Chapter 7: Proportions and Similarity

Objective: I will review proportions, properties of similar polygons and triangles.

mA+ mB+ mC = 180o Triangle Sum Thm.

2x + 3x + 4x = 180o

9x = 180o

x = 20o

mA = 40o

mB = 60o

mC = 80o

• The angle measures in ABC are in the extended ratio of 2:3:4. Find the measure of the three angles.

A

C

B

2x3x

4x

7.2 : Similar Polygons

• Similar polygons have:• Congruent corresponding angles

• Proportional corresponding sides

• Scale factor: the ratio of corresponding sides

A

B

C D

EL

M

N O

P

Polygon ABCDE ~ Polygon LMNOP

NO

CD

LM

AB

Ex:

Writing Similarity Statements• Decide if the polygons are similar. If they are, write

a similarity statement.

A B

C

D

6

12

9

15

W

X

Z

Y

10

8

6

4

2

3

4

6

WY

AB

2

3

6

9

YZ

BC

2

3

8

12

ZX

CD

2

3

10

15

XW

DA

A W

B Y

C Z

D X

All corr. sides are

proportionate and

all corr. angles are

ABCD ~ WYZX

7.3: Similar Triangles

• Similar triangles have congruent corresponding angles and proportional corresponding sides

A

B

C

Y

X

Z

ABC ~ XYZ

angle A angle X

angle B angle Y

angle C angle Z

YZ

BC

XZ

AC

XY

AB

7.3: Similar Triangles

• Triangles are similar if you show:• Any 2 pairs of corresponding sides are proportional and the included angles

are congruent (SAS Similarity)

A

B

C

R

S

T

18

12 6

4

7.3: Similar Triangles

• Triangles are similar if you show:• All 3 pairs of corresponding sides are proportional (SSS Similarity)

A

B

C

R

S

T

10

14

6

7

5

3

7.3: Similar Triangles

• Triangles are similar if you show:• Any 2 pairs of corresponding angles are congruent (AA Similarity)

A

B

C

R

S

T

7.4 : Parallel Lines and Proportional Parts

• If a line is parallel to one side of a triangle and intersects the other two sides of the triangle, then it separates those sides into proportional parts.

A

BC

XY

XB

AX

YC

AY*If XY ll CB, then

I will review geometric mean, Pythagorean theorem, Trig, Angle of Depression/Elevation and Law of Sines.

Objectives

Chapter 8 Review

The Geometric Mean

“x” is the geometric mean between “a” and “b” if:

a

x b

x

or x ab

x2 = ab

√x2 = √ab

Take Notice: The term said to be the

geometric mean will always be cross-

multiplied w/ itself.

Take Notice: In a geometric mean problem,

there are only 3 variables to account for,

instead of four.

You try it

• Find the geometric mean between 2 and 18.

6

Find the value of each variable

1.

x

3

2

13x

Find the value of each variable

2.

6

4y

52y

Find the length of a diagonal of a rectangle with length 8 and width 4.

4.

4

8

8

4

Find the length of a diagonal of a rectangle with length 8 and width 4.

4.

8

4

54

Review

• We use c2 a2 + b2

•C2 = then we a right triangle

•C2 < then we have acute triangle

•C2 > then we have obtuse triangle

• Always make ‘c’ the largest number!!

45º-45º-90º Theorem

In a 45-45-90 triangle, the hypotenuse is 2

times the length of each leg.

x

x

45

a

Hypotenuse = √2 ∙ leg

45

x√2

2 x: 90º

x : 45º

x : 45º

White Board Practice

6

x

x

Hypotenuse = √2 * leg

6 = √2 x

23x

30º-60º-90º Theorem

In a 30-60-90 triangle, the hypotenuse is

twice as long as the shorter leg and the

longer leg is 3 times the shorter leg.

x2x

60

30

3

THE MEASUREMENTS OF THE PATTERN ARE

BASED ON THE LENGTH OF THE SHORT LEG

(OPPOSITE THE 30 DEGREE ANGLE)

x 2x : 90º

3 x : 60º

x : 30º

White Board Practice

5

y

x

60º

Hypotenuse = 2 ∙ short leg

Long leg = √3 ∙ short leg

10

35

y

x

White Board Practice

9

y

x60º

30º

y = 3√3

x = 6√3

SOH-CAH-TOASineOppositeHypotenuseCosineAdjacentHypotenuseTangentOppositeAdjacent

Find the measures of the missing sides x and y

23º

100

y

x ≈ 110

y ≈ 47

67º

x

White boards - Example 2

• Find xº correct to the nearest degree.

xº

30

18

x ≈ 37º

Find the measurement of angle x

Xº

68

10

37x

Check It Out! Example 2a

Solve the triangle. Round to the nearest tenth.

