# Chapter 7 233 Worked-Out

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Chapter 7 233 Worked-Out
```Chapter 7
Chapter 7 Maintaining Mathematical Proficiency (p. 361)
1 − (−3) 1 + 3 4
3−0
3
3
1 − (−3) 1 + 3 4
Slope of line b: — = — = —
0 − (−3)
3
3
3
3
4−1
Slope of line c: — = — = −—
4
−2 − 2 −4
6
3
−6
−4 − 2
Slope of line d: — = — = −— = −—
8
4
4 − (−4) 4 + 4
5. Slope of line a: — = — = —
1. 4(7 − x) = 16
4(7 − x)
4
16
4
—=—
7−x=
4
−7
−7
−x = −3
−x
−1
Because the slopes of line a and line b are equal, a b.
Because the slopes of line c and line d are equal, c d.
3
4
Because — −— = −1, a ⊥ c, a ⊥ d, b ⊥ c, and b ⊥ d.
4
3
−3
−1
—=—
( )( )
x=3
1
−4 − (−2) −4 + 2 −2
4+2
6
3
4 − (−2)
−4
−4
−2 − 2
Slope of line b: — = — = — = 4
−3 − (−2) −3 + 2 −1
−3 − 1 −4
Slope of line c: — = — = 4
−1
2−3
−1
1
3−4
Slope of line d: — = — = −—
4
4
0 − (−4)
2. 7(1 − x) + 2 = −19
−2
6. Slope of line a: — = — = — = −—
−2
7(1 − x) = −21
7(1 − x)
7
−21
7
—=—
1 − x = −3
−1
−1
( )
−x = −4
1
Because 4 −— = −1, b ⊥ d and c ⊥ d.
4
Because the slopes of line b and line c are equal, b c.
−x −4
—=—
−1 −1
7. You can follow the order of operations with all of the other
x=4
operations in the equation and treat the operations in the
expression separately.
3. 3(x − 5) + 8(x − 5) = 22
11(x − 5) = 22
11(x − 5) 22
—=—
11
11
x−5= 2
+5
Chapter 7 Mathematical Thinking (p. 362)
1. false; There is no overlap between the set of trapezoids and
the set of kites.
2. true; There is no overlap between the set of kites and the set
+5
of parallelograms.
x=7
2 − (−2)
−2 − 4
2+2
−6
4
−6
2
3
3. false; There is area inside the parallelogram circle for
4. Slope of line a: — = — = — = −—
2
−4 − (−2) −4 + 2 −2
Slope of line b: — = — = — = −—
0 − (−3)
3
3
3
−3
−3
1
−3 − 0
Slope of line c: —= — = — = −—
3 − (−3) 3 + 3
6
2
4−0 4
Slope of line d: — = — = 2
3−1 2
Because the slopes of line a and line b are equal, a b.
1
Because −— (2) = −1, c ⊥ d.
2
rhombuses and other parallelograms that do not fall inside
the circle for rectangles.
4. true; All squares are inside the category of quadrilaterals.
are kites. All squares are parallelograms. All squares are
rectangles. Some rectangles are squares. No rhombuses
are trapezoids.
6.
( )
Parallelograms
Rhombuses
Squares
Trapezoids
Rectangles
Kites
Geometry
Worked-Out Solutions
233
Chapter 7
7.1 Explorations (p. 363)
1. a. The sum of the measures of the interior angles of
A
quadrilateral ABCD is 91° + 106° + 64° + 99° = 360°.
109.65°
C
56.31°
B
B
D
116.57°
77.47°
C
A
D
The sum of the measures of the exterior angles of
109.65° + 56.31° + 116.57° + 77.47° = 360°.
The sum of the measures of the interior angles of pentagon
EFGHI is 108° + 110° + 115° + 91° + 116° = 540°.
G
A 71.57°
H
F
45°
74.74°
E
90°
E
b.
I
Number of
sides, n
3
Sum of angle
measures, S
c.
C
78.69°
D
4
5
6
7
8
9
B
The sum of the measures of the exterior angles of
pentagon ABCDE is
45° + 71.57° + 74.74° + 78.69° + 90° = 360°.
180° 360° 540° 720° 900° 1080° 1260°
b.
y
(9, 1260)
(8, 1080)
(7, 900)
(6, 720)
(5, 540)
(4, 360)
(3, 180)
1200
1000
800
600
400
200
1 2 3 4 5 6 7 8 9 10 x
As the number of sides increases by 1, the sum of the
measures of the interior angles increases by 180°.
d. S = (n − 2)
⋅
180; Let n represent the number of sides of
the polygon. If you subtract 2 and multiply the difference
by 180°, then you get the sum of the measures of the
interior angles of a polygon.
Number of
sides, n
Sum of
exterior angle
measures, M
c.
y
500
450
400
350
300
250
200
150
100
50
0
3
4
5
6
7
8
9
360° 360° 360° 360° 360° 360° 360°
0 1 2 3 4 5 6 7 8 9 x
For any number of sides n, the sum of the measures of the
exterior angles is 360°. So, the line y = 360 can be drawn
through the points.
d. M = 360; The sum of the measures of the exterior angles
of any polygon is 360°.
3. The sum S of the measures of the interior angles of a
polygon is given by the equation S = (n − 2)180, where
n is the number of sides of the polygon. The sum M of the
measures of the exterior angles of a polygon is 360° for any
number of sides n.
234
Geometry
Worked-Out Solutions
Chapter 7
7.1 Monitoring Progress (pp. 364–367)
1. (n − 2)
⋅ 180° = (11 − 2) ⋅ 180°
= 9 ⋅ 180°
= 1620°
The sum of the measures of the interior angles of the
11-gon is 1620°.
2. (n − 2)
⋅ 180° = 1440°
⋅
(n − 2) 180° 1440°
—— = —
180°
180°
n−2=8
n = 10
The polygon has 10 sides, so it is a decagon.
3. x + 3x + 5x + 7x = 360°
16x = 360°
16x 360°
—=—
16
16
x = 22.5°
⋅
⋅
7x = 7 ⋅ 22.5° = 157.5°
3x = 3 22.5° = 67.5°
5x = 5 22.5° = 112.5°
The angle measures are 22.5°, 67.5°, 112.5°, and 157.5°.
4. The total sum of the angle measures of the polygon with
5 sides is 540°.
93° + 156° + 85° + 2x = 540°
7.1 Exercises (pp. 368-370)
Vocabulary and Core Concept Check
1. A segment connecting consecutive vertices is a side of the
polygon, not a diagonal.
2. The one that does not belong is “the sum of the measures of
the interior angles of a pentagon”. This sum is 540°, but in
each of the other three statements the sum is 360°.
Monitoring Progress and Modeling with Mathematics
3. (n − 2)
The sum of the measures of the interior angles in a nonagon
is 1260°.
4. (n − 2)
x = 103
m∠S = m∠T = 103°
5. Pentagon ABCDE is equilateral but not equiangular.
⋅ 180° = (14 − 2) ⋅ 180° = 12 ⋅ 180° = 2160°
The sum of the measures of the interior angles in a 14-gon
is 2160°.
5. (n − 2)
⋅ 180° = (16 − 2) ⋅ 180° = 14 ⋅ 180° = 2520°
The sum of the measures of the interior angles in a
16-gon is 2520°.
6. (n − 2)
⋅ 180° = (20 − 2) ⋅ 180° = 18 ⋅ 180° = 3240°
The sum of the measures of the interior angles in a 20-gon
is 3240°.
⋅ 180° = 720°
720°
(n − 2) ⋅ 180° —
=
——
180°
180°
7. (n − 2)
n−2=4
334 + 2x = 540
2x = 206
⋅ 180° = (9 − 2) ⋅ 180° = 7 ⋅ 180° = 1260°
n=6
The polygon has 6 sides, so it is a hexagon.
⋅ 180° = 1080°
1080°
(n − 2) ⋅ 180° —
=
——
180°
180°
8. (n − 2)
n−2=6
C
n=8
B
D
The polygon has 8 sides, so it is an octagon.
⋅ 180° = 2520°
2520°
(n − 2) ⋅ 180° —
=
——
180°
180°
9. (n − 2)
A
E
n − 2 = 14
6. The sum of the measures of the exterior angles is 360°.
The sum of the known exterior angle measures is
34° + 49° + 58° + 67° + 75° = 283°.
So, the measure of the exterior angle at the sixth vertex is
360° − 283° = 77°.
7. You can find the measure of each exterior angle by
subtracting the measure of the interior angle from 180°.
In Example 6, the measure of each exterior angle is
180° − 150° = 30°.
n = 16
The polygon has 16 sides, so it is a 16-gon.
⋅ 180° = 3240°
3240°
(n − 2) ⋅ 180° —
=
——
180°
180°
10. (n − 2)
n − 2 = 18
n = 20
The polygon has 20 sides, so it is a 20-gon.
Geometry
Worked-Out Solutions
235
Chapter 7
11. XYZW is a quadrilateral, therefore the sum of the measures
⋅
19. The polygon has 5 sides, therefore the sum of the measures
⋅
of the interior angles is (4 − 2) 180° = 360°.
of the interior angles is (5 − 2) 180° = 540°.
100° + 130° + 66° + x° = 360°
x° + x° + 164° + 102° + 90° = 540°
296 + x = 360
2x + 356 = 540
x = 64
2x = 184
12. HJKG is a quadrilateral, therefore the sum of the measures
⋅
of the interior angles is (4 − 2) 180° = 360°.
103° + 133° + 58° + x° = 360°
x = 92
m∠X = m∠Y = 92°
20. The polygon has 5 sides, therefore the sum of the measures
⋅
of the interior angles is (5 − 2) 180° = 540°.
294 + x = 360
x = 66
x° + 90° + x° + 119° + 47° = 540°
13. KLMN is a quadrilateral, therefore the sum of the measures
2x + 256 = 540
⋅
2x = 284
of the interior angles is (4 − 2) 180° = 360°.
x = 142
88° + 154° + x° + 29° = 360°
m∠X = m∠Y = 142°
271 + x = 360
x = 89
21. The polygon has 6 sides, therefore the sum of the measures
14. ABCD is a quadrilateral, therefore the sum of the measures
⋅
of the interior angles is (4 − 2) 180° = 360°.
⋅
of the interior angles is (6 − 2) 180° = 720°.
90° + 99° + 171° + x° + x° + 159° = 720°
2x + 519 = 720
x° + 92° + 68° + 101° = 360°
2x = 201
x + 261 = 360
x = 100.5
x = 99
15. The polygon has 6 sides, therefore the sum of the measures
⋅
of the interior angles is (6 − 2) 180° = 720°.
m∠X = m∠Y = 100.5°
22. The polygon has 6 sides, therefore the sum of the measures
102° + 146° + 120° + 124° + 158° + x° = 720°
650 + x = 720
⋅
of the interior angles is (6 − 2) 180° = 720°.
100° + x° + 110° + 149° + 91° + x° = 720°
x = 70
2x + 450 = 720
2x = 270
16. The polygon has 5 sides, therefore the sum of the measures
⋅
of the interior angles is (5 − 2) 180° = 540°.
x = 135
m∠X = m∠Y = 135°
86° + 140° + 138° + 59° + x° = 540°
423 + x = 540
23. 65° + x° + 78° + 106° = 360°
x = 117
x + 249 = 360
17. The polygon has 6 sides, therefore the sum of the measures
⋅
of the interior angles is (6 − 2) 180° = 720°.
x = 111
24. 48° + 59° + x° + x° + 58° + 39° + 50° = 360°
121° + 96° + 101° + 162° + 90° + x° = 720°
2x + 254 = 360
570 + x = 720
2x = 106
x = 150
x = 53
18. The polygon has 8 sides, therefore the sum of the measures
⋅
of the interior angles is (8 − 2) 180° = 1080°.
143° + 2x° + 152° + 116° + 125° + 140° + 139° + x° = 1080°
815 + 3x = 1080
3x = 265
1
x = 88—
3
25. 71° + 85° + 44° + 3x° + 2x° = 360°
5x + 200 = 360
5x = 160
x = 32
26. 40° + x° + 77° + 2x° + 45° = 360°
3x + 162 = 360
3x = 198
x = 66
236
Geometry
Worked-Out Solutions
Chapter 7
27. The sum of the measures of the interior angles of a pentagon
⋅
⋅
is (5 − 2) 180° = 3 180° = 540°.
540°
Each interior angle: — = 108°
5
360°
Each exterior angle: — = 72°
5
The measure of each interior angle of a regular pentagon is
108° and the measure of each exterior angle is 72°.
35.
n(180 − x) = 360
28. The sum of the measures of the interior angles of an 18-gon
⋅
⋅
n(180 − x)
180 − x
is (18 − 2) 180° = 16 180° = 2880°.
360
180 − x
360
n=—
180 − x
—=—
2880°
Each interior angle: — = 160°
18
360°
Each exterior angle: — = 20°
18
The measure of each interior angle of a regular 18-gon is
160° and the measure of each exterior angle is 20°.
The formula to find the number of sides n in a regular
polygon given the measure of one interior angle x°
360
is: n = —.
180 − x
29. The sum of the measures of the interior angles of a 45-gon
⋅
⋅
is (45 − 2) 180° = 43 180° = 7740°.
7740°
Each interior angle: — = 172°
45
360°
Each exterior angle: — = 8°
45
The measure of each interior angle of a regular 45-gon is
172° and the measure of each exterior angle is 8°.
30. The sum of the measures of the interior angles of a regular
⋅
⋅
90-gon is (90 − 2) 180° = 88 180° = 15,840°.
15,840°
Each interior angle: — = 176°
90
360°
Each exterior angle: — = 4°
90
The measure of each interior angle of a regular 90-gon is
176° and the measure of each exterior angle is 4°.
31. The measure of one interior angle of a regular pentagon was
found, but the measure of one exterior angle should be found
by dividing 360° by the number of angles.
360°
The correct response should be — = 72°.
5
32. There is only one exterior angle at each vertex. So, the
measure of one exterior angle is found by dividing 360° by
the number of vertices. The correct response should
360°
be — = 72°.
5
33. A regular hexagon has 6 sides. The sum of the measures of
⋅
the interior angles is (6 − 2) 180° = 720°. The measure of
720°
each interior angle is — = 120°.
6
⋅
⋅
n ⋅ 180 − 2 ⋅ 180 = nx
n ⋅ 180 − 360 = nx
n ⋅ 180 = nx + 360
n ⋅ 180 − nx = 360
(n − 2) 180°
n
(n − 2) 180 = nx
—— = x°
36.
360°
x° = —
n
360
n x=n —
n
n x = 360
⋅ ⋅
⋅
n ⋅ x 360
—=—
x
x
360
n=—
x
The formula to find the number of sides n in a regular
polygon given the measure of one exterior angle x° is:
360
n = —.
x
360°
360°
37. n = —— = — = 15
180° − 156°
24°
The number of sides of a polygon where each interior angle
has a measure of 156° is 15.
360°
180° − 165°
360°
15°
38. n = —— = — = 24
The number of sides of a polygon where each interior angle
has a measure of 165° is 24.
360°
9°
The number of sides of a polygon where each exterior angle
has a measure of 9° is 40.
39. n = — = 40
360°
6°
The number of sides of a polygon where each exterior angle
has a measure of 6° is 60.
40. n = — = 60
34. A regular decagon has 10 sides. The sum of the measures of
⋅
the interior angles is (10 − 2) 180° = 1440°. The measure
1440°
of each interior angle is — = 144°. The measure of each
10
360°
exterior angle is — = 36°.
10
Geometry
Worked-Out Solutions
237
Chapter 7
41. A, B;
47. Divide the quadrilateral into two triangles. The sum of the
measures of the interior angles of each triangle is 180°.
Therefore, the total sum of the interior angle measurements
for this quadrilateral is 2 180° = 360°.
360°
360°
A. n = —— = — = 20 ✓
180° − 162°
18°
360°
360°
B. n = —— = — = 40 ✓
180° − 171°
9°
360°
360°
C. n = — = — ≈ 3.43 ✗
180° − 75° 105°
360°
360°
D. n = — = — ≈ 2.57 ✗
180° − 40° 140°
⋅
Solving the equation found in Exercise 35 for n yields a
positive integer greater than or equal to 3 for A and B, but
not for C and D.
Divide the pentagon into three triangles. The sum of the
measures of the interior angles of each triangle is 180°.
Therefore, the total sum of the interior angle measurements
for this pentagon is 3 180° = 540°.
⋅
42. In a pentagon, when all the diagonals from one vertex are
drawn, the polygon is divided into three triangles. Because
the sum of the measures of the interior angles of each
triangle is 180°, the sum of the measures of the interior
angles of the pentagon is (5 − 2) 180° = 3 180° = 540°.
⋅
⋅
43. In a quadrilateral, when all the diagonals from one vertex are
drawn, the polygon is divided into two triangles. Because the
sum of the measures of the interior angles of each triangle
is 180°, the sum of the measures of the interior angles of the
quadrilateral is 2 180° = 360°.
Divide the hexagon into four triangles. The sum of the
measures of the interior angles of each triangle is 180°.
Therefore, the total sum of the interior angle measurements
for this hexagon is 4 180° = 720°.
