Prerequisite Skills (p. 504) 1 and 4 are vertical angles.

Chapter 8
Prerequisite Skills (p. 504)
2. The sum of the measures of the interior angles of a
convex n-gon is (n 2 2) + 1808.
1. Ž 1 and Ž 4 are vertical angles.
2. Ž 3 and Ž 5 are consecutive interior angles.
3. The lengths of the sides do not affect the sum of the
interior angle measures of a hexagon. Only the number of
sides affects the sum.
3. Ž 7 and Ž 3 are corresponding angles.
4. Ž 5 and Ž 4 are alternate interior angles.
8.1 Guided Practice (pp. 508–510)
5. mŽ A 1 mŽ B 1 mŽ C 5 1808
1. Use the Polygon Interior Angles Theorem. Substitute 11
x8 1 3x8 1 (4x 2 12)8 5 1808
for n.
8x 2 12 5 180
(n 2 2) + 1808 5 (11 2 2) + 1808 5 9 + 1808 5 16208
8x 5 192
2. Because the sum of the measures of the interior angles is
x 5 24
14408, set (n 2 2) + 1808 equal to 14408 and solve for n.
mŽ A 5 x8 5 248
(n 2 2) + 1808 5 14408
mŽ B 5 3x8 5 3(248) 5 728
n2258
mŽ C 5 (4x 2 12)8 5 (4(24) 2 12)8 5 848
6. Ž 3 and Ž 1 are corresponding angles, so
mŽ 1 5 mŽ 3 5 1058
Ž 1 and Ž 2 are alternate interior angles, so
mŽ 2 5 mŽ 1 5 1058
n 5 10
The polygon has 10 sides. It is a decagon.
3. mŽ T 5 mŽ S
mŽ P 1 mŽ Q 1 mŽ R 1 mŽ S 1 mŽ T 5 (n 2 2) + 1808
938 1 1568 1 858 1 mŽ T 1 mŽ T 5 (5 2 2) + 1808
7. Because Ž 1 and Ž 3 are corresponding angles,
mŽ 3 5 mŽ 1 5 988
8. Ž 4 is congruent to the supplement of Ž 3 because they
are corresponding angles. The supplement of Ž 3 is
congruent to the supplement of Ž 1 because they are
corresponding angles. So, mŽ 4 1 mŽ 1 5 1808.
3348 1 2mŽ T 5 5408
2mŽ T 5 2068
mŽ T 5 1038 5 mŽ S
4. Let x8 equal the measure of the fourth angle.
x8 1 898 1 1108 1 468 5 3608
mŽ 4 1 mŽ 1 5 1808
x 1 245 5 360
828 1 mŽ 1 5 1808
x 5 115
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
mŽ 1 5 988
9. Ž 2 is congruent to the supplement of Ž 4 because they
are alternate interior angles. So, mŽ 4 1 mŽ 2 5 180.
5. Use the Polygon Exterior Angles Theorem.
x8 1 348 1 498 1 588 1 678 1 758 5 3608
x 1 283 5 360
mŽ 4 1 mŽ 2 5 1808
x 5 77
mŽ 4 1 102 5 1808
mŽ 4 5 788
6. If the angles form a linear pair, they are supplementary
so their sum is 1808. The measure of the interior angle
could have been subtracted from 1808 to find the measure
of the exterior angle. 1808 2 1508 5 308
Lesson 8.1
Investigating Geometry Activity 8.1 (p. 506)
8.1 Exercises (pp. 510–513)
STEP 3
Polygon
Number
of sides
Number of
triangles
Sum of measures
of interior angles
Triangle
3
1
1 + 1808 5 1808
Quadrilateral
4
2
2 + 1808 5 3608
Pentagon
5
3
3 + 1808 5 5408
Hexagon
6
4
4 + 1808 5 7208
1. The sum of the interior angles of a convex heptagon is
5 + 1808 5 9008. The sum of the interior angles of a
convex octagon is 6 + 1808 5 10808. As the number of
sides is increased by 1, so is the number that is multiplied
by 1808 to get the sum of the measures of the interior
angles.
Skill Practice
1.
2. There are 2n exterior angles in an n-gon. However, only
1 angle at each vertex, or n angles, is considered when
using the Polygon Exterior Angles Theorem.