Step 1 Find the third angle measure.

mK = 31° Solve for mK.

mH + mJ + mK = 180°

42° + 107° + mK = 180°Substitute 42° for mH

and 107° for mJ.

Check It Out! Example 2a Continued

Step 2 Find the unknown side lengths.

sin H sin Jh j

=sin K sin H

k h=

sin 42° sin 107°h 12

=sin 31° sin 42°

k 8.4=

h sin 107° = 12 sin 42° 8.4 sin 31° = k sin 42°

h = 12 sin 42°

sin 107°

h ≈ 8.4

k = 8.4 sin 31°

sin 42°

k ≈ 6.5Solve for the

unknown side.

Law of Sines.

Substitute.

Crossmultiply.

Chapter 10: CirclesObjective: I will review and apply theorems related to circles

Radius

Chord

Secant

Tangent—a line that intersects the circle in only one point

Tangent-Chord (or secant) TheoremIf a tangent and a chord intersect at a

point on a circle, then the measure of each angle formed is one half its intercepted arc

100oXY

mX = ½ (100o)

mX = 50o

mY = ½ (260o)

mY = 130o

Theorem 10.12• If 2 secants intersect in the interior of a circle, then the

measure of the angle formed is one half of the sum of the arcs intercepted by the angle and its vertical angle.

mX = ½ (100 + 40)

mX = ½ (140)

mX = 70o

100o

X

40o

Theorem 10.14• If 2 lines intersect on the exterior of a circle, then

the measure of the angle formed is one half of the difference of the 2 intercepted arcs.

• If two segments from the same external point are tangent to a circle they are

AC = AB

A

B

C

Chord Segment Theorem• If two chords intersect in the interior of a circle, then the product of

the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.

5 cm

4 cm

2 cm

10 cm

B

E

D

A

C

(AB)(BC) = (DB)(BE)

(2)(10) = (4)(5)

20 = 20

Secant Segment Theorem• If two secant segments share the same endpoint outside a circle,

then the product of the length of one secant segment and the length of its external segment equals the product of the length of the other secant segment and the length of its external segment.

D

BAC

E

(AB)(AC) = (AD)(AE)

Secant-Tangent Segment Theorem• If a secant segment and a tangent segment share an endpoint

outside a circle, then the product of the length of the secant segment and the length of its external segment equals the square of the length of the tangent segment.

(AC)(AD) = (AB)2

C

D

B

A

10.1 Circles and Circumference

• Name a circle by the letter at the center of the circle

• Diameter- segment that extends from one point on the circle to another point on the circle through the center point

• Radius- segment that extends from one point on the circle to the center point

• Chord- segment that extends from one point on the circle to another point on the circle

• Diameter=2 x radius (d=2r)

• Circumference: the distance around the circle• C=2πr or C= πd

10.2 Angles, Arcs and Chords

• 10.2• Semi-circle: half the circle (180 degrees)

• Minor arc: less than 180 degrees• Name with two letters

• Major arc: more than 180 degrees• Name with three letters

• Minor arc = central angle

• Arc length:

rarc

2360

• Find x and angle AZE

10.3 Arcs and Chords

• If two chords are congruent, then their arcs are also congruent

• In inscribed quadrilaterals, the opposite angles are supplementary

• If a radius or diameter is perpendicular to a chord, it bisects the chord and its arc

• If two chords are equidistant from the center of the circle, the chords are congruent

A

B

C

DE

F

If FE=BC, then arc FE =

arc BC

Quad. BCEF is an

inscribed polygon –

opposite angles are

supplementary

angles B + E = 180 &

angles F + C = 180

Diameter AD is

perpendicular to chord EC

– so chord EC and arc EC

are bisected

Circle W has a radius of 10 centimeters. Radius is

perpendicular to chord which is 16 centimeters

long.

Find JL.

A radius perpendicular to a chord bisects it.

Definition of segment bisector

Draw radius

Use the Pythagorean Theorem to find WJ.

Pythagorean Theorem

Simplify.

Subtract 64 from each side.

Take the square root of each side.

Segment addition

Subtract 6 from each side.

Answer: 4

10.4 Inscribed Angles

• Inscribed angle: an angle inside the circle with sides that are chords and a vertex on the edge of the circle

• Inscribed angle = ½ intercepted arc

• An inscribed right angle, always intercepts a semicircle

• If two or more inscribed angles intercept the same arc, they are congruent

A. Find mX.

The insignia shown is a quadrilateral inscribed in a circle. Find mS and mT.