⋅
⋅
44. yes; Because an interior angle and an adjacent exterior angle
of a polygon form a linear pair, you can use the Polygon
Exterior Angles Theorem (Thm. 7.2) to find the measure of
the exterior angles, and then you can subtract this value from
180° to find the interior angle measures of a regular polygon.
45. A hexagon has 6 angles. If 4 of the exterior angles have
Divide the heptagon into five triangles. The sum of the
measures of the interior angles of each triangle is 180°.
Therefore, the total sum of the interior angle measurements
for this heptagon is 5 180° = 900°.
⋅
a measure of x°, the other two each have a measure of
2(x + 48)°, and the total exterior angle sum is 360°, then:
4x° + 2[2(x + 48)]° = 360°
4x + 2[2x + 96] = 360
4x + 4x + 192 = 360
8x + 192 = 360
8x = 168
x = 21
⋅
2(x + 48) = 2(21 + 48) = 2 69 = 138
So, the measures of the exterior angles are 21°, 21°, 21°, 21°,
138°, and 138°.
46. yes; The measure of the angle where the polygon caves in is
When diagonals are drawn from the vertex of the concave
angle as shown, the polygon is divided into n − 2 triangles
whose interior angle measures have the same total as the sum
of the interior angle measures of the original polygon. So,
an expression to find the sum of the measures of the interior
angles for a concave polygon is (n − 2) 180°.
⋅
48. The base angles of △BPC are congruent exterior angles
of the regular octagon, each with a measure of 45°. So,
m∠BPC = 180° − 2(45°) = 90°.
greater than 180° but less than 360°.
F
G
D
H
C
A
238
Geometry
Worked-Out Solutions
E
B
P
Chapter 7
49. a. The formula for the number of sides n in a regular
polygon, where h(n) is the measure of any interior angle
⋅
(n − 2) 180°
is h(n) = ——.
n
(9 − 2) 180° 7 180° 1260°
b. h(9) = —— = — = — = 140°
9
9
9
(n − 2) 180°
c. 150° = ——
n
150n = (n − 2) 180
⋅
⋅
⋅
⋅
150n = 180n − 360
Maintaining Mathematical Proficiency
53. x° + 79° = 180°
x = 180 − 79
x = 101
54. x° + 113° = 180°
x = 180 − 113
x = 67
55. (8x − 16)° + (3x + 20)° = 180°
−30n = −360
11x + 4 = 180
−30n −360
—=—
−30
−30
11x = 176
x = 16
n = 12
56. (6x − 19)° + (3x + 10)° = 180°
When h(n) = 150°, n = 12.
9x − 9 = 180
d.
9x = 189
Number of sides
Measure of each interior angle
3
60°
4
90°
5
108°
6
120°
7
128.6°
8
135°
h(n)
140
120
100
80
60
(8, 135)
(7, 128.6)
(6, 120)
(5, 108)
(4, 90)
x = 21
7.2 Explorations (p. 371)
1. a. Check students’ work; Construct ⃖⃗
AB and a line parallel
to ⃖⃗
AB through point C. Construct ⃖⃗
BC and a line parallel
to ⃖⃗
BC through point A. Construct a point D at the
intersection of the line drawn parallel to ⃖⃗
AB and the line
—, BC
—, CD
—, and
drawn parallel to ⃖⃗
BC. Finally, construct AB
—
DA by removing the rest of the parallel lines drawn.
b. Check students’ work (For sample in text,
m∠A = m∠C = 63.43° and m∠B = m∠D = 116.57°.);
Opposite angles are congruent, and consecutive
angles are supplementary.
c. Check students’ work (For sample in text, AB = CD =
2.24 and BC = AD = 4.); Opposite sides are congruent.
d. Check students’ work; Opposite angles of a parallelogram
(3, 60)
are congruent. Consecutive angles of a parallelogram are
supplementary. Opposite sides of a parallelogram
are congruent.
40
20
1 2 3 4 5 6 7 8 n
The value of h(n) increases on a curve that gets less steep
as n increases.
50. no; The interior angles are supplements of the adjacent
exterior angles, and because the exterior angles have
different values, the supplements will be different as well.
51. In a convex n-gon, the sum of the measures of the n interior
⋅
angles is (n − 2) 180° using the Polygon Interior Angles
Theorem (Thm. 7.1). Because each of the n interior angles
forms a linear pair with its corresponding exterior angle,
you know that the sum of the measures of the n interior and
exterior angles is 180n°. Subtracting the sum of the interior
angle measures from the sum of the measures of the linear
pairs gives you 180n° − [(n − 2) 180°] = 360°.
2. a. Check students’ work.
b. Check students’ work.
c. Check students’ work (For sample in text,
AE = CE = 1.58 and BE = DE = 2.55.); Point E
— and BD
—.
bisects AC
d. The diagonals of a parallelogram bisect each other.
3. A parallelogram is a quadrilateral where both pairs of
opposite sides are congruent and parallel, opposite angles
are congruent, consecutive angles are supplementary, and the
diagonals bisect each other.
⋅
540°
180°
diagonals, the new polygon will need 3 more sides.
52. In order to have — = 3 more triangles formed by the
Geometry
Worked-Out Solutions
239
Chapter 7
7.2 Monitoring Progress (pp. 373–375)
5. By the Parallelogram Diagonals Theorem (Thm. 7.6), the
diagonals of a parallelogram bisect each other.
1. m∠G = m∠E
) ( )
) ( )
FG = HE
4 6
1+3 5+1
—: —
, — = —, — = (2, 3)
Midpoint of TV
2 2
2
2
3
+
3
−2
+
6
—: —, — = —4, —6 = (2, 3)
Midpoint of SU
2 2
2
2
HE = 8
The coordinates of the intersection of the diagonals are (2, 3).
(
(
m∠E = 60°
m∠G = 60°
FG = 8
In parallelogram GHEF, FG = 8 and m∠G = 60°.
2. m∠J = m∠L
—
2
2−4
5−2
3
Starting at C, go up 2 units and left 3 units. So, the
coordinates of D are (0, 1).
6. Slope of AB = — = −—
2x° = 50°
4
x = 25
y
A
2
JK = ML
B
D
18 = y + 3
−2
15 = y
−2
In parallelogram JKLM, x = 25 and y = 15.
−4
C
4
x
3. m∠BCD + m∠ADC = 180°
m∠BCD + 2m∠BCD = 180°
7.2 Exercises (pp. 376-378)
3m∠BCD = 180°
Vocabulary and Core Concept Check
m∠BCD = 60°
1. In order to be a quadrilateral, a polygon must have 4 sides,
4. Given ABCD and GDEF are parallelograms.
Prove ∠C and ∠F are supplementary angles.
C
E
B
A
240
and parallelograms always have 4 sides. In order to be a
parallelogram, a polygon must have 4 sides with opposite
sides parallel. Quadrilaterals always have 4 sides, but do not
always have opposite sides parallel.
D
2. The two angles that are consecutive to the given angle are
F
G
STATEMENTS
REASONS
1. ABCD and GDEF are
parallelograms.
1. Given
2. ∠C and ∠D are
supplementary angles.
2. Parallelogram
Consecutive Angles
Theorem (Thm. 7.5)
3. m∠C + m∠D = 180°
3. Definition of
supplementary angles
4. ∠D ≅ ∠F
4. Parallelogram Opposite
Angles Theorem
(Thm. 7.4)
supplementary to it. So, you can find each of their measures
by subtracting the measure of the given angle from 180°. The
angle opposite the given angle is congruent and therefore has
the same measure.
Monitoring Progress and Modeling with Mathematics
3. The Parallelogram Opposite Sides Theorem (Thm. 7.3)
applies here. Therefore, x = 9 and y = 15.
4. By the Parallelogram Opposite Sides Theorem (Thm. 7.3),
n = 12 and m + 1 = 6. Therefore, m = 5.
5. Parallelogram Opposite Sides Theorem (Thm. 7.3):
z − 8 = 20
z = 28
5. m∠D = m∠F
5. Definition of congruent
angles
6. m∠C + m∠F = 180°
6. Substitution Property of
Equality
(d − 21)° = 105°
7. ∠C and ∠F are
supplementary angles.
7. Definition of
supplementary angles
Therefore, z = 28 and d = 126.
Geometry
Worked-Out Solutions
Parallelogram Opposite Angles Theorem (Thm. 7.4):
d = 126
Chapter 7
6. Parallelogram Opposite Sides Theorem (Thm. 7.3):
16 − h = 7
18. (b − 10)° + (b + 10)° = 180°
2b = 180
−h = −9
b = 90
h=9
d° = (b + 10)°
Parallelogram Opposite Angles Theorem (Thm. 7.4):
(g + 4)° = 65°
d = 100
g = 61
Therefore, h = 9 and g = 61.
7. m∠A + m∠B = 180°
51° + m∠B = 180°
m∠B = 129°
8. m∠M + m∠N = 180°
95° + m∠N = 180°
m∠N = 85°
9. LM = 13; By the Parallelogram Opposite Sides Theorem
(Thm. 7.3), LM = QN.
10. LP = 7; By the Parallelogram Diagonals Theorem (Thm. 7.6),
LP = PN.
11. LQ = 8; By the Parallelogram Opposite Sides Theorem
(Thm. 7.3), LQ = MN.
12. By the Parallelogram Diagonals Theorem (Thm. 7.6),
MP = PQ.
⋅
d = 90 + 10
⋅
MQ = 2 MP = 2 8.2 = 16.4
13. Parallelogram Consecutive Angles Theorem (Thm. 7.5)
m∠LMN + m∠MLQ = 180°
m∠LMN + 100° = 180°
m∠LMN = 80°
14. Parallelogram Opposite Angles Theorem (Thm. 7.4)
m∠NQL = m∠NML
m∠NQL = 80°
c = (b − 10)°
c = 90 − 10
c = 80
So, b = 90, c = 80, and d = 100.
19. k + 4 = 11
k=7
m=8
So, k = 7 and m = 8.
20. 2u + 2 = 5u − 10
2u = 5u − 12
−3u = −12
−3u ____
−12
—=
−3
−3
u=4
v
3
v = 18
So, u = 4 and v = 18.
—=6
21. In a parallelogram, consecutive angles are supplementary;
Because quadrilateral STUV is a parallelogram, ∠S and ∠V
are supplementary. So, m∠V = 180° − 50° = 130°.
22. In a parallelogram, the diagonals bisect each other. So
— are congruent to each other; Because
the two parts of GJ
— ≅ FJ
—.
quadrilateral GHJK is a parallelogram, GF
23. Given ABCD and CEFD
B
are parallelograms.
— ≅ FE
—
Prove AB
A
15. Parallelogram Opposite Angles Theorem (Thm. 7.4)
m∠MNQ = m∠MLQ
m∠MNQ = 100°
16. Alternate Interior Angles Theorem (Thm. 3.2)
m∠LMQ = m∠NQM
m∠LMQ = 29°
17. n° + 70° = 180°
n = 110
C
D
E
F
STATEMENTS
REASONS
1. ABCD and CEFD are
parallelograms.
1. Given
— —— —
2. AB ≅ DC , DC ≅ FE
— —
3. AB ≅ FE
2. Parallelogram Opposite
Sides Theorem (Thm. 7.3)
3. Transitive Property of
Congruence (Thm. 2.1)
2m° = 70°
m = 35
So, n = 110 and m = 35.
Geometry
Worked-Out Solutions
241
Chapter 7
24. Given ABCD, EBGF, and
E
A
HJKD are parallelograms.
F
Prove ∠2 ≅ ∠3
2
H
D
1
28.
y
F
G
3
4
B
K
J
C
2
G
STATEMENTS
REASONS
1. ABCD, EBGF, and
HJKD are
parallelograms.
1. Given
2. ∠1 ≅ ∠2, ∠3 ≅ ∠4,
∠1 ≅ ∠4
2. Parallelogram Opposite
Angles Theorem
(Thm. 7.4)
3. ∠2 ≅ ∠4
3. Transitive Property of
Congruence (Thm. 2.2)
4. ∠2 ≅ ∠3
4. Transitive Property of
Congruence (Thm. 2.2)
−4
2
D
4 x
−4
7
7−0
—=—
= — = −7
Slope of FG
0 − 1 −1
Starting at D, go up 7 units and left 1 unit. So, the
coordinates of E are (−3, 3).
29.
4
y
F
E
25. By the Parallelogram Diagonals Theorem (Thm. 7.6), the
diagonals of a parallelogram bisect each other.
−2 + 4 5 + 0
2 5
—: —
, — = —, — = (1, 2.5)
Midpoint of WY
2
2
2 2
2+0 5+0
2 5
—: —
, — = —, — = (1, 2.5)
Midpoint of ZX
2
2
2 2
−4
) ( )
) ( )
(
(
4
E
G
−2
2
4 x
−2
D
−4
3
3
1 − (−2)
—=—
=—=—=3
Slope of ED
−3 − (−4) −3 + 4 1
Starting at F, go down 3 units and left 1 unit. So, the
coordinates of G are (2, 0).
The coordinates of the intersection of the diagonals are (1, 2.5).
26. By the Parallelogram Diagonals Theorem (Thm. 7.6), the
diagonals of a parallelogram bisect each other.
(
) ( )
—: 5 + (−5), 2 + (−1) = 0, 1 = (0, 0.5)
Midpoint of TR
( 2
) (2 2)
2
−1 + 1 3 + (−2)
0 1
—: —
, — = —, — = (0, 0.5)
Midpoint of QS
2
2
2 2
— —
— —
30.
6
y
F
E
2
The coordinates of the intersection of the diagonals are (0, 0.5).
G
2
27.
y
E
−2
F
2
−4
−2
D
G
2
4 x
−2
5−2
3
—=—
Slope of ED
= — = −3
−1 − 0 −1
Starting at G, go up 3 units and left 1 unit. So, the
coordinates of F are (3, 3).
4
8 x
D
6
6−0
—=—
= — = −2
Slope of FG
5 − 8 −3
Starting at E, go down 6 units and right 3 units. So, the
coordinates of D are (2, −2).
31. x° + 0.25x° + x° + 0.25x° = 360°
2.5x = 360
x = 144
0.25x° = 0.25(144°) = 36°
The angles are 36° and 144°.
32. x° + (4x + 50)° + x° + (4x + 50)° = 360°
10x + 100 = 360
10x = 260
⋅
x = 26
(4x + 50)° = (4 26 + 50)° = 154°
The angles are 154° and 26°.
242
Geometry
Worked-Out Solutions
Chapter 7
33. The quadrilateral could not be a parallelogram because ∠A
and ∠C are opposite angles, but m∠A ≠ m∠C.
37. Given ABCD is a parallelogram.
B
A
m∠J + m∠K = 180°
34.
STATEMENTS
(3x + 7)° + (5x − 11)° = 180°
8x − 4 = 180
1.
8x = 184
x = 23
⋅
⋅
2.
m∠J = (3 23 + 7)° = 76°
m∠K = (5 23 − 11)° = 104°
3.
B
A
C
D
When you fold the parallelogram so that vertex A is on
vertex C, the fold will pass through the point where the
diagonals intersect, which demonstrates that this point of
—. Similarly, when
intersection is also the midpoint of AC
you fold the parallelogram so that vertex B is on vertex D,
the fold will pass through the point where the diagonals
intersect, which demonstrates that this point of intersection is
—.
also the midpoint of BD
r:
A
C
1
D
E
F
G
m∠1 = m∠G because corresponding parts of congruent
— ⃖⃗
—. So,
figures are congruent. Note that AB
CD FG
m∠1 = m∠BDE and m∠GDE = m∠G because they are
pairs of alternate interior angles, and m∠1 = m∠GDE
by the Transitive Property of Equality. Also, by the Angle
Addition Postulate (Post. 1.4), m∠2 = m∠BDE + m∠GDE.
By substituting, you get m∠2 = m∠1 + m∠1 = 2m∠1.
D
REASONS
ABCD is a
parallelogram.
— DC
—, BC
—
AB
∠BDA ≅ ∠DBC,
∠DBA ≅ ∠BDC
1.
Given
2.
Definition of
parallelogram
3.
Alternate Interior
Angles Theorem (Thm.
3.2)
4.
— ≅ BD
—
BD
4.
Reflexive Property of
Congruence (Thm. 2.1)
5.
△ABD ≅ △CDB
5.
ASA Congruence
Theorem (Thm. 5.10)
6.
∠A ≅ ∠C
6.
Corresponding parts of
congruent triangles are
congruent.
7.
m∠BDA = m∠DBC,
m∠DBA = m∠BDC
7.
Definition of congruent
angles
8.
m∠B = m∠DBC +
m∠DBA
m∠D = m∠BDA +
m∠BDC
8.
Postulate (Post. 1.4)
9.
m∠D = m∠DBC +
m∠DBA
9.
Substitution Property of
Equality
B
2
10. m∠D = m∠B
10. Transitive Property of
Equality
11. ∠D ≅ ∠B
11. Definition of congruent
angles
Q
x°
38. Given PQRS is a parallelogram.
Prove x° + y° = 180°
P
y°
STATEMENTS
REASONS
1. PQRS is a parallelogram.
1. Given
— PS
—
2. QR
C
Prove ∠A ≅ ∠C, ∠B ≅ ∠D
y°
R
x°
S
2. Definition of
parallelogram
3. ∠Q and ∠P are
supplementary.
