Download Chapter 6 Homework

Chapter6
Lesson 1: Polygons
A polygon is a closed two-dimensional figure with each side intersecting another side at its
endpoint. A polygon is named using a capital letter at each vertex of the polygon.
A
B
E
C
D
Polygon ABCDE
Not a polygon
Open figure
Not a polygon
Sides don’t intersect at endpoints
The measure of each interior angle of a convex polygon is less than 180°. A concave polygon has
an angle that is greater than 180°.
Convex polygon
Concave polygon
Polygons are named according to the number of sides they have.
Polygon Name
Triangle
Quadrilateral
Pentagon
Hexagon
Heptagon
Octagon
Nonagon
Decagon
Dodecagon
n-gon
Number of Sides
3
4
5
6
7
8
9
10
12
n
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There are four important terms used to describe polygons: vertex, side, interior angle, and diagonal.
A
A
1
E
D
2
3
4
5
B
E
B
C
D
C
Vertices: A, B, C, D, E
Sides: AB, BC, CD, DE, EA
Interior Angles: 1, 2, 3, 4, 5
Diagonals: AC, AD, BD, BE, CE
The more sides a polygon has, the more diagonals it can have. The number of diagonals of a
polygon can be calculated with this formula:
If a polygon has N sides, then ½ N(N-3) diagonals can be drawn.
If you draw all possible diagonals from one vertex of the polygon, you will divide the shape into
triangles. You know that the sum of the interior angles of a triangle is 180°, so you can easily
determine the sum of the interior angles of any polygon by counting the number of triangles that
were formed and multiplying your answer by 180°.
Number of sides = 6
Number of triangles = 4
Sum of interior angles = 4 x 180° = 720°
Sum of interior angles of a polygon = 180°(N-2)
It’s easy to prove that the sum of the exterior angles of any polygon (one exterior angle at each
vertex) is always 360°.
2
Given: A polygon with n vertices.
1
N
Informal Proof:
At each vertex, the sum of the interior and the exterior angle is 180°.
If you add up all the angles it is:
Sum of interior angles + Sum of exterior angles = 180N
Substitution:
180(N – 2) + Sum of exterior angles = 180N
Distribution:
180N – 360 + Sum of exterior angles = 180N
Subtraction:
-360 + Sum of exterior angles = 0
Addition:
Sum of exterior angles = 360
2|Page
Chapter 6 Homework
Lesson 1:
1. Draw a convex polygon that has six sides.
2. Draw a concave polygon that has eight sides.
3. Find the number of diagonals for a polygon having:
a. 7 sides
b. 10 sides
c. 12 sides
d. 20 sides
4. Draw a large hexagon and label its vertices A, B, C, D, E, and F. Draw all possible diagonals for
the shape. List the names of the diagonals.
5. Find the sum of the interior angles and the sum of the exterior angles of a polygon having:
a. 4 sides
b. 5 sides
c. 8 sides
d. 9 sides
e. 15 sides
6. Find the number of sides of a polygon and the name of the polygon if the sum of the interior
angles is:
a. 900°
b. 1440°
c. 2160°
d. 2700°
7. Find the measure of the missing angle of each of the following figures.
a. Quadrilateral: 42°, 75°, 118°
b. Pentagon: 116°, 138°, 94°, 88°
c. Hexagon: 95°, 154°, 80°, 145°, 76°
8. Find the number of sides of a polygon if the sum of the interior angles is 3 times as great as
the sum of the exterior angles.
9. The measures of the interior angles of a polygon are given. Solve for x.
a. Pentagon: (x + 3), (2x – 1), (3x + 8), (x – 10), (x + 4)
b. Octagon: x, 2x, 3x, 4x, 5x, 6x, 7x, 8x
3|Page
Lesson 2: Ratios and Proportions
A ratio of two numbers a and b (b≠0) is the quotient of the numbers. The ratio may be written
three different ways:
- As a fraction
-
With a colon
a:b
-
With words
a to b
Ratios, just like fractions, should be written in simplest form. For example, 6 to 8 is expressed as
follows:
=
or
3:4
or
3 to 4
A proportion is an equation that states that two ratios are equal:
=
(b≠0 and d≠0)
The largest and smallest values are called the extremes. The other two values are called the means.