A. Find x.

B. Find x.

A. Find mQPS.

A.

B.

10.7 Special Segments in a Circle

• Two Chords• seg1 x seg2 = seg1 x seg2

• Two Secants• outer segment x whole secant =

outer segment x whole secant

• Secant and Tangent• outer segment x whole secant = tangent squared

*Add the segments to get the whole secant

A. Find x.

B. Find x.

Find x.

LM is tangent to the circle. Find x. Round to the nearest tenth.

Find x. Assume that segments that appear to be tangent are tangent.

Areas

Example

A= ½ bhA= ½ (30)(10)A= ½ (300)A= 150 km

2

Parallelogram

• A parallelogram is a quadrilateral where the opposite sides are congruent and parallel.

• A rectangle is a type of parallelogram, but we often see parallelograms that are not rectangles (parallelograms without right angles).

Area of a Parallelogram

• Any side of a parallelogram can be considered a base. The height of a parallelogram is the perpendicular distance between opposite bases.

Find the area of rhombus RSTU.

Draw diagonal SU, and label the intersection

of the diagonals point X.

To find the area, you need to know the

lengths of both diagonals.

Example

| 27 cm |

10 cm

24 cm

Split the shape into a rectangle and triangle.

The rectangle is 24cm long and 10 cm wide.

The triangle has a base of 3 cm and a height of 10

cm.

Solution

Rectangle

A = lwA = 24(10)A = 240 cm

2

TriangleA = ½ bhA = ½ (3)(10)A = ½ (30)A = 15 cm

2

Total FigureA = A1 + A2

A = 240 + 15 = 255 cm2

Area of rectangle:

Find the shaded area. Round to the nearest tenth, if necessary.

A = lw = 37.5(22.5)

= 843.75 m2

Area of triangle:

= 937.5 m2

Total shaded area is about 1781.3 m2.

ANSWER 63 m2

Find the area of the trapezoid.

1.

Use the Area of a Trapezoid

Find the value of b2 given that the area of the

trapezoid is 96 square meters.

ANSWER The value of b2 is 15 meters.

Find the area of each shaded region.

1.

Surface Area and Volume

Objective: I will find the surface area and

volume of prisms, pyramids, cylinders, cones

spheres and composite figures

2. Find the volume and surface area of the right solid.

22 2SA r rH 22 (2) 2 (2)(6)SA

2 (4) 2 (12)SA

8 24SA

32SA cm2

1. Find the volume and surface area of the right solid.

SA = 2B + PH

SA = 2(30) + (30)(10)

P = 5 + 12 + 13

P = 30

SA = 60 + 300

SA = 360 cm2

1

2B bh

1(12)(5)

2B

30B

c2 = a2 + b2

c2 = (5)2 + (12)2

c2 = 25 + 144

c2 = 169

c = 13

Find the volume of the solid.

2.

Step 2 Find the volume of the composite figure.

Example 5 Continued

Find the surface area and volume of the composite figure. Give your answer in terms of .

The volume of the composite figure is the sum of the volume of the hemisphere and the volume of the cylinder.

The volume of the composite figure is 144 + 324 = 468 in3.

Find the volume of the pyramid. height h = 8 mapothem a = 4 mside s = 6 m

Area of base =

Exercise #2

h

as

Volume = 1/3 (area of base) (height)

= 1/3 ( 60m2)(8m)

= 160 m3

= ½ (5)(6)(4)

= 60 m2

Review

Transformations

and Vectors

Objective: I will review translations,

reflections, rotations, dilations and vectors

Writing a Rule9

8

7

6

5

4

3

2

1

0 1 2 3 4 5 6 7 8 9

Right 4 (positive change in x)

Down 3

(negative

change in y)A

A’

B

B’

C

C’

Writing a Rule

Can be written as:

R4, D3

(Right 4, Down 3)

Rule: (x,y) (x+4, y-3)

Example 3: Write a rule that describes the translation below

Point A (2, -1) Al (-2, 2)

Point B (4, -1) Bl (0, 2)

Point C (4, -4) Cl (0, -1)

Point D (2, -4) Dl (-2, -1)

Rule (x, y) (x – 4, y + 3)

Example 4: Write a rule that describes each translation below.

a.) 3 units left and 5 units up b.) 2 units right and 1 unit down

Rule (x, y) (x – 3, y + 5)Rule (x, y) (x + 2, y – 1)

Line of ReflectionThe line you

reflect a figure across

Ex: X or Y axis

X - axis

p. 625

p.