3. Consecutive Interior
Angles Theorem
(Thm. 3.4)
4. x° + y° = 180°
4. Definition of
supplementary angles
Geometry
Worked-Out Solutions
243
Chapter 7
39.
y + 14
Q
P
43. m∠USV + m∠TSU = m∠TUV
(x2)° + 32° = 12x°
x2
−2x + 37
x−5
− 12x + 32 = 0
(x − 8)(x − 4) = 0
(x − 8) = 0
M
4y + 5
N
MQ = NP
x=8
m∠USV = (x2)° = (82)° = 64°
(x − 4) = 0
–2x + 37 = x – 5
x=4
–3x = −42
m∠USV = (x2)° = (42)° = 16°
x = 14
⋅
MQ = –2 14 + 37 = –28 + 37 = 9
C
44. yes;
NP = 14 – 5 = 9
y + 14 = 4y + 5
C′
–3y = –9
y=3
QP = 3 + 14 = 17
⋅
MN = 4 3 + 5 = 12 + 5 = 17
The perimeter of MNPQ is 17 + 9 + 17 + 9 = 52 units.
40. — = —
M
N
3x
4x
A
D
P
14x = 28
41. no; Two parallelograms with congruent corresponding sides
may or may not have congruent corresponding angles.
42. a. decreases; Because ∠P and ∠Q are supplementary, as
m∠Q increases, m∠P must decrease so that their total is
still 180°.
2
4
6 x
2
4 − 6 −2
— =—
= — = −—
Slope of YX
3
6−3
3
Starting at W, go down 2 units and right 3 units. So, the
coordinates of E are (4, 0).
6
y
X
4
2
c. The mirror gets closer to the wall; As m∠Q decreases,
Geometry
Worked-Out Solutions
W
Let E be the fourth vertex.
skinnier, which means that Q and S get farther apart and
QS increases.
244
Y
−2
b. increases; As m∠Q decreases, the parallelogram gets
the parallelograms get skinnier, which means that P,
R, and the other corresponding vertices all get closer
together. So, the distance between the mirror and the
wall gets smaller.
X
4
−2
x=2
LM = 4 2 = 8 units
y
2
4x + 3x + 4x + 3x = 28
⋅
B
Let D be the fourth vertex.
2
2
6−4
—=—
= — = −—
Slope of XY
3
3 − 6 −3
Starting at W, go up 2 units and left 3 units. So, the
coordinates of D are (−2, 4).
4x
3x
F
Any triangle, such as △ABC, can be partitioned into four
congruent triangles by drawing the midsegment triangle,
such as △DEF. Then, one triangle, such as △CDE, can
be rotated 180° about a vertex, such as D, to create a
parallelogram as shown.
6
L
E
45. Three parallelograms can be created.
4x
3x
LM
MN
D
E′
QP = MN
Y
W
E
−2
2
4
6 x
−2
Chapter 7
Let F be the fourth vertex.
6−2 4
—=—
=—=2
Slope of XW
3−1 2
Starting at Y, go up 4 units and right 2 units. So, the
coordinates of F are (8, 8).
8
y
12. Subtraction Property
of Equality
13. ∠EPF is a right angle.
13. Definition of right
angle
— —
14. EK ⊥ FJ
14. Definition of
perpendicular lines
F
X
6
— ≅ KM
—
Prove HK
Y
— and MQ
—, such that KP
— GJ
— and MQ
— JL
—, thus
Construct KP
KPGJ is a parallelogram and MQJL is a parallelogram.
W
2
— —
47. Given ⃖⃗
GH ⃖⃗
JK ⃖⃗
LM, GJ ≅ JL
4
2
12. m∠EPF = 90°
4
6
8 x
—
— bisects ∠EFG,
FJ
and EFGH is
a parallelogram.
— ⊥ FJ
—
Prove EK
STATEMENTS
—
1. EK bisects ∠FEH
— bisects
and FJ
∠EFG. EFGH is a
parallelogram.
F
E
46. Given EK bisects ∠FEH,
G
J
P
H
J
K
L
P
Q
H
K
M
G
STATEMENTS
REASONS
1. Given
2. m∠PEH = m∠PEF,
m∠PFE = m∠PFG
2. Definition of angle
bisector
3. m∠HEF = m∠PEH
+ m∠PEF,
m∠EFG = m∠PFE +
m∠PFG
Postulate (Post. 1.4)
4. m∠HEF = m∠PEF +
m∠PEF, m∠EFG =
m∠PFE + m∠PFE
4. Substitution Property
of Equality
5. m∠HEF =
2(m∠PEF),
m∠EFG = 2(m∠PFE)
5. Distributive Property
6. m∠HEF + m∠EFG
= 180°
1.
2.
—≅—
⃖⃗
GH ⃖⃗
JK ⃖⃗
LM, GJ
JL
— and QM
—
Construct PK
— GL
—.
⃖⃗ QM
such that PK
REASONS
1. Given
2. Construction
3. GPKJ and JQML are
parallelograms.
3. Definition of
parallelogram
4. ∠GHK ≅ ∠JKM,
∠PKQ ≅ ∠QML
4. Corresponding
Angles Theorem
(Thm. 3.1)
— —— —
5. GJ ≅ PK , JL ≅ QM
— —
6. PK ≅ QM
5. Parallelogram
Opposite Sides
Theorem (Thm. 7.3)
6. Transitive Property of
Congruence
(Thm. 2.1)
7. ∠HPK ≅ ∠PKQ,
∠KQM ≅ ∠QML
6. Parallelogram
Consecutive Angles
Theorem (Thm. 7.5)
7. Alternate Interior
Angles Theorem
(Thm. 3.2)
8. ∠HPK ≅ ∠QML
7. 2(m∠PEF) +
2(m∠PFE) = 180°
7. Substitution Property
of Equality
8. Transitive Property of
Congruence
(Thm. 2.2)
9. ∠HPK ≅ ∠KQM
8. 2(m∠PEF + m∠PFE)
= 180°
8. Distributive Property
9. Transitive Property of
Congruence
(Thm. 2.2)
9. m∠PEF + m∠PFE
= 90°
9. Division Property of
Equality
10. △PHK ≅ △QKM
10. AAS Congruence
Theorem (Thm. 5.11)
10. m∠PEF + m∠PFE +
m∠EPF = 180°
10. Triangle Sum Theorem
(Thm. 5.1)
11. 90° + m∠EPF = 180°
11. Substitution Property
of Equality
— —
11. HK ≅ KM
11. Corresponding sides
of congruent triangles
are congruent.
Geometry
Worked-Out Solutions
245
Chapter 7
Maintaining Mathematical Proficiency
48. yes; ℓ m by the Alternate Interior Angles Converse
(Thm. 3.6).
49. yes;ℓ m by the Alternate Exterior Angles Converse
(Thm. 3.7).
2. By the Parallelogram Opposite Angles Converse (Thm. 7.8), if
both pairs of opposite angles of a quadrilateral are congruent,
then the quadrilateral is a parallelogram. So, solve
3x − 32 = 2x for x and 4y = y + 87 for y.
3x − 32 = 2x
−32 = −x
x = 32
50. no; By the Consecutive Interior Angles Converse (Thm. 3.8),
consecutive interior angles need to be supplementary for
lines to be parallel, and the consecutive interior angles are
not supplementary.
7.3 Explorations (p. 379)
1. a. Check students’ work.
—
4
4
7−3
−8 − (−12) −8 + 12 4
4
4
3 − (−1)
—=—
=—=—=1
−5 − (−9) −5 + 9 4
4
4
4
3 − (−1)
— = ——
= — = — = −—
Slope of AB
−12 − (−9) −12 + 9 −3
3
4
4
4
7−3
—
Slope of CD = — = — = — = −—
−8 − (−5) −8 + 5 −3
3
b. yes; Slope of BC = —— = — = — = 1
— equals the slope of AD
—, BC
—.
Because the slope of BC
—
—
—
—
Because the slope of AB equals the slope of CD , AB CD .
So, quadrilateral ABCD is a parallelogram.
c. Check students’ work. If the opposite sides of a
parallelogram.
d. If a quadrilateral is a parallelogram, then its opposite
sides are congruent. The converse is true. This is the
Parallelogram Opposite Sides Theorem (Thm. 7.3).
2. a. Check students’ work.
b. Yes, the quadrilateral is a parallelogram. The opposite
sides have the same slope, so they are parallel.
c. Check students’ work. If the opposite angles of a
parallelogram.
d. If a quadrilateral is a parallelogram, then its opposite
angles are congruent. The converse is true. This is the
Parallelogram Opposite Angles Theorem (Thm. 7.4).
3. To prove a quadrilateral is a parallelogram, show that the
opposite sides are congruent or that the opposite angles are
congruent.
4. In the figure, m∠A = m∠C and ∠B ≅ ∠D. Because the
opposite angles are congruent, you can conclude that ABCD
is a parallelogram.
7.3 Monitoring Progress (pp. 381–384)
1. WXYZ is a parallelogram because opposite angles are
congruent; ∠W ≅ ∠Y and ∠X ≅ ∠Z. So, m∠Z = 138°.
246
Geometry
Worked-Out Solutions
y + 87 = 4y
87 = 3y
29 = y
So, x = 32 and y = 29.
3. Opposite Sides Parallel and Congruent Theorem (Thm. 7.9)
4. Parallelogram Opposite Sides Converse (Thm. 7.7)
5. Parallelogram Opposite Angles Converse (Thm. 7.8)
6. By the Parallelogram Diagonals Converse (Thm. 7.10), if
the diagonals of a quadrilateral bisect each other, then the
quadrilateral is a parallelogram. So, solve 2x = 10 − 3x for x.
2x = 10 − 3x
5x = 10
x=2
So, x = 2.
—
−3 − 3 −6
3−2
1
−6
−6
−5 − 1
—
Slope of JK = — = — = — = −6
−3 − (−4) −3 + 4
1
7. Slope of LM = — = — = −6
— equals the slope of JK
—, LM
— JK
—.
Because the slope of LM
——
——
—
LM = √ (3 − 2)2 + (−3 − 3)2 = √ (1)2 + (−6)2 = √37
———
——
—
JK = √(−3−(−4))2 + (−5 − 1)2 = √ (1)2 + (−6)2 = √ 37
— ≅ JK
—.
Because LM = JK = √ 37 , LM
—
— and LM
— are congruent and parallel, which means that
So, JK
JKLM is a parallelogram by the Opposite Sides Parallel and
Congruent Theorem (Thm. 7.9).
8. Sample answer: Find the slopes of all four sides and show
that opposite sides are parallel. Another way is to find the
point of intersection of the diagonals and show that the
diagonals bisect each other.
7.3 Exercises (pp. 385–388)
Vocabulary and Core Concept Check
1. yes; If all four sides are congruent, then both pairs of opposite
sides are congruent. So, the quadrilateral is a parallelogram by
the Parallelogram Opposite Sides Converse (Thm. 7.7).
Chapter 7
2. The statement that is different is “Construct a quadrilateral
14. By the Opposite Sides Parallel and Congruent Theorem
with one pair of parallel sides”.
(Thm. 7.9):
2x + 3 = x + 7
x+3=7
x=4
So, x = 4.
Monitoring Progress and Modeling with Mathematics
3. Parallelogram Opposite Angles Converse (Thm. 7.8)
15. By the Opposite Sides Parallel and Congruent Theorem
(Thm. 7.9):
4. Parallelogram Opposite Sides Converse (Thm. 7.7)
3x + 5 = 5x − 9
5. Parallelogram Diagonals Converse (Thm. 7.10)
−2x + 5 = −9
−2x = −14
6. Parallelogram Opposite Angles Converse (Thm. 7.8)
x=7
7. Opposite Sides Parallel and Congruent Theorem (Thm. 7.9)
8. Parallelogram Diagonals Converse (Thm. 7.10)
So, x = 7.
16. By the Parallelogram Diagonals Converse (Thm. 7.10):
6x = 3x + 2
9. x = 114 and y = 66 by the Parallelogram Opposite Angles
3x = 2
2
x=—
3
Converse (Thm. 7.8).
10. x = 16 and y = 9, by the Parallelogram Opposite Sides
Converse (Thm. 7.7).
11. By the Parallelogram Opposite Sides Converse (Thm. 7.7):
4x + 6 = 7x − 3
6 = 3x − 3
2
So, x = —.
3
17.
10
8
6
4
9 = 3x
A
3=x
−2
4y − 3 = 3y + 1
y−3=1
y=4
So, x = 3 and y = 4.
12. By the Parallelogram Opposite Angles Converse (Thm. 7.8):
(4x + 13)° = (5x − 12)°
13 = x − 12
25 = x
(4y + 7)° = (3x − 8)°
⋅
4y + 7 = 3 25 − 8
4y + 7 = 75 − 8
y
B
C
D
2 4 6 8 10 12 x
−4
0
4−4
—=—
=—=0
Slope of BC
12 − 4 8
1−1 0
—=—
=—=0
8−0 8
— equals the slope of AD
—, so BC
—.
The slope of BC
——
—
——
—
—
BC = √ (12 − 4)2 + (4 − 4)2 = √ (8)2 + (0)2 = √64 = 8
—
√(8 − 0)2 + (1 − 1)2 = √(8)2 + (0)2 = √64 = 8
—. BC
— are opposite
Because BC = AD = 8, BC
sides that are both congruent and parallel. So, ABCD is a
parallelogram by the Opposite Sides Parallel and Congruent
Theorem (Thm. 7.9).
4y = 60
y = 15
So, x = 25 and y = 15.
13. By the Parallelogram Diagonals Converse (Thm. 7.10):
4x + 2 = 5x − 6
2=x−6
8=x
So, x = 8.
Geometry
Worked-Out Solutions
247
Chapter 7
18.
20.
y
5
4
3
F
−5 −4 −3
2
1
N
−5
−1
−3 −2 −1
1 2 3 4 5 x
−3
−4
−5
R
H
4
4−0
—=—
= — = undefined
Slope of EF
−3 − (−3) 0
−5 − (−1) −4
—=—
= — = undefined
Slope of GH
3−3
0
— equals the slope of GH
—, so EF
— GH
—.
The slope of EF
———
—
—
EF = √( −3 − (−3) )2 + (4 − 0)2 = √ 02 + 42 = √16 = 4
———
——
GH = √(3 − 3)2 + (−5 − (−1))2 = √ (0)2 + (−4)2
—
= √ 16 = 4
— ≅ GH
—. EF
— and GH
— are opposite
Because EF = GH = 4, EF
sides that are both congruent and parallel. So, EFGH is a
parallelogram by the Opposite Sides Parallel and Congruent
Theorem (Thm. 7.9).
8
K
y
L
4
2
J
−8 −6 −4 −2
3 x
−4
4
4−0
—=—
=—
Slope of NP
0 − (−5) 5
4
0−4
—=—
= −—
Slope of PQ
3−0
3
−4 − 0 −4 4
—=—
=—=—
Slope of QR
−2 − 3 −5 5
−4
4
−4 − 0
—=—
= — = −—
Slope of NR
−2 − (−5) −2 + 5
3
— equals the slope of QR
—, so NP
— QR
—.
The slope of NP
— equals the slope of NR
—, so PQ
— NR
—.
The slope of PQ
Because both pairs of opposite sides are parallel, NPQR
is a parallelogram by definition.
21. In order to be a parallelogram, the quadrilateral must
have two pairs of opposite sides that are congruent, not
consecutive sides. DEFG is not a parallelogram.
22. In order to determine that JKLM is a parallelogram by the
Opposite Sides Parallel and Congruent Theorem (Thm. 7.9),
— KL
—. There is not enough
you would need to know that JM
information provided to determine whether JKLM is a
parallelogram.
M
2 4 6 8 x
−4
−6
−8
23. The diagonals must bisect each other, so solve for x using
either 2x + 1 = x + 6 or 4x − 2 = 3x + 3. Also, the
opposite sides must be congruent, so solve for x using either
3x + 1 = 4x − 4 or 3x + 10 = 5x.
−1
1
6−7
—=—
= — = −—
Slope of KL
3 − (−5) 3 + 5
8
−1
1
2−3
—
Slope of JM = — = — = −—
6 − (−2) 6 + 2
8
2x + 1 = x + 6
— equals the slope of LM
—, so JK
— LM
—.
The slope of JK
———
——
JK = √(−5 − (−2))2 + (7 − 3)2 = √ (−5 + 2)2 + 42
—
Q
1
−2
G
19.
y
3
1
E
P
—
= √9 + 16 = √ 25 = 5
——
——
—
LM = √(6 − 3)2 + (2 − 6)2 = √(3)2 + (−4)2 = √9 + 16
—
= √25 = 5
— ≅ LM
—. JK
— and LM
— are opposite
Because JK = LM = 5, JK
sides that are both congruent and parallel. So, JKLM is a
parallelogram by the Opposite Sides Parallel and Congruent
Theorem (Thm. 7.9).
x+1=6
x=5
4x − 2 = 3x + 3
x−2=3
x=5
3x + 10 = 5x
10 = 2x
x=5
3x + 1 = 4x − 4
1=x−4
5=x
So, x = 5.
