Chapter 6
Chapter 6 Maintaining Mathematical Proficiency (p. 299)
1. Slope perpendicular to y =
1
—3 x
4. An altitude of a triangle is the perpendicular segment from
a vertex to the opposite side or to the line that contains the
opposite side.
− 5 is −3.
y = mx + b
y = −3x + b
5. A midsegment of a triangle is a segment that connects the
⋅
midpoints of two sides of the triangle.
1 = −3 3 + b
1 = −9 + b
6.1 Explorations (p. 301)
10 = b
An equation of the line is y = −3x + 10.
1. a. Check students’ work.
b. Check students’ work.
c. Check students’ work (for sample in text, CA ≈ 1.97,
2. Slope perpendicular to y = −x − 5 is 1.
— and CB
— have the
CB ≈ 1.97); For all locations of C, CA
same measure.
y = mx + b
y=x+b
d. Every point on the perpendicular bisector of a segment is
⋅
−3 = 1 4 + b
equidistant from the endpoints of the segment.
−3 = 4 + b
2. a. Check students’ work.
−7 = b
b. Check students’ work.
An equation of the line is y = x − 7.
c. Check students’ work (for sample in text, DE ≈ 1.24,
1
3. Slope perpendicular to y = −4x + 13 is —4 .
y = mx + b
1
y = —x + b
4
1
−2 = — (−1) + b
4
1
−2 = −— + b
4
1
4(−2) = 4 −— + 4b
4
−8 = −1 + 4b
⋅
( )
−7 = 4b
−7
—=b
4
An equation of the line is y = —14x − —74 .
4. w ≥ −3 and w ≤ 8, or −3 ≤ w ≤ 8
DF ≈ 1.24); For all locations of D on the angle bisector,
— and FD
— have the same measure.
ED
d. Every point on an angle bisector is equidistant from both
sides of the angle.
3. Any point on the perpendicular bisector of a segment is
equidistant from the endpoints of the segment. Any point on
the angle bisector is equidistant from the sides of the angle.
4. The distance point D is from ⃗
AB is 5 units, which is the same
as the distance D is from ⃗
AC. Point D is in the angle bisector,
so it is equidistant from either side of the angle. Therefore,
— ≅ DF
—.
DE
6.1 Monitoring Progress (pp. 303–305)
1. WZ = YZ
WZ = 13.75
5. m > 0 and m < 11, or 0 < m < 11
6. s ≤ 5 or s > 2
7. d < 12 or d ≥ −7
8. yes; As with Exercises 6 and 7, if the graphs of the two
inequalities overlap going in opposite directions and the variable
only has to make one or the other true, then every number on
the number line makes the compound inequality true.
Chapter 6 Mathematical Practices (p. 300)
1. A perpendicular bisector is perpendicular to a side of the
triangle at its midpoint.
2. An angle bisector divides an angle of the triangle into two
congruent adjacent angles.
3. A median of a triangle is a segment from a vertex to the
midpoint of the opposite side.
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By the Perpendicular Bisector Theorem, WZ = 13.75.
2.
WZ = YZ
4n − 13 = n + 17
3n = 30
n = 10
YZ = n + 17
YZ = 10 + 17
YZ = 27
By the Perpendicular Bisector Theorem, YZ = 27.
1
3. WX = —2 WY
WX = —12 (14.8)
WX = 7.4
4. DA = 6.9 by the Angle Bisector Theorem.
Geometry
Worked-Out Solutions
185
Chapter 6
5.
— —
—
—
is on the perpendicular bisector of VW . So, by the
6. UW = 55; Because VD ≅ WD and ⃖⃗
UX ⊥ VW , point U
AD = CD
3z + 7 = 2z + 11
z + 7 = 11
z=4
CD = 2z + 11 = 2 4 + 11 = 8 + 11 = 19
Perpendicular Bisector Theorem (Thm. 6.1), VU = WU.
VU = UW
⋅
9x + 1 = 7x + 13
6. Because AD = CD, ⃗
BD is the angle bisector of ∠ ABC and
m∠ ABC = 2m∠ CBD. Therefore, m∠ ABC = 2(39) = 78°.
7. no; In order to use the Converse of the Angle Bisector
— would have to be perpendicular
Theorem (Thm. 6.4), PS
—
to ⃗
QP, and RS would have to be perpendicular to ⃗
QR.
−1 − (−5) −1 + 5 4
3 − (−1)
3+1
4
The slope of the perpendicular line is m = −1.