626

Reflect a Figure in the Line y = x

Quadrilateral ABCD with vertices A(1, 1), B(3, 2), C(4, –1), and D(2, –3). Graph ABCD and its image under reflection of the line y = x.

Interchange the x- and y-coordinates of each vertex.

(x, y) → (y, x)

A(1, 1) → A'(1, 1)B(3, 2) → B'(2, 3)C(4, –1) → C'(–1, 4)D(2, –3) → D'(–3, 2)

Answer:

In the diagram to the left you will

notice that triangle ABC is reflected

over the y-axis and all of the points are

the same distance away from the y-

axis.

Therefore triangle AlBlCl is a reflection

of triangle ABC

Example 1: Draw all lines of reflection for the figures below. This is a

line where if you were to fold the two figures over it they would line up.

How many does each figure have?

a.) b.)

1 6

Rotation in a Coordinate Plane

Triangle ABC has vertices A(1, 0), B(3, 3), C(5, 0).

Rotate ∆ABC 90° counterclockwise about the origin.

Rotations Around the Origin

x

y

A

B

C

3

–3

Graph the pre-image coordinates.

The coordinates of the image of

triangle A’B’C’ are A’(0, 1), B’(-3,3),

C (0.5).

Remember: A 90 degree rotation x and y change places, then pay attention to the

characteristics of the quadrants.

C’

B’

A’

Triangle ABC has vertices A(1, 0), B(3, 3), C(5, 0).

Rotate ∆ABC 90° lockwise about the origin.

Rotations Around the Origin

x

y

A

B

C

3

–3

C’

B’

A’

Graph the pre-image coordinates.

The coordinates of the image of

triangle A’B’C’ are A’(0,-1), B’(3,-3),

C’(0,-5).

Example 1C: Identifying line of symmetry

Yes; four lines

of symmetry

Tell whether the figure has line symmetry. If so,

copy the shape and draw all lines of symmetry.

Tell whether each figure has line symmetry. If so,

copy the shape and draw all lines of symmetry.

Check It Out! Example 1

yes; two lines of

symmetrya.

b.yes; one line of

symmetry

Rotational Symmetry

Rotational Symmetry – if a figure can be rotated less than

360° and the image and pre-image are indistinguishable

(regular polygons are a great example)

Order: 3 4 6 8

Magnitude: 120° 90° 60° 45°

Remember Order = n (number of sides)

Magnitude = 360 / Order

Example 2: Identifying Rotational Symmetry

Tell whether each figure has rotational symmetry. If so,

give the angle of rotational symmetry and the order of

the symmetry.

no rotational

symmetry

yes; 180°;

order: 2

yes; 90°;

order: 4

A. B.

C.

Check It Out! Example 2

Tell whether each figure has rotational symmetry. If so,

give the angle of rotational symmetry and the order of

the symmetry.

yes; 120°;

order: 3

yes; 180°;

order: 2

no rotational

symmetry

a. b. c.

3

C

P

R

P'

R'

QQ'

Reduction/Enlargement

• The dilation is a reduction if 0 < k < 1 and it is an enlargement if k > 1.

6

REDUCTION: CP’

CP

3

6

1

2= =

2

P'

C

Q'

R'R

P

Q

5

ENLARGEMENT: CP’

CP

5

2=

Because ∆PQR ~ ∆P’Q’R’

P’Q’

PQ Is equal to the scale factor

of the dilation.

Ex. 1: Identifying Dilations

• Identify the dilation and find its scale factor.

2

3

C

P

P'

REDUCTION: CP’

CP

2

3=

The scale factor is k =

This is a reduction.

2

3

Ex. 1B -- Enlargement

• Identify the dilation and find its scale factor.

ENLARGEMENT: CP 1=

The scale factor is k =

This is an enlargement.

2

1

CP’ 2= 2

= 22

1

P'

C

P

Ex. 2: Dilation in a coordinate plane

• Draw a dilation of rectangle ABCD with A(2, 2), B(6, 2), C(6, 4), and D(2, 4). Use the origin as the center and use a scale factor of ½. How does the perimeter of the preimage compare to the perimeter of the image?

SOLUTION:

8

6

4

2

-2

5 10 15

D' C'

B'A'

D

C

BA

Because the center of the dilation

is the origin, you can find the

image of each vertex by

multiplying is coordinates by the

scale factor

A(2, 2) A’(1, 1)

B(6, 2) B’(3, 1)

C(6, 4) C’(3, 2)

D(2, 4) D’(1, 2)

Write a Vector in Component Form

• Write the component form of

• Find the magnitude

• Find the direction relative to west .

Operations with Vectors

Solve Algebraically

Find each of the following for

and . Check your answers graphically.

A.

Check Graphically