248
Geometry
Worked-Out Solutions
Chapter 7
24. yes; By the Consecutive Interior Angles Converse (Thm. 3.8),
— ZY
—. Because WX
— and ZY
— are also congruent, WXYZ is a
WX
parallelogram by the Opposite Sides Parallel and Congruent
Theorem (Thm. 7.9).
25. A quadrilateral is a parallelogram if and only if both pairs of
opposite sides are congruent.
26. A quadrilateral is a parallelogram if and only if both pairs of
opposite angles are congruent.
bisect each other.
28. Sample answer: Draw two horizontal segments that are the
same length and connect the endpoints.
−1
by the Parallelogram Opposite Sides Converse (Thm. 7.7).
b. Because m∠JKL = 60°, m∠JML = 60° by the
Parallelogram Opposite Angles Converse (Thm. 7.8).
m∠KJM = 180° − 60° = 120° by the Parallelogram
Consecutive Angles Theorem (Thm. 7.5).
m∠KLM = 120° by the Parallelogram Opposite Angles
Converse (Thm. 7.8).
35. You can use the Alternate Interior Angles Converse
— BC
— and BC
— are both
(Thm. 3.6) to show that AD
congruent and parallel. So, ABCD is a parallelogram by the
Opposite Sides Parallel and Congruent Theorem (Thm. 7.9).
36. You can use the Alternate Interior Angles Converse
— DC
— BC
—. Because both
(Thm. 3.6) to show that AB
pairs of opposite sides are parallel, ABCD is a parallelogram
by definition.
y
5
4
3
2
1
— —
c. Transitive Property of Parallel Lines (Thm. 3.9)
27. A quadrilateral is a parallelogram if and only if the diagonals
6
— —
34. a. Because JK ≅ LM and KL ≅ JM , JKLM is a parallelogram
A
D
B
37. Use the Corresponding Angles Converse to show that
— BC
— and the Alternate Interior Angles Converse to show
— DC
—. So, ABCD is a parallelogram by definition.
that AB
C
1 2 3 4 5 6 7 x
−2
29. Check students’ work. Because the diagonals bisect
each other, this quadrilateral is a parallelogram by the
Parallelogram Diagonals Converse (Thm. 7.10).
— —
— —
30. both; If you show that QR TS and QT RS , then QRST is a
— ≅ TS
— and
parallelogram by definition. If you show that QR
—
—
QT ≅ RS , then QRST is a parallelogram by the Parallelogram
Opposite Sides Converse (Thm. 7.7).
38. By the Parallelogram Opposite Sides Theorem (Thm. 7.3),
— ≅ LK
—. Also, you can use the Linear Pair Postulate
JM
(Thm. 2.8) and the Congruent Supplements Theorem (Thm.
2.4) to show that ∠GJM ≅ ∠HLK. Because ∠JGM and
∠LHK are congruent right angles, you can now state that
△MGJ ≅ △KHL by the AAS Congruence Theorem
(Thm. 5.11).
8
8
33. a. Because m∠AEF = 63° and ∠EAF is a right angle
(m∠EAF = 90°), m∠AFE = 90° − 63° = 27°.
b. From the diagram, ∠AFE ≅ ∠DFG. So, m∠DFG = 27°.
Because m∠FDG = 90°, m∠FGD = 90° − 27° = 63°.
c. From the diagram, ∠AFE ≅ ∠GHC ≅ ∠EHB. So,
m∠GHC = m∠EHB = 27°.
d. yes; ∠HEF ≅ ∠HGF because they both are adjacent
to two congruent angles that together add up to 180°,
and ∠EHG ≅ ∠GFE for the same reason. So, EFGH is
a parallelogram by the Parallelogram Opposite Angles
Converse (Thm. 7.8).
Geometry
Worked-Out Solutions
249
Chapter 7
39. Given ∠A ≅ ∠C, ∠B ≅ ∠D
B
C
—
A
Prove JKLM is a parallelogram.
D
REASONS
1. ∠A ≅ ∠C, ∠B ≅ ∠D
1. Given
L
P
J
STATEMENTS
REASONS
1.
1. Given
— and KM
—
Diagonals JL
M
bisect each other.
2. Let m∠A = m∠C =
x° and m∠B = m∠D
= y°.
2. Definition of
congruent angles
— —— —
2. JP ≅ LP , KP ≅ MP
3. m∠A + m∠B + m∠C
+ m∠D = x° + y° +
x° + y° = 360°
3. Corollary to the
Polygon Interior
Angles Theorem
(Cor. 7.1)
3. ∠KPL ≅ ∠MPJ
3. Vertical Angles
Congruence Theorem
(Thm. 2.6)
4. 2(x°) + 2(y°) = 360°
4. △KPL ≅ △MPJ
4. Simplify.
4. SAS Congruence Theorem
(Thm. 5.5)
5. 2(x° + y°) = 360°
5. Distributive Property
6. x° + y° = 180°
6. Division Property of
Equality
5. ∠MKL ≅ ∠KMJ,
— ≅ MJ
—
KL
5. Corresponding parts of
congruent triangles are
congruent.
7. m∠A + m∠B = 180°,
m∠A + m∠D = 180°
7. Substitution Property
of Equality
8. ∠A and ∠B are
supplementary. ∠A and
∠D are supplementary.
8. Definition of
supplementary angles
— —— —
9. BC AD , AB DC
9. Consecutive Interior
Angles Converse
(Thm. 3.8)
10. ABCD is a
parallelogram.
— —— —
Q
R
P
STATEMENTS
REASONS
1.
1. Given
2. ∠RQS ≅ ∠PSQ
3.
— ≅ QS
—
QS
S
2. Alternate Interior Angles
Theorem (Thm. 3.2)
4. SAS Congruence Theorem
(Thm. 5.5)
5. ∠QSR ≅ ∠SQP
5. Corresponding parts of
congruent triangles are
congruent.
7. PQRS is a
parallelogram.
6. Alternate Interior Angles
Converse (Thm. 3.6)
7. JKLM is a
parallelogram.
7. Opposite Sides Parallel
and Congruent Theorem
(Thm. 7.9)
42. Given DEBF is a parallelogram. AE = CF
Prove ABCD is a parallelogram.
A
D
F
E
B
C
REASONS
1. DEBF is a parallelogram,
AE = CF
— —— —
2. DE ≅ BF , FD ≅ EB
6. Alternate Interior Angles
Converse (Thm. 3.6)
7. Definition of parallelogram
Geometry
Worked-Out Solutions
1. Given
2. Parallelogram
Opposite Sides
Theorem (Thm. 7.3)
3. ∠DFB ≅ ∠DEB
3. Parallelogram
Opposite Angles
Theorem (Thm. 7.4)
4. ∠AED and ∠DEB form
a linear pair. ∠CFB and
∠DFB form a linear pair.
4. Definition of linear
pair
5. ∠AED and ∠DEB
are supplementary.
∠CFB and ∠DFB are
supplementary.
5. Linear Pair Postulate
(Post. 2.8)
6. ∠AED ≅ ∠CFB
6. Congruent
Supplements
Theorem (Thm. 2.4)
3. Reflexive Property of
Conguence (Thm. 2.1)
4. △PQS ≅ △RSQ
— —
6. QP RS
2. Definition of segment
bisector
STATEMENTS
Prove PQRS is a parallelogram.
— PS
—, QR
— ≅ PS
—
QR
— —
6. KL MJ
10. Definition of
parallelogram
40. Given QR PS , QR ≅ PS
250
K
bisect each other.
Prove ABCD is a parallelogram.
STATEMENTS
—
41. Given Diagonals JL and KM
Chapter 7
— —
7. AE ≅ CF
8. △AED ≅ △CFB
— —
7. Definition of
congruent segments
9. Corresponding
parts of congruent
triangles are
congruent.
Postulate (Post. 1.2)
11. FD = EB
11. Definition of
congruent segments
12. AB = CF + FD
12. Substitution
Property of Equality
13. AB = DC
13. Transitive Property
of Equality
15. ABCD is a parallelogram.
A
B
8. SAS Congruence
Theorem (Thm. 5.5)
10. AB = AE + EB,
DC = CF + FD
— —
14. AB ≅ DC
46. Draw the first parallelogram:
14. Definition of
congruent segments
D
C
A
B
D
C
Then rotate ABCD −120° about B:
A
B
D
C
47. Converse of the Parallelogram Consecutive Angles Theorem:
If every pair of consecutive angles of a quadrilateral are
supplementary, then the quadrilateral is a parallelogram.
15. Parallelogram
Opposite Sides
Converse (Thm. 7.7)
B
A
43. no; The fourth angle will be 113° because of the Corollary
to the Polygon Interior Angles Theorem (Cor. 7.1), but these
could also be the angle measures of an isosceles trapezoid
with base angles that are each 67°.
44. The segments that remain parallel as the stand is folded are
— EF
— BC
—, AE
— DF
—, and BE
— CF
—.
45. By the Parallelogram Opposite Sides Theorem (Thm. 7.3),
— ≅ CD
—. Also, ∠ABE and ∠CDF are congruent alternate
AB
— and CD
—. Then, you
interior angles of parallel segments AB
can use the Segment Addition Postulate (Post. 1.2), the
Substitution Property of Equality, and the Reflexive Property
— ≅ BE
—. So,
of Congruence (Thm. 2.1) to show that DF
△ABE ≅ △CDF by the SAS Congruence Theorem
(Thm. 5.5), which means that AE = CF = 8 because
corresponding parts of congruent triangles are congruent.
C
D
In ABCD, you are given that ∠A and ∠B are supplementary,
and ∠B and ∠C are supplementary. So, m∠A = m∠C.
Also, ∠B and ∠C are supplementary, and ∠C and ∠D
are supplementary. So, m∠B = m∠D. So, ABCD is a
parallelogram by the Parallelogram Opposite Angles
Converse (Thm. 7.8).
48. Given ABCD is a parallelogram and ∠A is a right angle.
Prove ∠B, ∠C, and ∠D are right angles.
y
D(0, d)
C(b, d)
B(b, 0)
A(0, 0)
x
By the definition of a right angle, m∠A = 90°. Because
ABCD is a parallelogram, and opposite angles of a
parallelogram are congruent, m∠A = m∠C = 90°. Because
consecutive angles of a parallelogram are supplementary,
∠C and ∠B are supplementary, and ∠C and ∠D are
supplementary. So, 90° + m∠B = 180° and 90° + m∠D =
180°. This gives you m∠B = m∠D = 90°. So, ∠B, ∠C,
and ∠D are right angles.
Geometry
Worked-Out Solutions
251
Chapter 7
49. Given quadrilateral ABCD with midpoints E, F, G, and H
that are joined to form a quadrilateral, you can construct
—. Then FG
— is a midsegment of △BCD, and EH
—
diagonal BD
is a midsegment of △DAB. So, by the Triangle Midsegment
— BD
—, FG =—1BD, EH
— BD
—, and
Theorem (Thm. 6.8), FG
2
1
EH = —2BD. So, by the Transitive Property of Parallel Lines
— FG
— and by the Transitive Property of
(Thm. 3.9), EH
Equality, EH = FG. Because one pair of opposite sides is
both congruent and parallel, EFGH is a parallelogram by the
Opposite Sides Parallel and Congruent Theorem (Thm. 7.9).
B
F
E
parallelogram are congruent; yes; You could start by setting
the two parts of either diagonal equal to each other by the
Parallelogram Diagonals Theorem (Thm. 7.6).
7.1–7.3 Quiz (p. 390)
1. 115° + 95° + 70° + x° = 360°
280 + x = 360
x = 80
2. (5 − 2)
C
⋅ 180° = 3 ⋅ 180° = 540°
60° + 120° + 150° + 75° + x° = 540°
405 + x = 540
G
A
H
x = 135
D
—
—
△EBC, and FJ is a midsegment of △EAD. So, by the
— BC
—,
Triangle Midsegment Theorem (Thm. 6.8), GH
1
1
—
—
GH = —BC, FJ AD , and FJ = — AD. Also, by the
3. x° + 60° + 30° + 72° + 46° + 55° = 360°
Parallelogram Opposite Sides Theorem (Thm. 7.3) and the
— are congruent and
definition of a parallelogram, BC
parallel. So, by the Transitive Property of Parallel Lines
— FJ
— GH
— BC
— and by the Transitive
Property of Equality, —12 BC = GH = FJ = —12 AD. Because
one pair of opposite sides is both congruent and parallel,
FGHJ is a parallelogram by the Opposite Sides Parallel and
Congruent Theorem (Thm. 7.9).
4. Interior angle = —— = —
50. Based on the given information, GH is a midsegment of
2
2
51. The quadrilateral is a parallelogram by the definition of a
parallelogram (a quadrilateral with both pairs of opposite
sides parallel).
52. The quadrilateral is a rectangle by the definition of a
rectangle (a quadrilateral with four right angles).
53. The quadrilateral is a square by the definition of a square
(a quadrilateral with four right angles and four congruent
sides).
54. The quadrilateral is a rhombus by the definition of a rhombus
(a quadrilateral with four congruent sides).
7.1–7.3 What Did You Learn? (p. 389)
1. The relationship between the 540° increase and the answer is
⋅
that the interior angle value added is 540° or 3 180°.
2. By the Parallelogram Diagonals Theorem (Thm. 7.6), the
diagonals of a parallelogram bisect each other. So, the
diagonals will have the same midpoint, and it will also be
the point where the diagonals intersect. Therefore, with
any parallelogram, you can find the midpoint of either
diagonal, and it will be the coordinates of the intersection
of the diagonals; Instead of this method, you could also
find the equations of the lines that define each diagonal,
set them equal to each other and solve for the values of the
coordinates where the lines intersect.
Geometry
Worked-Out Solutions
x + 263 = 360
x = 97
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
(10 − 2) 180° 8 180°
10
10
1440°
= — = 144°
10
360°
Exterior angle = — = 36°
10
In a regular decagon, the measure of each interior angle is
144° and the measure of each exterior angle is 36°.
(15 − 2) 180° 13 180°
15
15
2340°
= — = 156°
15
360°
Exterior angle = — = 24°
15
In a regular 15-gon, the measure of each interior angle is
156° and the measure of each exterior angle is 24°.
5. Interior angle = —— = —
Maintaining Mathematical Proficiency
252
3. Sample answer: Solve 5x = 3x + 10; opposite sides of a
(24 − 2) 180° 22 180°
24
24
3960°
= — = 165°
24
360°
Exterior angle = — = 15°
24
In a regular 24-gon, the measure of each interior angle is
165° and the measure of each exterior angle is 15°.
6. Interior angle = —— = —
(60 − 2) 180° 58 180°
60
60
10,440°
= — = 174°
60
360°
Exterior angle = — = 6°
60
In a regular 60-gon, the measure of each interior angle is
174° and the measure of each exterior angle is 6°.
7. Interior angle = —— = —
8. CD = 16; By the Parallelogram Opposite Sides Theorem
(Thm. 7.3), AB = CD.
Chapter 7
9. AD = 7; By the Parallelogram Opposite Sides Theorem
20.
W
7
6
10. AE = 7; By the Parallelogram Diagonals Theorem
4
3
(Thm. 7.6), AE = EC.
⋅ 10.2 = 20.4; By the Parallelogram Diagonals
Theorem (Thm. 7.6), BE = ED.
11. BD = 2
13. By the Parallelogram Consecutive Angles Theorem
(Thm. 7.5), ∠DAB and ∠ABC are supplementary. So,
m∠ABC = 180° − 120° = 60°.
14. By the Parallelogram Consecutive Angles Theorem
(Thm. 7.5), ∠DAB and ∠ADC are supplementary. So,
m∠ADC = 180° − 120° = 60°.
15. m∠ADB = 43°; By the Alternate Interior Angles Theorem
16. The quadrilateral is a parallelogram by the Opposite Sides
Parallel and Congruent Theorem (Thm. 7.9).
17. The quadrilateral is a parallelogram by the Parallelogram
Diagonals Converse (Thm. 7.10).
Z
−6 −5 −4
−1
Opposite Angles Converse (Thm. 7.8).
3
y
2
1
−7 −6 −5 −4 −3 −2−1
−3
Y
−4
−4
2
3−7
—=—
= — = — = −—
Slope of WX
3 − (−3) 3 + 3
6
3
−3 − 3 −6
—
Slope of XY = — = — = 3
1−3
−2
4
2
1 − (−3) 1 + 3
—
Slope of YZ = — = — = — = −—
−5 − 1
−6
−6
3
−6
−6
1
−
7
—=—=—=—=3
Slope of WZ
−5 − (−3) −5 + 3 −2
— equals the slope of YZ
—, WX
— YZ
— and
Because the slope of WX
—
—
—
—
because the slope of XY equals the slope of WZ , XY WZ .
Because both pairs of opposite sides are parallel, WXYZ
is a parallelogram by definition.
21. a. The stop sign is a regular octagon.
⋅
⋅
(8 − 2) 180° 6 180° 1080°
8
8
8
The measure of each interior angle is 135°.