3. A nonagon has 9 sides.
(n 2 2) + 1808 5 (9 2 2) + 1808 5 7 + 1808 5 12608
4. (n 2 2) + 1808 5 (14 2 2) + 1808 5 12 + 1808 5 21608
5. (n 2 2) + 1808 5 (16 2 2) + 1808 5 14 + 1808 5 25208
6. (n 2 2) + 1808 5 (20 2 2) + 1808 5 18 + 1808 5 32408
Geometry
Worked-Out Solution Key
233
Chapter 8,
continued
7. (n 2 2) + 1808 5 3608
18. B; x8 1 2x8 1 3x8 1 4x8 5 (n 2 2) + 1808
n2252
x 1 2x 1 3x 1 4x 5 (4 2 2) + 180
n54
10x 5 360
x 5 36
The polygon has 4 sides. It is a quadrilateral.
8. (n 2 2) + 1808 5 7208
Because x 5 368, then 4x8 5 1448
n2254
19. (n 2 2) + 1808 5 (5 2 2) + 1808 5 5408
n56
The measure of each interior angle is 540 4 5 5 1088.
The measure of each exterior angle is 360 4 5 5 728.
The polygon has 6 sides. It is a hexagon.
20. (n 2 2) + 1808 5 (18 2 2) + 1808 5 28808
9. (n 2 2) + 1808 5 19808
The measure of each interior angle is 28808 4 18 5 1608.
The measure of each exterior angle is 3608 4 18 5 208.
n 2 2 5 11
n 5 13
The polygon has 13 sides. It is a 13-gon.
21. (n 2 2) + 1808 5 (90 2 2) + 180 5 15,8408
The measure of each interior angle is 15,8408 4 90 5 1768.
The measure of each exterior angle is 3608 4 90 5 48.
10. (n 2 2) + 1808 5 23408
n 2 2 5 13
n 5 15
The polygon has 15 sides. It is a 15-gon.
6
12
10
JM
6 + JM 5 120
JM 5 20
x 1 423 5 540
}
The length of JM is 20.
12. x8 1 1218 1 968 1 1018 1 1628 1 908 5 (n 2 2) + 1808
x 1 121 1 96 1 101 1 162 1 90 5 (6 2 2) + 180
x 1 570 5 720
x 5 150
13. x8 1 1438 1 2x8 1 1528 1 1168 1 1258 1 1408 1 1398
5 (n 2 2) + 1808
23. The sides of each polygon are congruent, so the ratio
of corresponding sides will always be the same. The
measures of the angles in any regular pentagon are the
same because the measures do not depend on the side
length.
24. (n 2 2) + 1808 5 1568 + n
180n 2 360 5 156n
x 1 143 1 2x 1 152 1 116 1 125 1 140 1 139
5 (8 2 2) + 180
180n 5 360 1 156n
24n 5 360
3x 1 815 5 1080
n 5 15
3x 5 265
25. (98)n 5 3608
x ø 88.3
n 5 40
14. x8 1 788 1 1068 1 658 5 3608
26. The number of sides n, of a polygon can be calculated
x 1 249 5 360
with the Polygon Interior Angles Theorem. n must be a
positive, whole number.
x 5 111
15. x8 1 778 1 2x8 1 458 1 408 5 3608
a.
3x 1 162 5 360
(1658)n 5 (n 2 2) + 1808
(1718)n 5 (n 2 2) + 1808
165n 5 180n 2 360
171n 5 180n 2 360
3x 5 198
215n 5 2360
x 5 66
16. x8 1 x8 1 588 1 398 1 508 1 488 1 598 5 3608
2x 1 254 5 360
2x 5 106
x 5 53
17. The student’s error was thinking the sum of the exterior
angles of different polygons are different when in fact
this sum is always 3608.
The student should have claimed that the sum of the
interior angles of an octagon is greater than the sum of
the interior angles of a hexagon because it has more sides.
Geometry
Worked-Out Solution Key
b.
29n 5 2360
n 5 24; possible
n 5 40; possible
c.
d.
(758)n 5 (n 2 2) + 1808
(408)n 5 (n 2 2) + 1808
75n 5 180n 2 360
2105n 5 2360
n 5 3.43; not possible
40n 5 180n 2 360
2140n 5 2360
n 5 2.57; not possible
27. An increase of one in the number of sides of a polygon
results in an increase of 1808 in the sum of the measures
of the interior angles. If the sum is increased by 5408, the
increase in the number of sides is 5408 4 1808 5 3.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x 5 117
234
KL
JM
}5}
11. x8 1 868 1 1408 1 1388 1 598 5 (n 2 2) + 1808
x 1 86 1 140 1 138 1 59 5 (5 2 2) + 180
ST
RU
}5}
22.