8. Slope: m = — = — = — = 1
−1 + 3 −5 + (−1)
2 −6
midpoint = —, — = —, — = (1, −3)
2
2
2 2
y = mx + b
(
) (
)
⋅
⋅
y = −1 x + b
−3 = −1 (1) + b
−3 = −1 + b
−2 = b
So, an equation of the perpendicular bisector is y = −x − 2.
6.1 Exercises (pp. 306–308)
Vocabulary and Core Concept Check
1. Point C is in the interior of ∠DEF. If ∠DEC and ∠CEF are
⃗ is the bisector of ∠DEF.
congruent, then EC
2. The question that is different is: Is point B collinear with
X and Z? B is not collinear with X and Z. Because the two
segments containing points X and Z are congruent, B is the
same distance from both X and Z, point B is equidistant from
—.
X and Z, and point B is in the perpendicular bisector of XZ
Monitoring Progress and Modeling with Mathematics
3. GH = 4.6; Because GK = KJ and ⃖⃗
HK ⊥ ⃖⃗
GJ , point H is on
—. So, by the Perpendicular
the perpendicular bisector of GJ
Bisector Theorem (Thm. 6.1), GH = HJ = 4.6.
4. QR = 1.3; Because point T is equidistant from Q and S,
— by the
point T is on the perpendicular bisector of QS
Converse of the Perpendicular Bisector Theorem (Thm. 6.2).
So, by definition of segment bisector, QR = RS = 1.3.
5. AB = 15; Because ⃖⃗
DB ⊥ ⃖⃗
AC and point D is equidistant from
— by the
A and C, point D is on the perpendicular bisector of AC
Converse of the Perpendicular Bisector Theorem (Thm. 6.2).
By definition of segment bisector, AB = BC.
AB = BC
5x = 4x + 3
x=3
AB = 5 3 = 15
⋅
186
Geometry
Worked-Out Solutions
2x = 12
x=6
⋅
UW = 7 6 + 13 = 55
7. yes; Because point N is equidistant from L and M, point N is
— by the Converse of the
on the perpendicular bisector of LM
Perpendicular Bisector Theorem (Thm. 6.2). Because only
— at point K, ⃗
one line can be perpendicular to LM
NK must be
—
the perpendicular bisector of LM , and P is on ⃗
NK.
8. no; You would need to know that either LN = MN or
LP = MP.
9. no; You would need to know that ⃖⃗
PN ⊥ ⃖⃗
ML.
10. yes; Because point P is equidistant from L and M, point P is
— by the Converse of the
on the perpendicular bisector of LM
— ≅ MN
—,
Perpendicular Bisector Theorem (Thm. 6.2). Also, LN
—
so ⃗
PN is a bisector of LM . Because P can only be on one of
—.
the bisectors, ⃗
PN is the perpendicular bisector of LM
11. Because D is equidistant from ⃗
BC and ⃗
BA, ⃗
BD bisects
∠ ABC by the Converse of the Angle Bisector Theorem
(Thm. 6.4). So, m∠ ABD = m∠ CBD = 20°.
—
—
12. ⃗
QS is an angle bisector of ∠ PQR, PS ⊥ ⃗
QP, and SR ⊥ ⃗
QR.
So, by the Angle Bisector Theorem (Thm. 6.3), PS = RS = 12.
13. ⃗
JL bisects ∠ KJM.
Angle Bisector Theorem (Thm. 6.3)
m∠ KJL = m∠ MJK
Definition of angle bisector
7x = 3x + 16
4x = 16
x=4
⋅
m∠ KJL = 7x = 7 4 = 28°
⃗ bisects ∠ FEH.
14. EG
FG = GH
Angle Bisector Theorem (Thm. 6.3)
Converse of the Angle Bisector
Theorem (Thm. 6.4)
x + 11 = 3x + 1
−2x = −10
x=5
FG = 5 + 11 = 16
⃗ bisects
15. yes; Because H is equidistant from ⃗
EF and ⃗
EG, EH
∠ FEG by the Angle Bisector Theorem (Thm. 6.3).
16. no; Congruent segments connect H to both ⃗
EF and ⃗
EG, but
unless those segments are also perpendicular to ⃗
EF and ⃗
EG,
you cannot conclude that H is equidistant from ⃗
EF and ⃗
EG.
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Chapter 6
—
—
17. no; Because neither BD nor DC are marked as perpendicular
to ⃗
AB or ⃗
AC respectively, you cannot conclude that DB = DC.