360°
— = 45°
8
The measure of each exterior angle is 45°.
— —
— ≅ KL
— by the Parallelogram Opposite Sides
(Thm. 7.3). JM
Theorem (Thm. 7.3). ∠J ≅ ∠KLM by the Parallelogram
Opposite Angles Theorem (Thm. 7.4). ∠M ≅ ∠JKL by
the Parallelogram Opposite Angles Theorem (Thm. 7.4).
1 2 3 x
R
−3
−4
−5
−7
3 4 x
22. a. JK ≅ ML by the Parallelogram Opposite Sides Theorem
Q
T
1
−2
b. —— = — = — = 135°
18. The quadrilateral is a parallelogram by the Parallelogram
19.
X
2
1
12. m∠BCD = 120°; By the Parallelogram Opposite Angles
Theorem (Thm. 7.4), m∠DAB = m∠BCD.
y
— —
b. Because QT RS and QT = RS, QRST is a parallelogram
S
−2 − (−2) −2 + 2 0
—=—
=—=—=0
Slope of QR
3 − (−5)
3+5
8
0
−6 − (−6) −6 + 6
—
Slope of ST = — = — = — = 0
−7 − 1
−8
−8
— equals slope of ST
—, so QR
— ST
—.
The slope of QR
———
QR = √(3 − (−5))2 + (−2 − (−2))2
——
—
—
= √(3 + 5)2 + (−2 + 2)2 = √ 82 = √ 64 = 8
— ≅ ST
—. QR
— and ST
— are opposite
Because QR = ST = 8, QR
sides that are both congruent and parallel. So, QRST is a
parallelogram by the Opposite Sides Parallel and Congruent
Theorem (Thm. 7.9).
by the Opposite Sides Parallel and Congruent Theorem
(Thm. 7.9).
c. ST = 3 feet, because ST = QR by the Parallelogram
Opposite Sides Theorem (Thm. 7.3). m∠QTS = 123°,
because m∠QTS = m∠QRS by the Parallelogram
Opposite Angles Theorem (Thm. 7.4). m∠TQR = 57°,
because ∠TQR and ∠QTS are consecutive interior angles
and they are supplementary. So,
m∠TQR = 180° − 123° = 57°. Because ∠TSR and ∠TQR
are opposite angles by the Parallelogram Opposite Angles
Theorem (Thm. 7.4), m∠TSR = 57°.
Geometry
Worked-Out Solutions
253
Chapter 7
7.4 Explorations (p. 391)
7.4 Monitoring Progress (pp. 393–396)
because by definition, a square has four right angles.
— —
b. Check students’ work.
c.
— —
1. For any square JKLM, it is always true that JK ⊥ KL ,
1. a. Check students’ work.
2. For any rectangle EFGH, it is sometimes true that FG ≅ GH ,
because some rectangles are squares.
D
B
3. The quadrilateral is a square.
A
C
E
d. yes; yes; no; no; Because all points on a circle are the
— ≅ AE
— ≅ AC
—.
same distance from the center, AB
So, the diagonals of quadrilateral BDCE bisect each other,
which means it is a parallelogram by the Parallelogram
Diagonals Converse (Thm. 7.10). Because all 4 angles of
BDCE are right angles, it is a rectangle. BDCE is neither
— and EC
— are not
a rhombus nor a square because BD
—
—
necessarily the same length as BE and DC .
⋅
m∠BCD = 2 ⋅ 61° = 122° because each diagonal of a
rhombus bisects a pair of opposite angles.
29° = 58° because each diagonal of a
rhombus bisects a pair of opposite angles.
5. m∠EDG = 180° − 118° = 62° by the Parallelogram
e. Check students’ work. The quadrilateral formed by
Consecutive Angles Theorem (Thm. 7.5).
62°
m∠1 = — = 31° because each diagonal of a rhombus
2
bisects a pair of opposite angles.
the endpoints of two diameters is a rectangle (and a
parallelogram). In other words, a quadrilateral is a
rectangle if and only if its diagonals are congruent and
bisect each other.
m∠2 = m∠1 = 31° because each diagonal of a rhombus
bisects a pair of opposite angles.
m∠EFG = m∠EDG = 62° because opposite angles
of a parallelogram are congruent, and a rhombus is a
parallelogram.
2. a. Check students’ work.
b.
E
B
m∠3 = 31° because each diagonal of a rhombus bisects a
pair of opposite angles.
C
A
m∠4 = m∠3 = 31° because each diagonal of a rhombus
bisects a pair of opposite angles.
D
c. yes; no; yes; no; Because the diagonals bisect each other,
AEBD is a parallelogram by the Parallelogram Diagonals
Converse (Thm. 7.10). Because EB = BD = AD = AE,
AEBD is a rhombus. AEBD is neither a rectangle nor a
square because its angles are not necessarily right angles.
d. Check students’ work. A quadrilateral is a rhombus if
and only if the diagonals are perpendicular bisectors of
each other.
3. Because rectangles, rhombuses, and squares are all
parallelograms, their diagonals bisect each other by the
Parallelogram Diagonals Theorem (Thm. 7.6). The diagonals
of a rectangle are congruent. The diagonals of a rhombus are
perpendicular. The diagonals of a square are congruent and
perpendicular.
6. no; The quadrilateral might not be a parallelogram.
7.
QS = RT
4x − 15 = 3x + 8
x − 15 = 8
x = 23
Lengths of the diagonals:
⋅
⋅
QS = 4 23 − 15 = 92 − 15 = 77
RT = 3 23 + 8 = 69 + 8 = 77
4. yes; no; yes; no; RSTU is a parallelogram because the
diagonals bisect each other. RSTU is not a rectangle
because the diagonals are not congruent. RSTU is a
rhombus because the diagonals are perpendicular. RSTU is
not a square because the diagonals are not congruent.
5. A rectangle has congruent diagonals that bisect each other.
254
Geometry
Worked-Out Solutions
Chapter 7
—
y
—
7. JL is sometimes congruent to KM . Some rhombuses are
8. P(−5, 2), Q(0, 4), R(2, −1), S(−3, −3)
squares.
Q
J
K
M
L
P
1
−5
−3
3 x
−1
R
−3
S
8. ∠JKM is always congruent to ∠LKM. Each diagonal of a
———
PR = √(−5 −
2)2
+ (2 −
(−1))2
rhombus bisects a pair of opposite angles.
= √ 49 + 9 = √58
———
QS = √(0 −
(−3))2
+ (4 −
—
—
(−3))2
—
—
PQ = √(−5 − 0)2 + (2 − 4)2 = √ 25 + 4 = √29
——
—
—
QR = √(0 − 2)2 + (4 − (−1))2 = √ 4 + 25 = √29
— and QR
— are
Because PQ = QR, the adjacent sides PQ
congruent. So, PQRS is a square, a rectangle, and a rhombus.
1. Another name for an equilateral rectangle is a square.
2. If two consecutive sides of a parallelogram are congruent,
then the parallelogram is also a rhombus.
Monitoring Progress and Modeling with Mathematics
3. ∠L is sometimes congruent to ∠M. Some rhombuses are
squares.
M
L
parallelogram and the opposite angles of a parallelogram
are congruent.
M
K
L
—
—
5. JM is always congruent to KL . By definition, a rhombus is
a parallelogram, and opposite sides of a parallelogram are
congruent.
J
M
K
L
—
—
6. JK is always congruent to KL . By definition, a rhombus is a
parallelogram with four congruent sides.
J
M
9. square; All of the sides are congruent, and all of the angles
are congruent.
10. rectangle; Opposite sides are congruent and the angles
are 90°.
11. rectangle; Opposite sides are parallel and the angles are 90°.
13. A rhombus is a parallelogram with four congruent sides. So,
m∠1 = m∠FEG = 27°, by the Base Angles Theorem
(Thm. 5.6). By the Rhombus Opposite Angles Theorem
(Thm. 7.12), m∠2 = m∠1 = 27°. By the Rhombus
Diagonals Theorem (Thm. 7.11), m∠3 = 90°. By the
Rhombus Opposite Angles Theorem (Thm. 7.12),
m∠4 = m∠FEG = 27°. By the Rhombus Opposite Angles
Theorem (Thm. 7.12), m∠FED = 2 27° = 54°. By the
Parallelogram Consecutive Angles Theorem (Thm. 7.5),
m∠GFE = 180° − m∠FED = 180° − 54° = 126°. So,
m∠5 = m∠6 = 63°, by the Rhombus Opposite Angles
Theorem (Thm. 7.12).
⋅
14. By the Rhombus Diagonals Theorem (Thm. 7.11),
4. ∠K is always congruent to ∠M. A rhombus is a
J
L
are congruent.
Vocabulary and Core Concept Check
K
M
12. rhombus; Opposite angles are congruent and adjacent sides
7.4 Exercises (pp. 397–400)
J
K
= √ 9 + 49 = √58
Because PR = QS, the diagonals are congruent, so the
quadrilateral is either a square or rectangle.
——
J
—
—
m∠1 = 90°. By the Rhombus Opposite Angles Theorem
(Thm. 7.12), m∠EDG = 2 48° = 96°. By the
Parallelogram Consecutive Angles Theorem (Thm. 7.5),
m∠DGF = 180° − m∠EDG = 180° − 96° = 84°. So,
by the Rhombus Opposite Angles Theorem (Thm. 7.12),
m∠2 = m∠3 = 42°. By the Rhombus Opposite Angles
Theorem (Thm. 7.12), m∠4 = 48°. By the definition of a
— GF
—. So, m∠5 = 48°, by the Alternate
parallelogram, DE
Interior Angles Theorem (Thm. 3.2).
⋅
15. By the Parallelogram Consecutive Angles Theorem
(Thm. 7.5), m∠EDG = 180° − 106° = 74°. So, by the
Rhombus Opposite Angles Theorem (Thm. 7.12),
m∠1 = m∠2 = 37°. By the definition of a parallelogram,
— GF
—. So, m∠3 = 37°, by the Alternate Interior Angles
DE
Theorem (Thm. 3.2). By the Rhombus Opposite Angles
Theorem (Thm. 7.12), m∠4 = 37°. By the Parallelogram
Opposite Angles Theorem (Thm. 7.4), m∠5 = 106°.
K
L
Geometry
Worked-Out Solutions
255
Chapter 7
16. By the Rhombus Opposite Angles Theorem (Thm. 7.12),
⋅
m∠EDG = 2 72° = 144°. So, m∠1 = 180° − 144° = 36°,
by the Parallelogram Consecutive Angles Theorem
(Thm. 7.5). By the Triangle Sum Theorem (Thm. 5.1),
m∠1 + m∠2 + 72° = 180°. So, m∠2 + 108° = 180° and
m∠2 = 72°. By the Rhombus Opposite Angles Theorem
(Thm. 7.12), m∠3 = 72°. By the Rhombus Opposite Angles
Theorem (Thm. 7.12), m∠4 = 72°. By the Parallelogram
Opposite Angles Theorem (Thm. 7.4), m∠5 = 36°.
24. The quadrilateral is a rectangle. Opposite sides are congruent
and the angles are 90°.
25.
3x = 9
x=3
⋅
⋅
WY = 6 3 − 7 = 18 − 7 = 11
17. ∠W is always congruent to ∠X. All angles of a rectangle are
XZ = 3 3 + 2 = 9 + 2 = 11
congruent.
W
26.
X
WY = XZ
6x − 7 = 3x + 2
WY = XZ
14x + 10 = 11x + 22
3x = 12
Z
Y
x=4
—
—
18. WX is always congruent to YZ . Opposite sides of a rectangle
⋅
⋅
WY = 14 4 + 10 = 56 + 10 = 66
XZ = 11 4 + 22 = 44 + 22 = 66
are congruent.
W
X
27.
WY = XZ
24x − 8 = −18x + 13
Z
42x = 21
21 1
x=—=—
42 2
Y
—
—
19. WX is sometimes congruent to XY . Some rectangles are
squares.
W
X
Z
Y
—
⋅
⋅
1
WY = 24 — − 8 = 12 − 8 = 4
2
1
XZ = −18 — + 13 = −9 + 13 = 4
2
28.
—
20. WY is always congruent to XZ . The diagonals of a rectangle
are congruent.
W
X
Z
Y
—
—
21. WY is sometimes perpendicular to XZ . Some rectangles are
squares.
W
X
WY = XZ
16x + 2 = 36x − 6
−20x = −8
2
−8
x=—=—
−20 5
⋅
⋅
32 10 42
2
WY = 16 — + 2 = — + — = — = 8.4
5
5
5
5
72 30 42
2
XZ = 36 — − 6 = — − — = — = 8.4
5
5
5
5
29. Quadrilaterals that are equiangular are squares and
rectangles.
30. Quadrilaterals that are equiangular and equilateral are
squares.
Z
Y
22. ∠WXZ is sometimes congruent to ∠YXZ. Some rectangles
are squares.
W
X
31. Quadrilaterals where the diagonals are perpendicular are
squares and rhombuses.
32. Opposite sides are congruent are true for all parallelograms,
rectangles, rhombuses, and squares.
33. Diagonals bisect each other are true for all parallelograms,
rectangles, rhombuses, and squares.
34. Diagonals bisect opposite angles are true for rhombuses and
Z
Y
squares.
23. The quadrilateral is not a rectangle. All four angles are not
congruent.
256
Geometry
Worked-Out Solutions
Chapter 7
35. Diagonals do not necessarily bisect opposite angles of a
55.
J
m∠QSR = 90° − m∠PQS
y
4
rectangle. The sum of the two angles equals 90°.
K
2
x° = 90° − 58°
−4
x = 32
−2
2
−2
M
36. ∠QRP and ∠SQR should be complementary because they
−4
are the two acute angles of a right triangle.
m∠QRP = 90° − m∠SQR
Diagonals:
———
x° = 90° − 37°
———
(Thm. 7.12).
38. m∠AED = 90° by the Rhombus Diagonals Theorem
−2 − 3 −5 5
— =—
Slope of KM
=—=—
−3 − 0 −3 3
— ⊥ KM
—
JL
The diagonals are perpendicular and congruent, so JKLM is a
rectangle, a rhombus, and a square.
(Thm. 7.11).
⋅ 53°
= 180° − 106° = 74°
40. DB = 16 by the Parallelogram Diagonals Theorem
56.
J 8
41. AE = 6 by the Parallelogram Diagonals Theorem (Thm. 7.6).
42. AC = 12 by the Parallelogram Diagonals Theorem
(Thm. 7.6).
46. QP = 5 by the Parallelogram Diagonals Theorem
(Thm. 7.6).
47. RT = 10 by the Parallelogram Diagonals Theorem
(Thm. 7.6).
48. RP = 5 by the Parallelogram Diagonals Theorem (Thm. 7.6).
49. m∠MKN = 90° because the diagonals of a square are
perpendicular.
50. m∠LMK = 45° because the diagonals of a square bisect
opposite angles.
51. m∠LPK = 45° because the diagonals of a square bisect
K
N
−8
4
x
L −4
−8
44. m∠QRT = 34° by the Alternate Interior Angles Theorem
45. m∠SRT = 180° − 34° − 90° = 56°
y
4
M
43. m∠QTR = 90° − 34° = 56°
(Thm. 3.2).
—
KM = √ (0 − (−3))2 + (3 − (−2))2 = √(3)2 + (5)2
—
—
= √9 + 25 = √34
−3
3
−1 − 2
—=—
= — = −—
Slope of JL
1 − (−4)
5
5
37. m∠DAC = 53° by the Rhombus Opposite Angles Theorem
(Thm. 7.6).
——
JL = √ (1 − (−4))2 + (−1 − 2)2 = √(1 + 4)2 + (−3)2
—
—
= √25 + 9 = √34
x = 53
39. m∠ADC = 180° − 2
4 x
L
Diagonals:
———
—
JL = √ (−2 − (−2))2 + (7 − (−3))2 = √(0)2 + (10)2
—
= √100 = 10
——
KM = √ (−11 − 7)2 + (2 − 2)2
——
= √(−18)2 + 02
—
= √324
= 18
7 − (−3)
10
—=—
= ___ = undefined
Slope of JL
0
−2 − (−2)
0
2−2
—=—
=—=0
Slope of MK
−11 − 7 −18
— ⊥ MK
—
JL
The diagonals are perpendicular and not congruent, so JKLM
is a rhombus.
opposite angles.
52. KN = 1 because the diagonals of a square bisect each other.
53. LN = 2 because the diagonals of a square bisect each other.
54. MP = 2 because the diagonals of a square are congruent.
Geometry
Worked-Out Solutions
257
Chapter 7
57.
4
y
y
59.
8
2
M
J
4
L
−4
2
−8
ML = √(−2 −
(−2))2
+ (1 −
(−3))2
—
=√
+
——
———
—
——
—
——
—
—
——
—
—
—
= √64 = 8
9 − (−5) 14
—=—
= — = undefined
Slope of KM
1−1
0
0
2−2
—
—
—
=
=0
Slope of LJ =
−3 − 5 −8
—
—
KM ⊥ LJ
KJ = √(3 − 3)2 + (1 − (−3))2 = √ (0)2 + (4)2 = √ 16 = 4
JM = √(3 − (−2))2 + (1 − 1)2 = √ (5)2 + (0)2 = √ 25 = 5
1 − (−3)
4
—=—
Slope of ML
= — = undefined
−2 − (−2) 0
0
1−1
—=—
=—=0
Slope of JM
3 − (−2) 5
— ⊥ JM
—
ML
The sides are perpendicular and not congruent. So, JKLM is
a rectangle.