=
Extremes
Means
=
The product of the extremes (a · d) is always equal to the product of the means (b · c) for a true
proportion.
Example:
=
2 x 12 = 24
3 x 8 = 24
If the means are both the same, they are called a geometric mean.
= m is the geometric mean.
Example: Solve for the geometric mean between 3 and 27.
3
=
27
2
m = 3(27)
m2 = 81
m = √81
m=9
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9 is the geometric mean between 3 and 27.
Chapter 6 Homework
Lesson 2:
1. Find each of the following ratios using the measurements of this triangle.
B
a. AB: BC
b. BC:AC
4
7
A
9
C
2. The measures of segments DE and EF have a ratio of 2:7. If DE is 6 inches long, how long is
EF?
D
E
F
3. The measures of the interior angles of triangle WXY have a ratio of 3:5:7, respectively. Find
the measure of each angle.
4. The measures of consecutive angles of a parallelogram have the ratio of 2:3. Find the
measure of each angle of the parallelogram.
Solve for x.
5.
6.
7.
=
=
8.
.
9.
=
=
= Find the geometric mean between each of the extremes.
10. 4 and 16
11. 6 and 24
12. 3x and 12x3
Determine whether the pairs of ratios are true proportions.
13. and 14. and 5|Page
Lesson 3: Proportions in a Triangle
When a line parallel to one side of a triangle passes through the other two sides of the triangle, it
divides those sides into proportional segments.
In the figure below, BD is parallel to CE. The following segments are proportional:
!
A
!"
!
"
B
!"
D
C
"
=
#
%&'
#$
=
!&((&
=
#
%&'
$
)*&+,
=
#$
!&((&
$
)*&+,
=
%&'
!&((&
%&'
)*&+,
=
!&((&
)*&+,
E
Conversely, if a line divides two sides of a triangle into proportional segments, that line is parallel to
the third side of the triangle.
Chapter 6 Homework
Lesson 3:
H
1. If KW || EG, find the length of each indicated segment.
a. HK = 4, HW = 10, GW = 15, KE = ?
K
b. HE = 20, KE = 12, WG = 9, and HG = ?
c. KH = 7, KE = 14, HG = 12, and HW = ?
W
E
G
d. KH = 2x – 15, KE = x, HW = 1, HG = 4. Find KH and KE.
2. For each of the following segments lengths, determine whether TP || BC.
B
a. AT = 5, TB = 15, AP = 8, PC = 24.
b. TB = 9, AB = 18, AP = 6, PC = 6.
T
c. AT = 4, AB = 12, AP = 6, AC = 15.
C
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P
A
Lesson 4: Similar Polygons
Two polygons are similar polygons if their corresponding angles are congruent and their
corresponding sides are proportional. In the diagram below, polygon ABCD is similar to polygon
EFGH. The symbol for “similar” is “~”.
B
F
A
E
H
G
D
ABCD ~ EFGH
C
Since the polygons are similar, you know that their corresponding angles are congruent:
∠A ≅∠E, ∠B ≅∠F, ∠C ≅∠G, and ∠D ≅∠H.
You also know that their corresponding sides are proportional:
.1
13
-3
-.
=
=
=
02
24
/4
/0
The ratio of the lengths of two corresponding sides is called the scale factor. The triangles below
are similar. The ratio of their sides, or scale factor, is .
5
2
=
6
3
6
8
10
9
12
15
2
7
=
89
3
8:
2
=
8;
3
9
Scale factor = <
The ratio of the perimeters of similar polygons is equal to the scale factor.
Perimeter = 6 + 8 + 10 = 24
Perimeter = 9 + 12 + 15 = 36
9=
2
=
<5
3
7|Page
If two polygons are similar, then the ratio of any two corresponding lengths in the polygons is
equal to their scale factor. In the example below, ∆LMO ~ ∆RST and their scale factor is . Since
their altitudes will proportional to this ratio, you can solve for the length of RU.
>?
3
=
@A
5
R
L
3
12
=
@A
5
?