—
18. yes; D is on the angle bisector of ∠ BAC, DB ⊥ ⃗
AB and
— ⊥ ⃗
CD
AC. So, DB = DC by the Angle Bisector Theorem
(Thm. 6.3).
—
−1 − 5
7−1
−6
6
19. Slope of MN : m = — = — = −1
The slope of the perpendicular line is m = 1.
1 + 7 5 + (−1)
8 4
midpoint = —, — = —, — = (4, 2)
2
2
2 2
y = mx + b
(
) ( )
⋅
⋅
—
4
8
1 − (−7)
7
−4 − 10 −14
The slope of the perpendicular line is m = —74.
22. Slope of YZ : m = — = — = −—
10 + (−4) −7 + 1
6 −6
—= —
, — = —, — = (3, −3)
midpoint of YZ
2
2
2 2
y = mx + b
7
y = —x + b
4
7
−3 = — 3 + b
4
21
−3 = — + b
4
−12 = 21 + 4b
) (
(
)
⋅
y=1 x+b
−33 = 4b
2 = 1 (4) + b
33
−— = b
4
— is
An equation of the perpendicular bisector of YZ
7
33
y = —4 x − —
.
4
2=4+b
−2 = b
— is y = x − 2.
An equation of the perpendicular bisector of MN
—
12 3
12 − 0
6 − (−2)
8
2
2
The slope of the perpendicular line is m = −—3.
20. Slope of QR : m = — = — = —
−2 + 6 0 + 12
4 12
—= —
, — = —, — = (2, 6)
midpoint of QR
2
2
2 2
y = mx + b
2
y = −—x + b
3
2
6 = −— (2) + b
3
4
6 = −— + b
3
18 = −4 + 3b
(
) ( )
⋅
22 = 3b
—
—
23. Because DC is not necessarily congruent to EC , ⃖⃗
AB will
not necessarily pass through point C. The reasoning
—, ⃖⃗
should be: Because AD = AE, and ⃖⃗
AB ⊥ DE
AB is the
—
perpendicular bisector of DE .
—
24. Because BP is not necessarily perpendicular to ⃗
CB, you
do not have sufficient evidence to say that BP = AP. The
reasoning should be: By the Angle Bisector Theorem
(Thm. 6.3), point P is equidistant from ⃗
CB and ⃗
CA.
25. The Perpendicular Bisector Theorem (Thm. 6.1) will allow
— ≅ CD
—.
the conclusion AD
26. a. The relationship between ⃗
PG and ∠ APB is that ⃗
PG is the
angle bisector of ∠ APB.
b. m∠ APB gets larger. Covering the goal becomes more
22
3
—=b
22
b=—
3
— is
An equation of the perpendicular bisector of QR
2
22
y = −—3x + —
.
3
—
8−4
9 − (−3)
4
12
1
3
21. Slope of UV : m = — = — = —
3
The slope of the perpendicular line is m = −—1 = −3.
—=
midpoint of UV
y = mx + b
(
−3 + 9 4 + 8
6 12
—, — = —, — = (3, 6)
2
2
2 2
) ( )
y = −3x + b
⋅
6 = −3 3 + b
6 = −9 + b
15 = b
— is
An equation of the perpendicular bisector of UV
y = −3x + 15.
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difficult if the goalie remains at the same distance from
the puck on the perpendicular bisector. As the angle
increases, the goalie is farther away from each side of
the angle.
1
—
—
of XY . Draw two arcs of equal radii, using X and Y as
27. Draw XY , using a radius that is greater than —2 the distance
centers, so that the arcs intersect. Draw a line through both
intersections of the arcs.
Set a compass at 3 centimeters
Z
by its own scale or with a ruler.
3 cm
Retaining this setting, place the
3.9 cm
3.9 cm
compass point on the midpoint
— and mark the point on
of XY
the perpendicular bisector as
X 2.5 cm
2.5 cm Y
point Z. The distance between
— is 3 centimeters
point Z and XY
because of the compass setting
and in this example, XZ and YZ are both equal to
3.9 centimeters. This construction demonstrates the
Perpendicular Bisector Theorem (Thm. 6.1).
Geometry
Worked-Out Solutions
187
Chapter 6
28. Because every point on a compass arc is the same distance
from one endpoint, and every point on the other compass
arc with the same setting is the same distance from the other
endpoint, the line connecting the points where these arcs
intersect contains the points that are equidistant from both
endpoints. You know from the Converse of the Perpendicular
Bisector Theorem (Thm. 6.2) that the set of points that are
equidistant from both endpoints make up the perpendicular
bisector of the given segment.