K
The diagonals are perpendicular and not congruent, so JKLM
is a rhombus.
60.
y
4
y
J
−2
4
−2
−4
−2
2
x
M
x
M
−2
−4
K
L
2
——
LJ = √(−3 − 5)2 + (2 − 2)2 = √(−8)2 + (0)2
—
= √25 = 5
—
KM = √ (1 − 1)2 + (9 − (−5))2 = √(0)2 + (14)2
(4)2
LK = √(3 − (−2))2 + (−3 − (−3))2 = √ (5)2 + (0)2
4
M
Diagonals:
—
(0)2
8 x
= √196 = 14
= √16 = 4
J
4
−4
———
58.
−4
K
−4
Sides:
J
4 x
−2
L
K
Diagonals:
L
——
——
LJ = √(−1 − 5)2 + (2 − 2)2 = √ (−6)2 + (0)2
—
= √36 = 6
Diagonals:
———
——
MK = √(4 − (−3))2 + (−1 − 2)2 = √ (7)2 + (−3)2
—
—
= √49 + 9 = √58
———
——
JL = √ (−1 − 2)2 + (4 − (−3))2 = √ (−3)2 + (7)2
—
—
= √ 9 + 49 = √58
−1 − 2
—=—
3
= −__
Slope of MK
7
4 − (−3)
4 − (−3)
7
—=—
= −__
Slope of JL
3
−1 − 2
The diagonals are congruent and not perpendicular, so JKLM
is a rectangle.
——
—
KM = √ (2 − 2)2 + (5 − (−1))2 = √(0)2 + (6)2
—
= √36 = 6
2−2
0
—=—
=—=0
Slope of LJ
−1 − 5 −6
5 − (−1) 6
—=—
= — = undefined
Slope of KM
2−2
0
—
—
LJ ⊥ KM
The diagonals are perpendicular and congruent, so JKLM is a
rectangle, a rhombus, and a square.
61. ABCD is a rhombus, because the sides are congruent.
104° + x° = 180°
x = 76
y + 8 = 3y
8 = 2y
4=y
So, x = 76 and y = 4.
258
Geometry
Worked-Out Solutions
Chapter 7
62. PQRS is a square because all four angles are 90° and the
diagonals are perpendicular.
— —
72. Given ABCD is a parallelogram, AC ⊥ BD .
Prove ABCD is a rhombus.
5x° = (3x + 18)°
A
2x = 18
B
x=9
X
2y = 10
D
y=5
So, x = 9 and y = 5.
— — — —
63. a. HBDF is a rhombus because BD ≅ DF ≅ BH ≅ HF .
ACEG is a rectangle because ∠HAB, ∠BCD, ∠DEF, and
∠FGH are right angles.
b. AE = GC and AJ= JE = CJ = JG, because the
diagonals of a rectangle are congruent and bisect
each other.
64.
parallelogram
rectangle
rhombus
—
—
∠QRS, and SQ bisects ∠PSR and ∠RQP.
73. Given PQRS is a parallelogram, PR bisects ∠SPQ and
square
All of the shapes have 4 sides. So, quadrilateral is at the top
of the diagram. Because the rest all have two pairs of parallel
sides, they are all parallelograms. Then, parallelograms with
four right angles make one category (rectangle), while those
with four congruent sides make another (rhombus), and if a
parallelograms is both a rhombus and a rectangle, then it
is a square.
65. A square is always a rhombus. By the Square Corollary
(Cor. 7.4), a square is a rhombus.
66. A rectangle is sometimes a square. A rectangle with four
congruent sides is a square.
67. A rectangle always has congruent diagonals. The diagonals
of a rectangle are congruent by the Rectangle Diagonals
Theorem (Thm. 7.13).
68. The diagonals of a square always bisect its angles. A square
is a rhombus.
69. A rhombus sometimes has four congruent angles. Some
rhombuses are squares.
70. A rectangle sometimes has perpendicular diagonals. Some
rectangles are rhombuses.
71. Measure the diagonals to see if they are congruent. They
—
should be √12.5 ≈ 3.54 meters in length.
C
Because ABCD is a parallelogram, its diagonals bisect each
other by the Parallelogram Diagonals Theorem (Thm. 7.6).
— ≅ DX
— by the definition of segment bisector. Because
So, BX
— ⊥ BD
—, ∠DXC ≅ ∠BXC. By the Reflexive Property of
AC
— ≅ XC
—. So, △BXC ≅ △DXC
Congruence (Thm. 2.1), XC
—≅
by the SAS Congruence Theorem (Thm. 5.5). So, BC
—
DC because corresponding parts of congruent triangles are
— ≅ BC
— and DC
— ≅ AB
— because opposite
sides of a parallelogram are congruent. So, by the Transitive
— ≅ BC
— ≅ DC
—,
Property of Congruence (Thm. 2.1), AB
which means that by the Rhombus Corollary (Cor. 7.2),
ABCD is a rhombus.
Prove PQRS is a rhombus.
Q
P
R
T
S
STATEMENTS
REASONS
1. PQRS is a
—
parallelogram. PR
bisects ∠SPQ and
— bisects
∠QRS. SQ
∠PSR and ∠RQP.
1. Given
2. ∠SRT ≅ ∠QRT,
∠RQT ≅ ∠RST
2. Definition of angle
bisector
— ≅ TR
—
3. TR
3. Reflexive Property of
Congruence (Thm. 2.1)
4. △QRT ≅ △SRT
4. AAS Congruence
Theorem (Thm. 5.11)
— ≅ SR
—
5. QR
— ≅ PS
—, PQ
— ≅ SR
—
6. QR
— ≅ QR
— ≅ SR
— ≅ PQ
—
7. PS
8. PQRS is a rhombus.
5. Corresponding parts of
congruent triangles are
congruent.
6. Parallelogram Opposite
Sides Theorem (Thm. 7.3)
7. Transitive Property of
Congruence (Thm. 2.1)
8. Definition of rhombus
Geometry
Worked-Out Solutions
259
Chapter 7
74. Given WXYZ is a rhombus.
W
— bisects ∠ZWX and ∠XYZ.
Prove WY
— bisects ∠WZY and ∠YXW.
ZX
V
Z
STATEMENTS
REASONS
1. WXYZ is a rhombus.
1. Given
— ≅ XY
— ≅ YZ
— ≅ WZ
—
2. WX
— ≅ XV
—, YV
— ≅ YV
—,
3. XV
—
—
—
—
ZV ≅ ZV , and WV ≅ WV
4. WXYZ is a parallelogram.
— bisects WY
—. WY
— bisects
5. XZ
—.
XZ
— ≅ YV
—, XV
— ≅ ZV
—
6. WV
X
Y
2. Definition of a
rhombus
3. Reflexive Property
of Congruence
(Thm. 2.1)
4. Definition of a
rhombus
5. Parallelogram
Diagonals Theorem
(Thm. 7.6)
6. Definition of
segment bisector
7. △WXV ≅ △YXV ≅ △YZV
≅ △WZV
7. SSS Congurence
Theorem (Thm. 5.8)
8. ∠WXV ≅ ∠YXV,
∠XYV ≅ ∠ZYV,
∠YZV ≅ ∠WZV,
∠ZWV ≅ ∠XWV
8. Corresponding
parts of congruent
triangles are
congruent.
— bisects ∠ZWX and
9. WY
— bisects ∠WZY
∠XYZ. ZX
and ∠YXW.
9. Definition of angle
bisector
75. A diagonal will never divide a square into an equilateral
triangle because the diagonals of a square always create two
right triangles.
The conditional statement is true by the definition of a rhombus.
The converse is true because if a quadrilateral has four
congruent sides, then both pairs of opposite sides are
congruent. So, by the Parallelogram Opposite Sides
Converse (Thm. 7.7), it is a parallelogram with four
congruent sides, which is the definition of a rhombus.
82. Conditional statement: If a quadrilateral is a rectangle, then it
has four right angles.
Converse: If a quadrilateral has four right angles, then it is a
rectangle.
The conditional statement is true by the definition of a
rectangle. The converse is true because if a quadrilateral
has four right angles, then both pairs of opposite angles
are congruent. So, by the Parallelogram Opposite Angles
Converse (Thm. 7.8), it is a parallelogram with four right
angles, which is the definition of a rectangle.
83. Conditional statement: If a quadrilateral is a square, then it is
a rhombus and a rectangle.
Converse: If a quadrilateral is a rhombus and a rectangle,
then it is a square.
The conditional statement is true because if a quadrilateral is a
square, then by the definition of a square, it has four congruent
sides, which makes it a rhombus by the Rhombus Corollary
(Cor. 7.2), and it has four right angles, which makes it a
rectangle by the Rectangle Corollary (Cor. 7.3). The converse
is true because if a quadrilateral is a rhombus and a rectangle,
then by the Rhombus Corollary (Cor. 7.2), it has four
congruent sides, and by the Rectangle Corollary (Cor. 7.3), it
has four right angles. So, by the definition, it is a square.
84. no; If a rhombus is a square, then it is also a rectangle.
85. Given △XYZ ≅ △XWZ,
X
Y
∠XYW ≅ ∠ZWY
Prove WXYZ is a rhombus.
76. The diagonal of a rhombus can divide the rhombus into
two equilateral triangles. If the angles of a rhombus are
60°, 120°, 60°, and 120°, then the diagonal that bisects
the opposite 120° angles will divide the rhombus into two
equilateral triangles.
77. A square can be called a regular quadrilateral because it has
four congruent sides and four congruent angles.
78. Sample answer: You need to know whether the figure is a
parallelogram.
Z
STATEMENTS
REASONS
1. △XYZ ≅ △XWZ,
∠XYW ≅ ∠ZWY
1. Given
2. ∠YXZ ≅ ∠WXZ,
∠YZX ≅ ∠WZX,
— ≅ XW
—, YZ
— ≅ WZ
—
XY
2. Corresponding parts of
congruent triangles are
congruent.
from two rhombuses might not be congruent.
— bisects ∠WXY
3. XZ
and ∠WZY.
All squares are similar because corresponding angles of two
squares are congruent.
4. ∠XWY ≅ ∠XYW,
∠WYZ ≅ ∠ZWY
4. Base Angles Theorem
(Thm. 5.6)
5. ∠XYW ≅ ∠WYZ,
∠XWY ≅ ∠ZWY
5. Transitive Property of
Congruence (Thm. 2.2)
79. All rhombuses are not similar because corresponding angles
80. Because the line connecting a point with its preimage in a
reflection is always perpendicular to the line of reflection,
when a diagonal connecting two vertices is perpendicular to
the other diagonal, both can be a line of symmetry.
81. Conditional statement: If a quadrilateral is a rhombus, then it
has four congruent sides.
Converse: If a quadrilateral has four congruent sides, then it
is a rhombus.
260
W
Geometry
Worked-Out Solutions
3. Definition of angle
bisector
— bisects ∠XWZ and
6. WY
∠XYZ.
7. WXYZ is a rhombus.
6. Definition of angle
bisector
7. Rhombus Opposite Angles
Theorem (Thm. 7.12)
Chapter 7
— —— —
— ⊥ DC
—
86. Given BC ≅ AD , BC ⊥ DC ,
A
B
D
C
Prove ABCD is a rectangle.
88. Given PQRS is a parallelogram. P
Q
— ≅ SQ
—
PR
Prove PQRS is a rectangle.
STATEMENTS
S
REASONS
—, BC
— ⊥ DC
—,
1. BC
—
—
STATEMENTS
1. Given
—
2. BC
REASONS
1. PQRS is a parallelogram,
2. Lines Perpendicular
to a Transversal
Theorem
(Thm. 3.12)
3. ABCD is a parallelogram.
3. Opposite Sides
Parallel and
Congruent Theorem
(Thm. 7.9)
4. m∠DAB = m∠BCD,
4. Parallelogram
Opposite Angles
Theorem (Thm. 7.4)
R
— ≅ SQ
—
PR
— ≅ QR
—
2. PS
1. Given
2. Parallelogram
Opposite Sides
Theorem (Thm. 7.3)
— ≅ PQ
—
3. PQ
3. Reflexive Property
of Congruence
(Thm. 2.1)
4. △PQR ≅ △QPS
4. SSS Congruence
Theorem (Thm. 5.8)
5. ∠SPQ ≅ ∠RQP
5. Corresponding parts
of congruent triangles
are congruent.
5. m∠BCD = m∠ADC = 90°
5. Definition of
perpendicular lines
6. m∠DAB = m∠BCD =
6. Transitive Property
of Equality
6. m∠SPQ = m∠RQP
6. Definition of
congruent angles
7. ABCD has four right angles.
7. Definition of a right
angle
7. m∠SPQ + m∠RQP =
180°
8. ABCD is a rectangle.
8. Definition of a
rectangle
7. Parallelogram
Consecutive Angles
Theorem (Thm. 7.5)
8. 2m∠SPQ = 180° and
2m∠RQP = 180°
8. Substitution Property
of Equality
9. m∠SPQ = 90° and
m∠RQP = 90°
9. Division Property of
Equality
10. m∠RSP = 90° and
m∠QRS = 90°
10. Parallelogram
Opposite Angles
Theorem (Thm. 7.4)
11. ∠SPQ, ∠RQP, ∠RSP,
and ∠QRS are right
angles.
11. Definition of a right
angle
12. PQRS is a rectangle.
12. Definition of a
rectangle
87. Given PQRS is a rectangle.
P
— ≅ SQ
—
Prove PR
S
Q
R
STATEMENTS
REASONS
1. PQRS is a rectangle.
1. Given
2. PQRS is a
parallelogram.
2. Definition of a rectangle
— ≅ QR
—
3. PS
3. Parallelogram Opposite
Sides Theorem (Thm. 7.3)
4. ∠PQR and ∠QPS
are right angles.
4. Definition of a rectangle
5. ∠PQR ≅ ∠QPS
5. Right Angles Congruence
Theorem (Thm. 2.3)
— ≅ PQ
—
6. PQ
7. △PQR ≅ △QPS
— ≅ SQ
—
8. PR
6. Reflexive Property of
Congruence (Thm. 2.1)
7. SAS Congruence
Theorem (Thm. 5.5)
Maintaining Mathematical Proficiency
89. AE = EC
x = 10
1
ED = —CB
2
1
y = — 16 = 8
2
⋅
8. Corresponding parts of
congruent triangles are
congruent.
Geometry
Worked-Out Solutions
261
Chapter 7
1
2
1
7 = —x
2
14 = x
90. DE = —BC
2. If EG = FH, then by the Isosceles Trapezoid Diagonals
Theorem (Thm. 7.16) the trapezoid is an isosceles trapezoid.
3. Because m∠FGH = 110°, m∠GFE = 70° by the
Consecutive Interior Angles Theorem (Thm. 3.4). So,
∠HEF ≅ ∠GFE and the trapezoid is isosceles.
BD = DA
y=6
4. J
N
x=9
1
DE = — AC
2
1
13 = — y
2
26 = y
9 cm
K
12 cm
P
M
L
1 (JK + ML)
NP = __
⋅
2
1 (9 + ML)
12 = __
2
24 = 9 + ML
7.5 Explorations (p. 401)
15 = ML
1. a. Check students’ work.
So, ML = 15 centimeters.
c. If the base angles of a trapezoid are congruent, then the
trapezoid is isosceles.
5. Sample answer: Find the coordinates of Y and Z and
calculate the distance between the points.
2. a. Check students’ work.
b. ∠B ≅ ∠C
c. If a quadrilateral is a kite, then it has exactly one pair of
congruent opposite angles.
3. A trapezoid is a quadrilateral with exactly one pair of
parallel sides. A trapezoid that has congruent base angles is
isosceles. A kite is a quadrilateral with exactly one pair of
congruent opposite angles.
4. yes; When the base angles are congruent, the opposite sides
(
(
) ( )
) ( )
0 + 8 6 + 10
8 16
Midpoint Y = —, — = —, — = (4, 8)
2
2
2 2
2 + 12 2 + 2
14 4
Midpoint Z = —, — = —, — = (7, 2)
2
2
2 2
——
7.5 Monitoring Progress (pp. 402–406)
—
9−6
4 − (−5)
3
9
⋅
—
—
6. 3x° + 75° + 90° + 120° = 360°
3x + 285 = 360
3x = 75
⋅
x = 25
The angles are 3 25 = 75°, 75°, 90°, and 120°.
The value of x is 25. The measure of the congruent angles
is 75°.
1
3
1. Slope of AB = — = — = __
4−2
2 = __
1
—=—
= __
Slope of CD
4 − (−2) 6 3
7. Quadrilateral DEFG could be an isosceles trapezoid,
4
2−6
−4 = −—
— = __________
= ___
3
−2 − (−5)
3
8. Quadrilateral RSTU is a kite because it has two pairs of
parallelogram, rectangle, square, or rhombus.
consecutive congruent sides and the opposite sides are
not congruent.