12
O
12 (5) = 3(RU)
P
M
T
U
S
20 = RU
Chapter 6 Homework
Lesson 4:
1. List all pairs of congruent angles for similar polygons TRACK ~FIELD.
2. ∆CAT ~ ∆DOG.
a. Write an equation that shows their corresponding sides are proportional.
b. If CA = 8, AT = 14, CT = 18 and DG = 27, find the lengths of DO and OG.
3. Determine whether the pair of polygons is similar. Explain your reasoning.
7
a.
3
7
7
3
3
7
3
8
b.
4
4
4
6
4
6
8
c.
60°
50°
70°
60°
70°
8|Page
50°
4. JKLM ~ EFGH.
a. Find the scale factor of the similar shapes.
b. Find the values of x, y, and z.
c. Find the perimeter of each polygon.
J
3
H
30
G
60°
M
20
y
11
z°
K
x
L
E
8
F
5. The scale factor of two similar polygons is 3:5.
a. If the length of the shortest side of the smaller polygon is 24, what is the length of
the shortest side of the larger polygon?
b. If the length of the longest side of the larger polygon is 60, what is the length of the
longest side of the smaller polygon?
c. If the sum of the perimeters of both polygons is 208, find the perimeter of each
polygon.
K
S
15
B
A
10
J O
L
T
H R
6. ∆JKL ~ ∆RST
a. JB and RA are medians. The length of JB is 4 units longer than the length of RA.
How long are JB and RA?
b. KO and SH are altitudes. The sum of their lengths is 15. How long are KO and SH?
Bonus
AD, EF, and BC are perpendicular to DC. AD = 40 feet and BC = 25 feet. Find the length of EF.
A
B
E
D
F
C
9|Page
Lesson 5: Angle-Angle Theorem
The definition of similarity requires that we show that two polygons have equal corresponding
angles and proportional corresponding sides. That’s a lot of work! Fortunately, we can use a
shortcut to show that two triangles are similar. The shortcut is called the Angle-Angle Theorem.
Angle-Angle Theorem:
If two angles of one triangle are congruent to two angles of another
triangle, then the two angles are similar.
To prove that two triangles are similar, show that two angles of one triangle congruent two angles
of the other triangle. Then you can use the Angle-Angle Theorem to prove that they are similar.
A
B
C
Example:
Given: CB ⊥ BA, and CD ⊥ DE
Prove: ∆ABC ~ ∆EDC
D
1.
2.
3.
4.
5.
Statements
CB ⊥ BA, and CD ⊥ DE
∠ABC and ∠EDC are right
angles.
∠ABC ≅ ∠EDC
∠ACB ≅ ∠ECD
∆ABC ~ ∆EDC
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Reasons
Given
Perpendicular lines form right
angles.
All right angles are congruent.
Vertical angles are equal
AA Theorem
E
Chapter 6 Homework
Lesson 5:
1. Write the Angle-Angle (AA) Theorem.
Determine whether each pair of triangles is similar.
If they are, write a similarity statement (∆_ _ _ ~∆_ _ _)
2.
5.
3.
6.
50°
50°
5
3
6
4.
4
7.
6
105°
7
105° 3
4
For each case described below, determine if it is possible for ∆JKL and ∆XYZ to be similar. Explain
why or why not.
8. m∠J = 71°, m∠K = 52°, m∠X = 71°, and m∠Z = 57°
9. ∆JKL is a right triangle and m∠X + m∠Y = 150°
10. m∠J = 87° and m∠Y = 94°
11. m∠J + m∠K = 85° and m∠Y + m∠Z = 80°
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C
The figure to the right contains six similar isosceles triangles.
12. Name the six triangles that are similar.
F
A
D
E
B
13. What are the angle measures of these triangles?
14. What kind of triangle is ∆CDE?
15. The Roman general Marcellus is attacking the city of Syracuse. Archimedes is standing at Point
A by the city wall and he has place a catapult at Point C. A ditch is parallel to the city wall. The
Roman camp is at Point R.
a. Find the distance from Archimedes to the Roman camp (AR).
b. Find the distance from the Catapult to the Roman camp (CR).
R
ditch
40 m
C
60 m
102 m
city wall
A
H
A
16.
Given: HW || TA, and HY || AX
Prove: ∆WHY ~ ∆TAX
Statements
12 | P a g e
Y
Reasons
X
T
W
13 | P a g e