29. B; (3x − 9)° = 45°
x = 18
—
30. B; Slope of MN :
0
5−5
m=—=—=0
−1 − 7 −8
The slope of the perpendicular line is undefined.
7 + (−1) 5 + 5
6 10
—= —
, — = —, — = (3, 5)
midpoint of MN
2
2
2 2
The equation of a line that has a slope that is undefined
through the point (3, 5) is x = 3. So, the point that lies on the
perpendicular bisector is (3, 9).
(
) ( )
31. no; In isosceles triangles, for example, the ray that has an
endpoint of the vertex and passes through the base (the
opposite side of the vertex) is not only an angle bisector of
the vertex, but also a perpendicular bisector of the base.
Prove
REASONS
AD bisects ∠ BAC.
1. ⃗
1. Given
2. ∠ BAD ≅ ∠ CAD
2. Definition of angle
bisector
— ⊥ ⃗
— ⊥ ⃗
3. DB
AB, DC
AC
5. Right Angles Congruence
Theorem (Thm. 2.3)
6. —
AD ≅ —
AD
6. Reflexive Property of
Congruence (Thm. 2.1)
7. △ADB ≅ △ADC
7. AAS Congruence
Theorem (Thm. 5.11)
8. —
DB ≅ —
DC
8. Corresponding parts of
congruent triangles are
congruent.
9. DB = DC
b. Given
CA = CB
—.
Point C lies on the perpendicular bisector of AB
3. Given
4. ∠ABD and ∠ACD are 4. Definition of
right angles.
perpendicular lines
5. ∠ABD ≅ ∠ACD
3x = 54
32. Given
STATEMENTS
Prove
9. Definition of congruent
segments
B
BD = CD,
— ⊥ ⃗
DB
AB,
— ⊥ ⃗
DC
AC
⃗
AD bisects ∠ BAC.
D
A
C
C
A
B
P
⃖⃗ such that point P is on
Given isosceles △ACB, construct CP
— and CP
—. So, ∠ CPB and ∠ CPA are right angles by
⃖⃗ ⊥ AB
AB
the definition of perpendicular lines, and △CPB and △CPA
— ≅ BC
— and CP
— ≅ CP
—
are right triangles. Also, because AC
by the Reflexive Property of Congruence (Thm. 2.1),
△CPB ≅ △CPA by the HL Congruence Theorem (Thm. 5.9).
— ≅ BP
— because corresponding parts of congruent
So, AP
triangles are congruent, which means that point P is the
—, and ⃖⃗
—.
midpoint of AB
CP is the perpendicular bisector of AB
33. a. Given
Prove
— ⊥ ⃗
— ⊥ ⃗
⃗ bisects ∠ BAC, DB
AD
AB, DC
AC
DB = DC
STATEMENTS
REASONS
1.
1. Given
DB ⊥ ⃗
AB,
BD = CD, —
—
DC ⊥ ⃗
AC
2. ∠ ABD and ∠ ACD
are right angles.
2. Definition of
perpendicular lines
3. △ABD and △ACD
are right triangles.
3. Definition of a right
triangle
4. —
BD ≅ —
CD
4. Definition of congruent
segments
5. —
AD ≅ —
AD
5. Reflexive Property of
Congruence (Thm. 2.1)
6. △ABD ≅ △ACD
6. HL Congruence Theorem
(Thm. 5.9)
7. ∠ BAD ≅ ∠ CAD
7. Corresponding parts of
congruent triangles are
congruent.
B
D
A
8. ⃗
AD bisects ∠ BAC.
C
188
Geometry
Worked-Out Solutions
8. Definition of angle
bisector
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Chapter 6
34. a. Roosevelt School; Because the corner of Main and
3rd Street is exactly 2 blocks of the same length from each
hospital, and the two streets are perpendicular, 3rd Street
is the perpendicular bisector of the segment that connects
the two hospitals. Because Roosevelt school is on 3rd
Street, it is the same distance from both hospitals by the
Perpendicular Bisector Theorem (Thm. 6.1).
b. no; Because the corner of Maple and 2nd Street is
approximately the midpoint of the segment that connects
Wilson School to Roosevelt School, and 2nd Street is
perpendicular to Maple, 2nd Street is the perpendicular
bisector of the segment connecting Wilson and Roosevelt
Schools. By the contrapositive of the Converse of the
Perpendicular Bisector Theorem (Thm. 6.2), the Museum
is not equidistant from the two schools because it is not on
2nd Street.