Chapter 8,
continued
Problem Solving
28. (n 2 2) + 1808 5 (5 2 2) + 1808 5 3 + 1808 5 5408
1
36. a. h(n) 5 (n 2 2) + 1808 4 n 5 } (n 2 2) + 1808
n
1
b. h(n) 5 } (n 2 2) + 1808
n
The sum of the interior angle measures of the playing
field is 5408.
1
h(n) 5 }n(n 2 2) + 1808
1
1508 5 }n (n 2 2) + 1808
1
150n 5 (n 2 2) + 180
1
h(9) 5 }9 (9 2 2) + 180
29. (n 2 2) + 1808 5 (6 2 2) + 1808 5 4 + 1808 5 7208
The sum of the interior angle measures of the playing
field is 7208.
5 }9 (1260)
30. (n 2 2) + 1808 5 (6 2 2) + 1808 5 7208
150n 5 180n 2 360
5 140
230n 5 2360
Measure of one angle 5 7208 4 6 5 1208
n 5 12
31. Sum of interior angles
c.
(n 2 2) + 1808 5 (10 2 2) + 1808 5 14408
The measure of each interior angle is 14408 4 10 5 1448.
The measure of each exterior angle is 3608 4 10 5 368.
32. a. P
Q
R
T
Measure of interior angle (degrees)
The measure of each interior angle of the hexagon is 1208.
The value of h(n)
increases as the value
of n increases. The
graph shows that when
n increases, h(n) also
increases.
h
120
90
60
30
0
0
2
4
8 n
6
Sides of regular polygon
37. a.
S
b. (n 2 2) + 1808 5 (5 2 2) + 1808 5 5408
Polygon
Number
of sides
Number of
triangles
Sum of measures
of interior angles
Quadrilateral
4
2
2 + 1808 5 3608
Pentagon
5
3
3 + 1808 5 5408
Hexagon
6
4
4 + 1808 5 7208
Heptagon
7
5
5 + 1808 5 9008
c. mŽ P 1 mŽ Q 1 mŽ R 1 mŽ S 1 mŽT 5 5408
908 1 908 1 mŽ R 1 908 1 mŽ R 5 5408
2(mŽ R) 1 2708 5 5408
2mŽ R 5 2708
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
mŽ R 5 1358 5 mŽT
33. Draw all of the diagonals of ABCDE that have A as an
}
}
endpoint. The diagonals formed, AD and AC, divide
ABCDE into three triangles. By the Angle Addition
Postulate, mŽ CDE 5 mŽ CDA 1 mŽADE. Similarly
mŽ EAB 5 mŽ EAD 1 mŽ DAC 1 mŽ CAB and
mŽ BCD 5 mŽ BCA 1 mŽACD. The sum of the
measures of the interior angles of ABCDE is equal to the
sum of the measures of the angles of triangles n ADE,
n ACD, and n ABC. By the Triangle Sum Theorem,
the sum of the measures of the interior angles of each
triangle is 1808, so the measures of the interior angles of
ABCDE is 3 + 1808 5 (5 2 2) + 1808.
34. By the Polygon Interior Angles Theorem, the sum of the
measures of the interior angles of a regular polygon is
(n 2 2) + 1808. A quadrilateral has 4 sides, so the sum
of measures of the interior angles is (4 2 2) + 1808
5 2 + 1808 5 3608.
35. Let A be a convex n-gon. At each vertex, each interior
b. Using the results from the table in part (a), you can see
that the sum of the measures of the interior angles of
a concave polygon is given by s(n) 5 (n 2 2) + 1808
where n is the number of sides.
38.
H
A
B
G
P
F
C
E
D
The measure of one exterior angle of a regular octagon
at each vertex is 3608 4 8 5 458. This means Ž PBC and
Ž PCB each have a measure of 458. Because the sum of
the measures of the interior angles of a triangle is 1808,
the measure of Ž BPC is 1808 2 458 2 458 5 908.
angle and one of the exterior angles form a linear pair, so
the sum of their measures is 1808. Then the sum of the
measures of the interior angles and one exterior angle at
each vertex is n + 1808. By the Polygon Interior Angles
Theorem, the sum of the measures of the interior angles
of A is (n 2 2) + 1808. So the sum of the measures of the
exterior angles of A, one at each vertex, is
n + 1808 2 (n 2 2) + 1808 5 n + 1808 2 n + 1808 1 3608
5 3608.
