cos, or tan?

Page 99: Lesson 9.1 Using Similar Right Triangles

+5

6. Find the direction of the vector

36° northwest

?

-7

tan A =

m A = 36˚

5 7

oppo

site

adjacent

hypotenuse

hypotenuse?opposite leg?

adjacent leg?

sin, cos, or tan?

Use +7 because it is length

Page 100: Lesson 9.1 Using Similar Right Triangles

Step 3: Direction

Step 2: Magnitude

7. Find the magnitude and direction

of AB if A (1,2) and B (4, 6).We can do this without drawing the vector!= X , Yx2 – x1 , y2 – y1

Step 1: Component Form4 – 1 , 6 – 2 = 3 , 4

(X)2 + (Y)2| AB | =

| AB | = (3)2 + (4)2 25 5=tan A =

|Y| |X|

53 northeast

tan A =

4

3

Page 101: Lesson 9.1 Using Similar Right Triangles

LOOK at the component form!

How do we know if the vector is north or south, east or west

without sketching it?

+,+ is right and up northeast-,+ is left and up northwest

-,- is left and down southwest+,- is right and down southeast

Page 102: Lesson 9.1 Using Similar Right Triangles

+,+ right,up

northeast

-,+ left ,up

northwest

left ,down-,-

southwest

right,down +,-

southeast

It might help to think about a map of the US!

Page 103: Lesson 9.1 Using Similar Right Triangles

8. Find the magnitude and direction

of AB if A (-3,3) and B (4, -5).= X , Y

(X)2 + (Y)2

tan A =

|Y| |X|

49 southeast

x2 – x1 , y2 – y1

| AB | =

= 7 , - 84 – –3 , – 5 – 3(7)2 + (-8)2=

| AB | = 113 10.6

tan A =

87

Page 104: Lesson 9.1 Using Similar Right Triangles

Component Form= X , Y

Magnitude(X)2 + (Y)2

Direction

tan A =

|Y| |X|

A north/south - east/west

x2 – x1 , y2 – y1

| AB | =

Page 105: Lesson 9.1 Using Similar Right Triangles

Adding Vectors in Component Form

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2,-3 + 5,4 =

5,42,-3

7,1

7,19.

Pick any starting pointmove right 2 and down 3from that point, go right 5 and up 4

instead of taking this route, you could take a short cut

See a pattern?

Page 107: Lesson 9.1 Using Similar Right Triangles

9,-2

-3,5 6,3

Pick a starting pointmove left 3 and up 5from that point, go right 9 and down 2

short cut?

See a pattern?-3,5 + 9,-2 = 6,310.

Page 108: Lesson 9.1 Using Similar Right Triangles

12.

11.

3, 4 , 6,5 , 2,1u v w ------------- -

u v

u w

------------- - 9 , 1

1 , - 3

Page 109: Lesson 9.1 Using Similar Right Triangles

12 miles

44˚

x

sin 28° = 60 x

x 127.8 cos 22° =

300 x

x 323.56

x 16.68

tan 44° = x 12

x 11.59

12 miles

44˚

x

cos 44° = 12 x

Page 110: Lesson 9.1 Using Similar Right Triangles

x 116.6

tan 25° = x 250

x 31.2

tan 58° = 50 x

x 84.8

cos 32° = x 100

x 46.7

sin 40° = 30 x

Page 111: Lesson 9.1 Using Similar Right Triangles