35. a. y = x
b. y = −x
c. y = ∣ x ∣
— —
c. First, WV ≅ WV by the Reflexive Property of Congruence
— ≅ ZW
— and XV
— ≅ ZV
—,
(Thm. 2.1). Then, because XW
△WVX ≅ △WVZ by the SSS Congruence Theorem
(Thm. 5.8). So, ∠ VXW ≅ ∠ VZW because corresponding
parts of congruent triangles are congruent.
Maintaining Mathematical Proficiency
39. The triangle is isosceles because it has two congruent sides.
40. The triangle is scalene because no sides are congruent.
41. The triangle is equilateral because all sides are congruent.
42. The triangle is an acute triangle because all angles measure
less than 90°.
43. The triangle is a right triangle because one angle measures 90°.
44. The triangle is obtuse because one angle measure is greater
than 90°.
36. no; In spherical geometry, all intersecting lines meet in two
points which are equidistant from each other because they
are the two endpoints of a diameter of the circle.
— ≅ CD
— and AE
— ≅ CE
—
AD
—
—
AB ≅ CB
37. Given
Prove
6.2 Explorations (p. 309)
1. a–c. Sample answer:
5
3
A
2
D
A
0
−2
−1
C
0
C
1
2
3
4
5
6
7
−1
— ≅ CD
— and AE
— ≅ CE
—, by the Converse of the
Because AD
Perpendicular Bisector Theorem (Thm. 6.2), both points
—. So, ⃖⃗
D and E are on the perpendicular bisector of AC
DE is
—
—
—, then by the
the perpendicular bisector of AC . So, if AB ≅ CB
Converse of the Perpendicular Bisector Theorem (Thm. 6.2),
point B is also on ⃖⃗
DE. So, points D, E, and B are collinear.
Conversely, if points D, E, and B are collinear, then by the
Perpendicular Bisector Theorem (Thm. 6.2), point B is also
—. So, AB
— ≅ CB
—.
on the perpendicular bisector of AC
Prove
D
1
B
E
38. Given
B
4
— at point Y.
Plane P is a perpendicular bisector of XZ
—
—
—
—
a. XW ≅ ZW
b. XV ≅ ZV c. ∠ VZW ≅ ∠ VZW
−2
−3
a. The perpendicular bisectors of the sides of △ABC all
intersect at one point.
c. The circle passes through all three vertices of △ABC.
2. a–c. Sample answer:
5
A
B
3
D
2
1
X
Y
E
4
0
V
P
W
−2
−1
0
1
2
3
4
5
6
7
−1
C
−2
−3
Z
—
— at point Y, YW
— is a perpendicular bisector of
bisector of XZ
—
XZ by definition of a plane perpendicular to a line. So, by
— ≅ ZW
—.
the Perpendicular Bisector Theorem (Thm. 6.1), XW
a. Because YW is on plane P, and plane P is a perpendicular
—
b. Because YV is on plane P, and plane P is a perpendicular
— at point Y, YV
— is a perpendicular bisector of
bisector of XZ
—
XZ by definition of a plane perpendicular to a line. So, by
— ≅ ZV
—.
the Perpendicular Bisector Theorem (Thm. 6.1), XV
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a. The angle bisectors all intersect at one point.
c. distance ≈ 2.06; The circle passes through exactly one
point of each side of △ABC.
3. The perpendicular bisectors of the sides of a triangle meet
at a point that is the same distance from each vertex of the
triangle. The angle bisectors of a triangle meet at a point that
is the same distance from each side of the triangle.
Geometry
Worked-Out Solutions
189
Chapter 6
6.2 Monitoring Progress (pp. 311–314)
1. The pretzel distributor is located at point F, which is the
circumcenter of △ABE.
4.
QM = QN
3x + 8 = 7x + 2
−4x = −6
B
x = —46 = —23
⋅
⋅ ( — ) + 2 = — + 2 = 10.5 + 2 = 12.5
QM = 3x + 8 = 3 —32 + 8 = —92 + 8 = 4.5 + 8 = 12.5
QN = 7x + 2 = 7
F
A
2. Graph △RST.
S(−6, 5)
R(−2, 5)
5. Draw two angle bisectors and label the intersection of the
y
bisectors, the incenter, as L. Draw a perpendicular segment
from the incenter L to any one side of the triangle. Label that
point E. Draw a circle with center L and radius LE. It should
touch all sides of the triangle. The location of the lamppost is
at L.