Geometry
Worked-Out Solution Key
235
continued
Mixed Review for TAKS
4. KM 5 2 + KN 5 2 + 2 5 4
39. A;
5. mŽ JML 5 1808 2 mŽ KJM 5 180 2 110 5 70
The boundary line passes through the points (0, 3) and
(23, 0). Using these points, the slope of the boundary
line is:
y2 2 y1
023
23
8.2 Exercises (pp. 518–521)
Skill Practice
m5}
5}
5}
51
x 2x
23 2 0
23
2
6. mŽ KML 5 mŽ JML 2 mŽ JMK 5 70 2 30 5 40
1
1. That both pairs of opposite sides of a parallelogram
The y-intercept, b, is 3, so the equation of the boundary
line is y 5 x 1 3. Because the boundary line is solid, the
inequality is either aor q.
The half-plane that includes the point (0, 0) is shaded.
Because 0a0 1 3, the inequality isa. So, the inequality
yax 1 3 best describes the graph.
are parallel is a property included in its definition.
Other properties of parallelograms are their opposite
sides and angles are congruent, consecutive angles are
supplementary, and the diagonals bisect each other.
2. mŽ C 5 658 because Ž C is opposite ŽA in the
parallelogram. Consecutive angles in a parallelogram are
supplementary, so Ž B and Ž D each have a measure of
1808 2 658 5 1158.
40. J;
Let x 5 the number of movies Will has. Then Don has
(x 1 9) movies and Kyle has 3(x 1 9) movies. The
equation that can be used to find how many movies Will
has is: x 1 (x 1 9) 1 3(x 1 9) 5 46.
41. B;
$4750 2 $3500
3. x 5 9
y 5 15
4. n 5 12
m1156
m55
$1250
5}
5 $625
Rate of change 5 }}
2003 2 2001
2
5. a8 5 558
Because the year 2007 is 6 years from the year 2001,
a reasonable projection for the value of the investment
account in 2007 is $3500 1 6($625) 5 $3500 1 $3750
5 $7250.
7. 20 5 z 2 8
a 5 55
28 5 z
126 5 d
9. mŽA 1 mŽ B 5 1808
10. mŽ L 1 mŽ M 5 1808
518 1 mŽ B 5 1808
mŽ L 1 958 5 1808
mŽ B 5 1298
1198 1 mŽY 5 1808
If point A is dragged out, the angle measures of Ž A and Ž C
decrease, and the angle measures of Ž B and Ž D increase.
Similarly, when point B is dragged away from the figure,
the angle measures of Ž B and Ž D decrease while the angle
measures of Ž A and Ž C increase. Whether point A is dragged
or point B is dragged, Ž A and Ž C always have the same
measure and Ž B and Ž D always have the same measure.
mŽY 5 618
12.
P
1028
788
788
R
ŽR > ŽP and ŽS > ŽQ
mŽ R 1 mŽ S 5 1808
2. Opposite side lengths and opposite angle measures in a
parallelogram are always equal.
15 5 y
3. NM 5 KN
NM 5 2
Geometry
Worked-Out Solution Key
5 788 1 248
248 1 2mŽ S 5 1808
mŽ S 5 788
mŽ G 5 mŽ E
mŽ G 5 608
13. b 2 1 5 9
b 5 10
mŽ J 5 mŽ L
2x8 5 508
x 5 25
5 1028
2mŽ S 5 1568
8.2 Guided Practice (pp. 516–517)
18 5 y 1 3
mŽ R 5 mŽ S 1 248
(mŽ S 1 248) 1 mŽ S 5 1808
3. Answers will vary.
2. JK 5 LM
Q
1028
S
1. Both sets of opposite sides in the polygon are parallel.
FG 5 8
mŽ L 5 858
11. mŽ X 1 mŽY 5 1808
STEP 4
1. FG 5 HE
g 5 61
h59
}
The sides AB and DC remain parallel and their lengths remain
}
}
equal to each other. Similarly, AD and BC remain parallel and
their lengths remain equal to each other.
236
(g 1 4)8 5 658
2h 5 29
Investigating Geometry Activity 8.2 (p. 514)
}
p 5 60
1058 5 (d 2 21)8
8. 16 2 h 5 7
Lesson 8.2
STEP 3
6. 2p8 5 1208
14.
4m 5 16
m54
5a 5 15
a53
2n 5 9 2 n
3n 5 9
n53
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 8,