4 − 9 __
5
— = _____
Slope of CB
= = undefined
4−4 0
———
——
AD = √(−2 − (−5))2 + (2 − 6)2 = √ (−2 + 5)2 + (−4)2
=
—
√32
+ 16 = √9 + 16 = √25 = 5
CB = √(4 −
4)2
+ (4 −
9)2
9. Quadrilateral YVWX is a trapezoid because two sides are
parallel and the diagonals do not bisect each other.
—
—
——
=√
——
(0)2
+
(−5)2
—
= √25 = 5
— equals the slope of DC
—, and the slope of AD
—
The slope of AB
—
is not equal to the slope of BC . Because ABCD has exactly
one pair of parallel sides, it is a trapezoid. Also, AD = BC.
So, ABCD is an isosceles trapezoid.
262
—
= √9 5 = 3√5 units
are also congruent.
5. no; In a kite, only one pair of opposite angles is congruent.
—
YZ = √(4 − 7)2 + (8 − 2)2 = √ (−3)2 + 62 = √9 + 36
Geometry
Worked-Out Solutions
are not sufficient to give it a more specific name.
7.5 Exercises (pp. 407–410)
Vocabulary and Core Concept Check
1. A trapezoid has exactly one pair of parallel sides and a kite
has two pairs of consecutive congruent sides.
Chapter 7
2. The question that is different is “Is there enough information
— ≅ DC
— ?” There is not enough information
to prove that AB
—
—, but there is enough information to
to prove that AB ≅ DC
prove the other three.
Monitoring Progress and Modeling with Mathematics
9−3
6
6
—
3. Slope of YZ = __________ = _______ = __ = undefined
−3 − (−3) −3 + 3 0
8 − 4 __
— = _____
Slope of XW
= 4 = undefined
1−1 0
1
9−8
— = _______
Slope of YX
= −—
−3 − 1
4
3−4
−1 = __
1
— = _______
Slope of ZW
= ___
−3 − 1 −4 4
——
——
YX = √(−3 − 1)2 + (9 − 8)2 = √ (−4)2 + (1)2
—
—
= √16 + 1 = √ 17
——
——
ZW = √(−3 − 1)2 + (3 − 4)2 = √(−4)2 + (−1)2
—
—
= √16 + 1 = √ 17
— equals the slope of XW
—, and the slope of YX
—
The slope of YZ
—
is not equal to the slope of ZW . Because WXYZ has exactly
one pair of parallel sides, it is a trapezoid. Also, YX = ZW.
So, WXYZ is an isosceles trapezoid.
—
3−0
3
3
−3 − (−3) −3 + 3 0
1 − (−4) _____
5
— = ________
Slope of EF
= 1 + 4 = −__
−1 − 1
−2
2
2
3−1
2
— = __________
Slope of DE
= _______
= — = −1
−3 − (−1) −3 + 1 −2
0 − (−4) ___
— = ________
Slope of GF
= 4 = −1
−3 − 1
−4
4. Slope of DG = __________ = _______ = __ = undefined
———
——
DG = √(−3 − (−3))2 + (3 − 0)2 = √ (−3 + 3)2 + (3)2
—
= √9 = 3
———
——
EF = √(−1 − 1)2 + (1 − (−4))2 = √ (−2)2 + (1 + 4)2
—
—
= √4 + 25 = √ 29
— equals the slope of GF
—, and the slope of DG
—
The slope of DE
—
is not equal to the slope of EF . Because DEFG has exactly
one pair of parallel sides, it is a trapezoid. Also, DG ≠ EF.
So, DEFG is not an isosceles trapezoid.
—
4−4
5−0
0
5
5. Slope of NP = _____ = __ = 0
0−0
0
— = ________
Slope of MQ
= ___ = 0
8 − (−2) 10
4−0
4=2
— = ________
Slope of NM
= __
0 − (−2) 2
4−0
4
— = _____
Slope of PQ
= −__
3
5−8
——
—
—
NM = √ (0 − (−2))2 + (4 − 0)2 = √(2)2 + (4)2 = √4 + 16
—
—
= √ 20 = 2√5
——
——
—
PQ = √ (5 − 8)2 + (4 − 0)2 = √(−3)2 + (4)2 = √9 + 16
—
= √ 25 = 5
— is equal to the slope of MQ
—, and the slope
The slope of NP
— is not equal to the slope of PQ
—. Because MNPQ has
of NM
exactly one pair of parallel sides, it is a trapezoid. Also,
NM ≠ PQ. So, MNPQ is not an isosceles trapezoid.
—
9−9
8−1
0
7
6. Slope of HL = _____ = __ = 0
0
2 − 2 = __
— = _____
Slope of JK
=0
5−4 1
9−2
7
— = _____
Slope of HJ
= −__
1−4
3
7
9−2 —
— = _____
Slope of LK
=
8−5 3
——
——
HJ = √ (1 − 4)2 + (9 − 2)2 = √(−3)2 + (7)2
—
—
= √9 + 49 = √58
——
—
LK = √(8 − 5)2 + (9 − 2)2 = √ (3)2 + (7)2
—
—
= √9 + 49 = √ 58
— equals the slope of JK
—, and the slope of HJ
—
The slope of HL
—
is not equal to the slope of LK . Because HJKL has exactly
one pair of parallel sides, it is a trapezoid. Also, HJ = LK.
So, HJKL is an isosceles trapezoid.
7. m∠K + m∠L = 180°
118° + m∠L = 180°
m∠L = 180° − 118° = 62°
Quadrilateral JKLM is isosceles, so m∠J = m∠K = 118°
and m∠M = m∠L = 62°.
8. m∠R + m∠S = 180°
m∠R + 82° = 180°
m∠R = 180° − 82° = 78°
Quadrilateral QRST is isosceles, so m∠S = m∠R = 82° and
m∠Q = m∠T = 98°.
1
2
1 28
MN = __
2
MN = 14
9. MN = __(10 + 18)
⋅
— is 14.
The length of midsegment MN
Geometry
Worked-Out Solutions
263
Chapter 7
1
2
1 133
MN = __
2
MN = 66.5
10. MN = __(76 + 57)
15. m∠G + m∠H + m∠E + m∠F = 360°
x° + 100° + x° + 40° = 360°
⋅
2x + 140 = 360
— is 66.5.
The length of midsegment MN
2x = 220
x = 110
1 (AB + DC)
11. MN = __
So, m∠G = 110°.
2
1 (AB + 10)
7 = __
16. m∠G + m∠H + m∠E + m∠F = 360°
2
x° + 90° + x° + 150° = 360°
14 = AB + 10
2x + 240 = 360
4 = AB
— is 4.
The length of midsegment AB
2x = 120
x = 60
1
2
1 (AB + 11.5)
18.7 = __
2
12. MN = __(AB + DC)
So, m∠G = 60°.
17. m∠G + m∠H + m∠E + m∠F = 360°
x° + 110° + 60° + 110° = 360°
37.4 = AB + 11.5
x + 280 = 360
25.9 = AB
x = 80
— is 25.9.
The length of midsegment AB
13.
D
So, m∠G = 80°.
y
18. m∠G + m∠H + m∠E + m∠F = 360°
x° + 90° + 110° + 90° = 360°
8
4
x + 290 = 360
C
A
12
8
−4
x = 70
x
So, m∠G = 70°.
B
1 (AB + DC), when you solve for DC, you
19. Because MN = __
——
AB = √(2 −
8)2
—
+ (0 −
—
(−4))2
——
=√
—
(−6)2
+
(4)2
= √36 + 16 = √ 52 = 2√13
——
CD = √ (12 −
0)2
—
+ (2 −
10)2
—
——
=√
(12)2
⋅
—
+
(−8)2
—
= √144 + 64 = √ 208 = √16 13 = 4√13
—
—
1 ( 4√—
1 ( 6√—
Midsegment = __
13 + 2√ 13 ) = __
13 ) = 3√13
2
2
should get
2
DC = 2(MN) − AB
= 2(8) − 14
= 16 − 14
= 2.
20. In the kite shown, ∠B ≅ ∠D. Find m∠A by subtracting the
measures of the other three angles from 360°.
14.
y
V
m∠A + m∠B + m∠C + m∠D = 360°
8
m∠A + 120° + 50° + 120° = 360°
S
m∠A = 360° − 50° − 2(120°)
m∠A = 70°
8
12
x
U
21. Quadrilateral JKLM is a rectangle because it is a
T
— —
———
22. Quadrilateral RSPQ is a trapezoid, because PS QR , and
TU = √(3 − (−2))2 + (−2 − (−4))2
——
—
—
= √(3 + 2)2 + (−2 + 4)2 = √25 + 4 = √ 29
———
—
SV = √ (13 − (−2))2 + (10 − 4)2 = √ (15)2 + (6)2
—
—
⋅
—
—
= √ 225 + 36 = √ 261 = √ 9 29 = 3√ 29
—
—
1 ( √—
1 ( 4√—
Midsegment = __
29 + 3√ 29 ) = __
29 ) = 2√29
2
2
264
Geometry
Worked-Out Solutions
∠QPS and ∠PSR are not congruent.
23. Quadrilateral ABCD is a square because it has four congruent
sides and four right angles.
24. Quadrilateral XYZW is a kite because it has two pairs
of consecutive congruent sides and opposite sides are
not congruent.
Chapter 7
25. no; Even though the diagonals are perpendicular, it does not
indicate that the quadrilateral is a rhombus. It could be a kite.
— ——
35. Given JL ≅ LN , KM is a midsegment of △JLN.
Prove Quadrilateral JKMN is an isosceles trapezoid.
26. no; A square has four right angles and the diagonals bisect
L
each other, but this could also describe a rectangle.
1
27. 12.5 = __( 3x + 1 + 15 )
K
M
J
2
25 = 3x + 16
N
STATEMENTS
9 = 3x
— ≅ LN
—, KM
— is a
1. JL
midsegment of △JLN.
3=x
REASONS
1. Given
1
2
1 (5x)
15 = __
2
— JN
—
2. KM
3. JKMN is a trapezoid.
3. Definition of trapezoid
30 = 5x
4. ∠LJN ≅ ∠LNJ
4. Base Angles Theorem
(Thm. 5.6)
5. JKMN is an isosceles
trapezoid.
5. Isosceles Trapezoid Base
Angles Converse
(Thm. 7.15)
28. 15 = __(3x + 2 + 2x − 2)
6=x
1
2
1 (8 + 20)
MQ = __
2
1 (28) = 14
MQ = __
2
29. MQ = __(NP + LR)
2. Triangle Midsegment
Theorem (Thm. 6.8)
— —— —
36. Given ABCD is a kite. AB ≅ CB , AD ≅ CD
— ≅ AE
—
Prove CE
1 (MQ + KS)
LR = __
2
1 (14 + KS)
20 = __
2
40 = 14 + KS
C
B
E
D
A
26 = KS
STATEMENTS
REASONS
The diameter of the bottom layer of the cake is 26 inches.
1. ABCD is a kite,
— ≅ CD
—, AB
— ≅ CB
—
1. Given
30. The length of the stick from X to W is 18 inches, and the
length of the stick from W to Z is 29 inches. A kite is a
quadrilateral that has two pairs of consecutive congruent
sides, but opposite sides are not congruent.
31. For ABCD to be an isosceles trapezoid, ∠A ≅ ∠D or
∠B ≅ ∠C. If a trapezoid has a pair of congruent base
angles, then it is an isosceles trapezoid.
— —
32. Sample answer: For ABCD to be an kite, BC ≅ DC . Then
△ABC ≅ △ADC and ABCD has two pairs of consecutive
congruent sides.
— —
33. Sample answer: For ABCD to be a parallelogram, BE ≅ DE .
Then the diagonals bisect each other.
— —
34. Sample answer: For ABCD to be a square, AB ≅ BC . A
rectangle with a pair of congruent adjacent sides is a square.
— ≅ BD
—, ED
— ≅ ED
—
2. BD
2. Reflexive Property of
Congruence (Thm. 2.1)
3. △ABD ≅ △CBD
3. SSS Congruence Theorem
(Thm. 5.8)
4. Corresponding parts of
congruent triangles are
congruent.
5. △CED ≅ △AED
5. SAS Congruence Theorem
(Thm. 5.5)
— ≅ AE
—
6. CE
6. Corresponding parts of
congruent triangles are
congruent.
37. RSTU is a kite for which S is any point on ⃖⃗
UV such that
UV ≠ SV and SV ≠ 0.
Geometry
Worked-Out Solutions
265
Chapter 7
38.
y
B
−8
40. Given ABCD is a trapezoid.
4
D
−4
C
A
3 − (−13)
16
16
—=—
= _______ = ___
Slope of BC
3
−3 − (−6) −3 + 6
−2 − (−13) −2 + 13 ___
—
___________
Slope of CD =
= — = 11
12
6 − (−6)
6+6
1
2
42. y = __(b1 + b2)
1 (2x + 7 + 2x − 5)
y = __
2
——
AD = √(6 − 4)2 + (−2 − 5)2 = √ (2)2 + (−7)2
1 (4x + 2)
y = __
2
y = 2x + 1
—
= √4 + 49 = √ 53
——
——
AB = √(4 − (−3))2 + (5 − 3)2 = √ (4 + 3)2 + (2)2
—
—
= √72 + 4 = √ 49 + 4 = √ 53
43. a. A
B
D
C
rectangle; The diagonals are
congruent, but not perpendicular.
———
BC = √(−3 − (−6))2 + (3 − (−13))2
—
——
D
41. no; It could be a square or rectangle.
There are no parallel segments.
——
E
—,
Given trapezoid ABCD with ∠A ≅ ∠D and BC
—
—
construct CE parallel to BA . Then, ABCE is a parallelogram
— ≅ EC
—. ∠A ≅ ∠CED by the
by definition, so AB
Corresponding Angles Theorem (Thm. 3.1), so
∠CED ≅ ∠D by the Transitive Property of Congruence
(Thm. 2.2). Then by the Converse of the Base Angles
— ≅ DC
—. So, AB
— ≅ DC
— by the
Theorem (Thm. 5.7), EC
Transitive Property of Congruence (Thm. 2.1), and trapezoid
ABCD is isosceles.
7
— = _______
−2 − 5 = −__
2
6−4
2
— = ________
5 − 3 = __
Slope of AB
4 − (−3) 7
—
C
Prove ABCD is an isosceles trapezoid.
8 x
−12
—
B
—
∠A ≅ ∠D, BC
A
—
= √(−3 + 6)2 + (16)2 = √9 + 256 = √ 265
———
CD = √(6 −
(−6))2
+ (−2 −
(−13))2
——
= √(6 +
6)2
b.
——
+ (−2 + 13)2 = √(12)2 + (11)2
—
A
B
—
= √ 144 + 121 = √265
Consecutive sides are equal. So, by the definition of a kite,
B
39. Given ABCD is an isosceles trapezoid.
—
BC
Prove ∠A ≅ ∠D, ∠B ≅ ∠BCD
Geometry
Worked-Out Solutions
D
44. Given QRST is an isosceles trapezoid.
C
Q
Prove ∠TQS ≅ ∠SRT
A
E
D
—, construct CE
—
Given isosceles trapezoid ABCD with BC
—
parallel to BA . Then, ABCE is a parallelogram by definition,
— ≅ EC
—. Because AB
— ≅ CD
— by the definition of an
so AB
— ≅ CD
— by the Transitive Property
isosceles trapezoid, CE
of Congruence (Thm. 2.1). So, ∠CED ≅ ∠D by the Base
Angles Theorem (Thm. 5.6) and ∠A ≅ ∠CED by the
Corresponding Angles Theorem (Thm. 3.1). So, ∠A ≅ ∠D
by the Transitive Property of Congruence (Thm. 2.2). Next,
by the Consecutive Interior Angles Theorem (Thm. 3.4),
∠B and ∠A are supplementary and so are ∠BCD and ∠D.
So, ∠B ≅ ∠BCD by the Congruent Supplements Theorem
(Thm. 2.4).
266
C
rhombus; The diagonals are
perpendicular, but not congruent.
R
T
S
STATEMENTS
REASONS
1. QRST is an isosceles
trapezoid.
1. Given
2. ∠QTS ≅ ∠RST
2. Isosceles Trapezoid
Base Angles Theorem
(Thm. 7.14)
— ≅ RS
—
3. QT
— ≅ TS
—
4. TS
3. Definition of an isosceles
trapezoid
4. Reflexive Property of
Congruence (Thm. 2.1)
5. △QST ≅ △RTS
5. SAS Congruence
Theorem (Thm. 5.5)
6. ∠TQS ≅ ∠SRT
6. Corresponding parts of
congruent triangles are
congruent.
Chapter 7
—
—
45. a. yes; Because AQ is not parallel to BP , ∠ABX ≅ ∠BAX
— PQ
—.
and AB
b. Because 360° ÷ 12 = 30°, m∠AXB = 30°. Because
180° − 30°
∠ABX ≅ ∠BAX, m∠ABX = m∠BAX = __________
2
= 75°. So, m∠AQP = m∠BPQ = 180° − 75° = 105°.