4
(−4, 2)
y=2
−6
x
T(−2, −1)
x = −4
−2 + (−6) 5 + 5
−8 10
—= —
, — = —, — = (−4, 5)
midpoint of RS
2
2
2 2
5
+
(−1)
−2
+
(−2)
−4 4
— = —, — = —
, — = (−2, 2)
midpoint of RT
2
2
2 2
—
The equation of the perpendicular bisector of RS through
its midpoint (−4, 5) is x = −4, and the equation of the
— through its midpoint (−2, 2) is
perpendicular bisector of RT
y = 2. The point of intersection of the two perpendicular
bisectors is (−4, 2). So, the coordinates of the circumcenter
of △RST is (−4, 2).
(
(
) ( )
) ( )
3. Graph △WXY.
−2
2
(0, −1)
6.2 Exercises (pp. 315–318)
Vocabulary and Core Concept Check
1. When three or more lines, rays, or segments intersect in
the same point, they are called concurrent lines, rays, or
segments.
Monitoring Progress and Modeling with Mathematics
4 x
3. Because G is the circumcenter of △ABC, AG = BG = CG.
y = −1
Therefore, because AG = 9, BG = 9.
4. Because G is the circumcenter of △ABC, AG = BG = CG.
x=0
Therefore, because GC = 11, GA = 11.
Y(1, −6)
−1 + 1 4 + 4
0 8
—= —
, — = —, — = (0, 4)
midpoint of WX
2
2
2 2
1 + 1 4 + (−6)
2 −2
—= —
, — = —, — = (1, −1)
midpoint of XY
2
2
2 2
— through its
The equation of the perpendicular bisector of WX
midpoint (0, 4) is x = 0, and the equation of the perpendicular
— through its midpoint (1, −1) is y = −1. The
bisector of XY
point of intersection of the two perpendicular bisectors is
(0, −1). So, the coordinates of the circumcenter of △WXY
are (0, −1).
(
(
190
E
because it shows the incenter of the triangle. The other three
show the circumcenter.
−4
−6
L
2. The triangle that does not belong is the fourth triangle
y X(1, 4)
W(−1, 4)
−4
21
2
By the Incenter Theorem, QM = QN = QP and
QP = 12.5 units.
E
6
3
2
) ( )
) ( )
Geometry
Worked-Out Solutions
5. Because P is the incenter of △XYZ, PA = PB = PC.
Therefore, because PC = 9, PB = 9.
6. Because P is the incenter of △XYZ, KP = HP = FP.
Therefore, because KP = 15, HP = 15.
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Chapter 6
7. Graph △ABC.
y
9. Graph △HJK.
y
x=5
12
C(8, 10)
H(−10, 7)
(5, 8)
4
4
A(2, 6) B(8, 6)
K(−2, 3)
y=8
4
8
12
−12
x
(
(
) ( )
) ( )
4
x
E(−1, −1)
(
(
)
)
⋅
5 = 1 (−8) + b
— is m = 2.
The slope of the line perpendicular to HK
y = −5
−12
−10 + (−6) 7 + 3
— = ——
,—
midpoint of JH
2
2
−16 10
= —, — = (−8, 5)
2 2
y = mx + b
— is y = x + 13.
The equation of the line perpendicular to JH
1
4
7−3
—: m = ——
= — = −—
Slope of HK
2
−10 − (−2) −8
D(−7, −1)
F(−7, −9)
4
7−3
Slope of JH: m = —— = — = −1
−10 − (−6) −4
— is m = 1.
The slope of the line perpendicular to JH
13 = b
y
−4
x
−4
5 = −8 + b
8. Graph △DEF.
−8
4
J(−6, 3)
2+8 6+6
10 12
—= —
, — = —, — = (5, 6)
midpoint of AB
2
2
2 2
6
+
10
8
+
8
16 16
— = —, — = —
, — = (8, 8)
midpoint of BC
2
2
2 2
— through
The equation of the perpendicular bisector of AB
its midpoint (5, 6) is x = 5, and the equation of the
— through its midpoint (8, 8) is
perpendicular bisector of BC
y = 8. The point of intersection of the two perpendicular
bisectors is (5, 8). So, the coordinates of the circumcenter of
△ABC are (5, 8).
x = −4
8
(−4, −5)
−7 + (−1) −1 + (−1)
—= —
,—
midpoint of DE
2
2
−8 −2
= —, — = (−4, −1)
2 2
−7 + (−7) −1 + (−9)
—
midpoint of DF = —, —
2
2
−14 −10
= —, — = (−7, −5)
2
2
— through
The equation of the perpendicular bisector of DE
its midpoint (−4, −1) is x = −4, and the equation of
— through its midpoint
the perpendicular bisector of DF
(−7, −5) is y = −5. The point of intersection of the two
perpendicular bisectors is (−4, −5). So, the coordinates of
the circumcenter of △DEF are (−4, −5).
