chapter 7: proportions and similarity · •inscribed angle: an angle inside the circle with sides...
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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