1
2
1 (PQ + RS)
MN = __
2
1 (PQ + 5PQ)
MN = __
2
1 (6PQ)
MN = __
2
46. A; Midsegment: MN = __ (b1 + b2)
47. Given EFGH is a kite.
F
Prove ∠E ≅ ∠G, ∠F ≠ ∠H
E
Property of Congruence (Thm. 2.1). So, △FGH ≅ △FEH by
the SSS Congruence Theorem (Thm. 5.8), and ∠E ≅ ∠G
because corresponding parts of congruent triangles are
congruent. Next, assume temporarily that ∠F ≅ ∠H. Then
EFGH is a parallelogram by the Parallelogram Opposite
Angles Converse (Thm. 7.8), and opposite sides are
congruent. However, this contradicts the definition of a kite,
which says that opposite sides cannot be congruent. So, the
assumption cannot be true and ∠F is not congruent to ∠H.
48. a. ABCD is a trapezoid.
b. DEFG is an isosceles trapezoid.
—
— is the midsegment of △ADF.
GE
— CD
—, BE
— AF
—, and BE = __1 (CD + AF)
Prove BE
49. a. Given BG is the midsegment of △ACD.
2
A
C
G
B
E
F
D
G
A
D
G
A
E
F
By the Triangle Midsegment Theorem (Thm. 6.8),
— CD
—, BG
— = —1CD, GE
— AF
—, and GE = —1AF. By the
BG
2
2
Transitive Property of Parallel Lines (Thm. 3.9),
— BE
— AF
—. Also, by the Segment Addition Postulate
CD
(Post. 1.2), BE = BG + GE. So, by the Substitution
Property of Equality, BE = —12 CD + —12AF = —12(CD + AF).
50. no; A concave kite and a convex kite can have congruent
corresponding sides and a pair of congruent corresponding
angles, but the kites are not congruent.
L
M
STATEMENTS
REASONS
1. JKLM is an isosceles
— JM
—,
trapezoid, KL
—
—
JK ≅ LM
1. Given
2. ∠JKL ≅ ∠MLK
2. Isosceles Trapezoid
Base Angles Theorem
(Thm. 7.14)
3. Reflexive Property of
Congruence (Thm. 2.1)
4. SAS Congruence
Theorem (Thm. 5.5)
5. Corresponding parts of
congruent triangles are
congruent.
H
— ≅ FG
— and EH
— ≅ GH
—, construct
Given kite EFGH with EF
—, which is congruent to itself by the Reflexive
diagonal FH
K
J
— ≅ KM
—
5. JL
G
— ≅ FG
—, EH
— ≅ GH
—
EF
B
— ≅ KM
—
Prove JL
4. △JKL ≅ △MLK
3PQ 3
MN = ____
Ratio: ____
=—
5PQ 5
RS
D
— JM
—, JK
— ≅ LM
—
KL
— ≅ KL
—
3. KL
MN = 3PQ
C
51. a. Given JKLM is an isosceles trapezoid.
b. If the diagonals of a trapezoid are congruent, then the
— JM
—
trapezoid is isosceles. Let JKLM be a trapezoid, KL
—
—
and JL ≅ KM . Construct line segments through K and L
— as shown below.
perpendicular to JM
J
K
L
A
B
M
— JM
—, ∠AKL and ∠KLB are right angles, so
Because KL
— ≅ BL
—. Then △JLB ≅ △MKA
KLBA is a rectangle and AK
by the HL Congruence Theorem (Thm. 5.9). So,
— ≅ JM
— by the Reflexive Property of
∠LJB ≅ ∠KMA. JM
Congruence (Thm. 2.1). So, △KJM ≅ △LMJ by the SAS
Congruence Theorem (Thm. 5.5). Then ∠KJM ≅ ∠LMJ,
and the trapezoid is isosceles by the Isosceles Trapezoid
Base Angles Converse (Thm. 7.15).
— —
—
—
—
midpoint of KL , G is the midpoint of KM , and H is the
—. By the definition of midsegment, EF
— is a
midpoint of JM
—
— is
midsegment of △JKL, FG is a midsegment of △KML, GH
—
a midsegment of △JKM, and EH is a midsegment of △JML.
— JK
—,
By the Triangle Midsegment Theorem (Thm. 6.8), EF
—
—
—
—
—
—
—
—
FG LM , GH JK , and EH LM . You know that EF GH
— EH
— by the Transitive Property of Parallel Lines
and FG
52. You are given JK ≅ LM , E is the midpoint of JL , F is the
(Thm. 3.9). EFGH is a parallelogram by the definition of a
parallelogram. By the Triangle Midsegment Theorem
(Thm. 6.8), EF = —12 JK, FG = —12 LM, GH = —12 JK, and
EH = —12 LM. You can conclude JK = LM by the definition
of congruent segments. Then by the Substitution Property
of Equality, FG = —12 JK and EH = —12 JK. It follows from the
Transitive Property of Equality that EF = FG = GH = EH.
— ≅ FG
—≅
Then by the definition of congruent segments, EF
—
—
GH ≅ EH . Therefore, by the definition of a rhombus, EFGH
is a rhombus.
Geometry
Worked-Out Solutions
267
Chapter 7
Maintaining Mathematical Proficiency
7. c + 5 = 11
53. Sample answer: A similarity transformation that maps the
c=6
blue preimage to the green image is a translation 1 unit right
followed by a dilation with a scale factor of 2.
d + 4 = 14
d = 10
54. Sample answer: A similarity transformation that maps the
blue preimage to the green image is a reflection in the y-axis
followed by a dilation with a scale factor of —13 .
—
= (−2, −1)
7.4–7.5 What Did You Learn? (p. 411)
(
triangles, △DEF and △DGF.
2. If one type of quadrilateral is under another in the diagram,
9.
4
the length of the given base, and then either add or subtract
that amount to the midsegment to find the other base.
⋅
⋅
30-gon is (30 − 2) 180° = 28 180° = 5040°.
The measure of each interior angle is
(30 − 2) 180° 28 180° 5040°
—— = — = — = 168°.
30
30
30
360°
The measure of each exterior angle is — = 12°.
30
⋅
⋅
2. The sum of the measures of the interior angles is
⋅
⋅
(6 − 2) 180° = 4 180° = 720°.
K
2
587 + x = 720
x = 133
⋅
x° + 160° + 2x° + 125° + 110° + 112° + 147° = 900°
3x + 654 = 900
3x = 246
x = 82
13x = 195
x = 15
5. a° + 101° = 180°
a = 79
M
L
−4
3 − (−3) 3 + 3
—=—
Slope of KL
= — = −6
5−6
−1
Starting at J, go down 6 units and right 1 unit. So, the
coordinates of M are (2, −2).
10. Parallelogram Opposite Sides Converse (Thm. 7.7)
13. By the Parallelogram Opposite Sides Converse (Thm. 7.7):
(7 − 2) 180° = 5 180° = 900°.
13x + 165 = 360
−2
6 x
12. Parallelogram Opposite Angles Converse (Thm. 7.8)
3. The sum of the measures of the interior angles is
4. 49° + 7x° + 83° + 33° + 6x° = 360°
4
11. Parallelogram Diagonals Converse (Thm. 7.10)
120° + 97° + 130° + 150° + 90° + x° = 720°
⋅
y J
−2
1. The sum of the measures of the interior angles of a regular
)
The coordinates of the intersection of the diagonals are
(−2, −1).
then a quadrilateral from the lower category will always fit
into the category above it.
Chapter 7 Review (pp. 412–414)
) (
−8 + 4 1 + (−3)
−4 −2
—: —
, — = —, —
Midpoint of QS
2
2
2 2
= (−2, −1)
1. They are congruent base angles of congruent isosceles
3. Find the difference between the length of the midsegment and
( −62+ 2 −32+ 1 ) ( −42 −22 )
8. Midpoint of TR : —, — = —, —
4x + 7 = 12x − 1
−8x + 7 = −1
−8x = −8
x=1
y + 1 = 3y − 11
−2y + 1 = −11
−2y = −12
y=6
The quadrilateral is a parallelogram when x = 1 and y = 6.
14. By the Parallelogram Diagonals Converse (Thm. 7.10):
6x − 8 = 4x
−8 = −2x
b = 101
6. a − 10 = 18
a = 28
4=x
The quadrilateral is a parallelogram when x = 4.
(b + 16)° = 103°
b = 87
268
Geometry
Worked-Out Solutions
Chapter 7
15.
8
y
20.
X
y
J
8
W
K
6
4
4
M
2
2
Y
−4
2
4 x
−2
Z
−4
9−7
2
2
4
2
——
—
—
JK = √(5 − 9)2 + (8 − 6)2 = √(−4)2 + 22 = √16 + 4
—
—
—
= √20 = √4 5 = 2√5
⋅
——
—
—
KL = √(9 − 7)2 + (6 − 2)2 = √ 22 + 42 = √4 + 16
⋅
—
—
—
= √20 = √4 5 = 2√5
Because JK = KL, and JKLM is a rectangle, which is also
a parallelogram, opposite sides are equal. So,
JK = KL = LM = MJ. By the definitions of a rhombus and a
square, quadrilateral JKLM is also a rhombus and a square.
congruent sides and four right angles.
21. m∠Z = m∠Y = 58°
m∠X = 180° − 58° = 122°
−2x + 34 = 3x − 26
m∠W = m∠X = 122°
−5x + 34 = −26
⋅
5−9
( )
18. The special quadrilateral is a square because it has four
XZ = 3 12 − 26 = 36 − 26 = 10
x
3 − 5 −2
1
Because the product of the two slopes is −— ( 2 ) = −1,
2
—⊥ KL
—, KL
— ⊥ LM
—, LM
—⊥ MJ
—,
there are four right angles (JK
—
—
and MJ ⊥ JK ). So, quadrilateral JKLM is a rectangle.
two pairs of parallel sides.
⋅
8
7−3
17. The special quadrilateral is a parallelogram because it has
WY = −2 12 + 34 = −24 + 34 = 10
6
4 − 8 −4
—=—
=—=2
Slope of MJ
congruent sides.
x = 12
4
1
2 − 4 −2
—=—
= — = −—
Slope of LM
16. The special quadrilateral is a rhombus because it has four
−5x = −60
2
6−2 4
—=—
=—=2
Slope of KL
— equals the slope of YZ
—, so WX
— YZ
—. The
The slope of WX
—
—
—
—
slope of XY equals the slope of WZ , so XY WZ . Because both
pairs of opposite sides are parallel, WXYZ is a parallelogram
by definition.
WY = XZ
0
1
8−6
2
— =—
= — = −—
Slope of JK
2
8−6
—=—
=—
Slope of WX
2 − (−1) 3
0 − 8 −8
—=—
=—=8
Slope of XY
1 − 2 −1
−2 − 0 −2 2
—=—
=—=—
Slope of YZ
−2 − 1 −3 3
−8
−8
−2 − 6
—=—
=—=—=8
Slope of WZ
−2 − (−1) −2 + 1 −1
19.
L
0
23.
y
10
⋅
1
52 = 26.
2
1
2
22. The length of the midsegment is —(39 + 13) = —
J
8
K
6
4
2
0
M
0
L
2
4
6
8
10 x
——
KL = √(10 −
—
8)2
—
—
—
—
+ (6 − 2)2 = √ 22 + 42 = √4 + 16
⋅
—
—
= √20 = √ 4 5 = 2√5
——
JM = √ (6 − 2)2 + (10 − 2)2 = √42 + 82 = √ 16 + 64
—
⋅
—
—
= √80 = √ 16 5 = 4√5
—
—
—
1
1
1
Midsegment = —(KL + JM) = —(2√ 5 + 4√ 5 ) = —(6√ 5 )
2
2
2
—
= 3√ 5
Geometry
Worked-Out Solutions
269
Chapter 7
24. 7x° + 65° + 85° + 105° = 360°
8.
y
7x + 255 = 360
10
7x = 105
8
x = 15
Q
6
The two congruent angles are 105°.
R
S
4
25. yes; By the Isosceles Trapezoid Base Angles Converse
(Thm. 7.15), if a trapezoid has a pair of congruent base
angles, then the trapezoid is isosceles.
2
0
26. The quadrilateral is a trapezoid because it has exactly one
P
0
2
4
6
8
x
11 − 1 10
—=—
= — = undefined
Slope of RP
pair of parallel sides.
5−5
0
6−6 0
—=—
=—=0
Slope of SQ
27. The quadrilateral is a rhombus because it has four congruent
sides.
9−1
angles.
8
—⊥ SQ
—, and quadrilateral PQRS is a rhombus.
So, RP
28. The quadrilateral is a rectangle because it has four right
9. yes; the diagonals bisect each other.
Chapter 7 Test (p. 415)
1. The diagonals bisect each other, so r = 6 and s = 3.5.
2. In a parallelogram, opposite angles are equal. So, b = 101.
Consecutive interior angles are supplementary, so
a = 180 − 101 = 79.
3. In a parallelogram, opposite sides are congruent. So, p = 5
—
11. yes; m∠Y = 360° − (120° + 120° + 60°) = 360° −
300° = 60°. Because opposite angles are congruent, the
12. If one angle in a parallelogram is a right angle, then
consecutive angles are supplementary. So, the parallelogram
is a rectangle.
and q = 6 + 3 = 9.
4. If consecutive interior angles are supplementary, then the
lines that form those angles are parallel. So, the quadrilateral
is a trapezoid.
13. Show that a quadrilateral is a parallelogram with four
congruent sides and four right angles, or show that a
quadrilateral is both a rectangle and a rhombus.
5. The quadrilateral is a kite because it has two pairs of
consecutive congruent sides, but opposite sides are not
congruent.
—
10. no; JK and ML might not be parallel.
14.
4
6. The quadrilateral is an isosceles trapezoid because it has
2
y
congruent base angles.
7. 3x° + 5(2x + 7)° = 360°
3x + 10x + 35 = 360
E
−4
M
J
⋅
x = 25
2x + 7 = 2 25 + 7 = 50 + 7 = 57
The measures of the exterior angles of the octagon are 25°,
25°, 25°, 57°, 57°, 57°, 57°, and 57°.
4 x
−2
13x + 35 = 360
13x = 325
L
K
−4
—
3−0 3
4−2 2
Starting at L, go down 3 units and left 2 units. So, the
coordinates of M are (2, 0).
2 2
— 4 + (−2) 3 + (−1)
b. Midpoint of JL = —, — = —, — = (1, 1)
2
2 2
2
0+2 2+0
2 2
—= —
, — = —, — = (1, 1)
Midpoint of KM
2
2
2 2
The coordinates of the intersection of the diagonals are
(1, 1).
1
1
15. Midsegment = —(6 + 15) = —(21) = 10.5
2
2
The middle shelf will have a diameter of 10.5 inches.
a. Slope of JK = — = —
(
⋅
) ( )
) ( )
(
⋅
⋅
(n − 2) 180° (5 − 2) 180° 3 180° 540°
5
5
5
n
The measure of each interior angle of a regular pentagon
is 108°.
16. —— = —— = — = — = 108°
270
Geometry
Worked-Out Solutions
Chapter 7
17. Design AB = DC and AD = BC, then ABCD is a
parallelogram.
120°
7. G; Because the sum of the measures of the exterior angles
is 360°, you can find the measure of each exterior angle in
a regular polygon by taking 360° divided by the number of
sides. So, it follows that you can reverse this process and do
the inverse of division. The product of the number of sides
and the measure of each exterior angle is a constant 360°.
120°
30°
Chapter 7 Standards Assessment (pp. 416–417)
—
—
—
AB , and ED is also the radius of the inscribed circle drawn
1. B; Step 3 shows the construction of ED perpendicular to
in Step 4.
2. H; Because ∠ABC and ∠DCB are formed by two pairs of
congruent angles, it can be shown that ∠ABC ≅ ∠DCB. So,
ABCD is an isosceles trapezoid by the Isosceles Trapezoid
Base Angles Converse (Thm. 7.15).
3. 73; Consecutive angles of a parallelogram are supplementary.
So, the measure of the consecutive angle is
180° − 107° = 73°.
—
4. B; Because R and S have the same y-coordinate, RS is a
vertical segment. So, the perpendicular bisector is going to
be horizontal, and every point on the perpendicular bisector
—, the
will have the same x-coordinate. In order to bisect RS
x-coordinate of each point (x, y) must be the average of
−6 + 0 −6
−6 and 0 : x = — = — = −3.
2
2
5. G; Statement II is the condition of the Parallelogram
Opposite Angles Converse (Thm. 7.8), so it is enough to
prove that a quadrilateral is a parallelogram. Statement III
is the condition of the Parallelogram Diagonals Converse
(Thm. 7.10), so it is enough to prove that a quadrilateral
is a parallelogram. Statements I and IV can also occur in
trapezoids.
6. C; Find the slope of the line through (x1, y1) = (−3, 0) and
(x2, y2) = (6, 3).
3
3 1
y2 − y1
3−0
m=—
=—=—=—=—
x2 − x1 6− (−3) 6 + 3 9 3
The line parallel to the line shown has the same slope. So,
1
use m = — and b = 4 to write an equation of the parallel line
3
in slope-intercept form.
y = mx + b
1
y = —x + 4
3