(
(
(
(
)
)
)
)
−10 + (−2) 7 + 3
— = ——
,—
midpoint of HK
2
2
−12 10
= —, — = (−6, 5)
2 2
y = mx + b
(
(
)
)
⋅
5 = 2 (−6) + b
5 = −12 + b
17 = b
— is y = 2x + 17.
The equation of the line perpendicular to HK
−6 + (−2) 3 + 3
—= —
,—
midpoint of JK
2
2
−8 6
= —, — = (−4, 3)
2 2
— through its
The equation of the perpendicular bisector of JK
midpoint (−4, 3) is x = −4.
(
)
(
)
The intersection of y = x + 13 and x = −4:
y = −4 + 13 = 9, the point of intersection is (−4, 9).
The intersection of y = 2x + 17 and x = −4:
y = 2(−4) + 17 = 9, the point of intersection is (−4, 9).
So, the coordinates of the circumcenter of △HJK are (−4, 9).
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Geometry
Worked-Out Solutions
191
Chapter 6
10. Graph △LMN.
11.
2
−2
4
3x = 9
8 x
6
x = —93 = 3
N(8, −6)
M(5, −3)
NE = 3x + 7 = 3 3 + 7 = 16
By the Incenter Theorem, ND = NE = NF and NF = 16 units.
L(3, −6)
−3 − (−6) −3 + 6 3
—: m = —
=—=—
Slope of LM
5−3
2
2
2
—
The slope of the line perpendicular to LM is m = −—3 .
—=
midpoint of LM
(
(
−x = −6
NG = x + 3 = 6 + 3 = 9
)
⋅
NH = 2x − 3 = 2 6 − 3 = 12 − 3 = 9
By the Incenter Theorem, NG = NH = NJ and NJ = 9 units.
13.
3x = 12
x=4
−11 = 6b
11
−— = b
6
11
— is y = −—2x − —
The equation of the line perpendicular to LM
.
3
6
—:
Slope of MN
⋅
NK = 2x − 2 = 2 4 − 2 = 8 − 2 = 6
NL = −x + 10 = −4 + 10 = 6
By the Incenter Theorem, NK = NL = NM and NM = 6 units.
14. NQ = NR
−6 − (−3) −6 + 3 −3
m = — = — = — = −1
8−5
3
3
— is m = 1.
The slope of the line perpendicular to MN
5 + 8 −3 + (−6)
13 9
—= —
, — = —, −—
midpoint of MN
2
2
2
2
y = mx + b
9 13
−— = — + b
2
2
9
13
2 −— = 2 — + 2b
2
2
−9 = 13 + 2b
) (
2x = 3x − 2
−1x = −2
x=2
)
( ) ( )
⋅
NQ = 2x = 2 2 = 4
−11 = b
— is y = x − 11.
The equation of the line perpendicular to MN
−6 − (−6) −6 + 6 0
—: m = —
=—=—=0
Slope of LN
8−3
5
5
3 + 8 −6 + (−6)
—
midpoint of LN = —, —
2
2
11 −12
11
= —, — = —, −6
2 2
2
11
— is x = —
The equation of the line perpendicular to LN
.
2
2
11
11
The intersection of y = −—3 x − —
and
x
=
:
—
6
2
2
( )
)
) (
22
)
33
11
11
11
11
—
y = −—3 —
−—
= −—
= −—
6 −—
6 = − 2 , the point of
2
6
6
(
11
)
11
intersection is —
, −—
2 . So, the coordinates of the
2
11
11
circumcenter of △LMN are —
, −—
2 .
2
(
Geometry
Worked-Out Solutions
)
⋅
NR = 3x − 2 = 3 2 − 2 = 6 − 2 = 4
By the Incenter Theorem, NQ = NR = NS and NS = 4 units.
15.
PX = PY
3x + 2 = 4x − 8
−1x = −10
−22 = 2b
(
(
NK = NL
2x − 2 = −x + 10
⋅
(
NG = NH
x + 3 = 2x − 3
)
⋅
( ) ( )
12.
x=6
3 + 5 −6 + (−3)
—, —
2
2
8 9
= —, −— = (4, −4.5)
2 2
y = mx + b
2
9
−— = −— 4 + b
2
3
9
8
−— = −— + b
2
3
9
8
6 −— = 6 −— + 6 b
2
3
−27 = −16 + 6b
192
⋅
⋅
ND = 6x − 2 = 6 3 − 2 = 18 − 2 = 16
−4
−6
ND = NE
6x − 2 = 3x + 7
y
x = 10
⋅
⋅
PX = 3x + 2 = 3 10 + 2 = 30 + 2 = 32
PY = 4x − 8 = 4 10 − 8 = 40 − 8 = 32
By the Incenter Theorem, PX = PY = PZ and PZ = 32 units.
16.
PX = PZ
4x + 3 = 6x − 11
−2x = −14
x=7
⋅
⋅
PX = 4x + 3 = 4 7 + 3 = 28 + 3 = 31
PZ = 6x − 11 = 6 7 − 11 = 42 − 11 = 31
By the Circumcenter Theorem, PX = PY = PZ and PY = 31 units.
Copyright © Big Ideas Learning, LLC
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Chapter 6
17. Sample answer:
22. Construct an angle bisector of ∠BCA and ∠ABC. Label
4
the intersection of the two angle bisectors as D. Draw a
—, label that point E. Using D
perpendicular line from D to AB
—
as the center and DE as the radius, construct a circle. Point
D is the incenter of △ABC.
A
3
2
D
1
B
0
−1
5
C
0
1
2
3
5
4
6
4
−1
A
3
−2
E
2
18. Sample answer:
5
−1
A
3
D
0
1
B
2
3
5C
4
6
−1
2
3
5
4
6
the intersection of the two angle bisectors as D. Draw a
—, label that point E. Using D
perpendicular line from D to AB
— as the radius, construct a circle. Point
as the center and DE
D is the incenter of △ABC.
5
3
A
4
2
3
1
A
E
2
D
0
B
1
6
4
−1
0
23. Construct an angle bisector of ∠ABC and ∠CAB. Label
1
0
B
−1
2
19. Sample answer:
C
0
4
−1
D
1
0
1
C
2
3
5
4
6
1
D
0
−1
0
1
2
C3
B
4
5
6
7
8
9
−1
−2
24. Construct an angle bisector of ∠ABC and ∠BCA. Label
20. Sample answer:
5
4
the intersection of the two angle bisectors as D. Draw a
—, label that point E. Using D
perpendicular line from D to BC
—
as the center and DE as the radius, construct a circle. Point
D is the incenter of △ABC.
A
3
2
D
5
1
B
4
C
0
−1
0
1
2
3
4
5
6
A
3
−1
2
21. Construct an angle bisector of ∠ABC and ∠BCA. Label
the intersection of the two angle bisectors as D. Draw a
—, label that point E. Using D
perpendicular line from D to AC
—
as the center and DE as the radius, construct a circle. Point
D is the incenter of △ABC.
E
2
0
1
2
C
3
4
5
6
−1
26. Because point T is the intersection of the perpendicular
D
1
C
0
B
B
— and GF
— are not necessarily
is the incenter. But, because GD
perpendicular to a side of the triangle, there is not sufficient
— and GF
— are congruent.
evidence to conclude that GD
Point G is equidistant from the sides of the triangle.
A
3
−1
E
0
−1
25. Because point G is the intersection of the angle bisectors, it
5
4
D
1
0
1
2
3
4
5
−1
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All rights reserved.
6
bisectors, it is the circumcenter and is equidistant from the
vertices of the triangle, not necessarily the sides.
TU = TW = TY
Geometry
Worked-Out Solutions
193
Chapter 6
27. You could copy the positions of the three houses, and
connect the points to draw a triangle. Then draw the three
perpendicular bisectors of the triangle. The point where the
perpendicular bisectors meet, the circumcenter, should be the
location of the meeting place.
33. Graph △ABC.
12
A
D
C
B
B(6, 6)
8
A(2, 5)
4
28. To find the location of the fountain, use the Incenter
Theorem. The incenter of the triangle is equidistance from
the sides of the triangle. That point will be the same distance
from each edge of the koi pond. So, place the fountain at the
incenter of the pond.
y
C(12, 3)
−4
4
8
12
x
−4
6−5 1
—: m = —
=—
Slope of AB
6−2 4
— is m = −4.
The slope of the line perpendicular to AB
2+6 5+6
8 11
—= —
, — = —, — = (4, 5.5)
midpoint of AB
2
2
2 2
y = mx + b
(
) ( )
⋅
5.5 = −4 4 + b
29. The circumcenter of a scalene triangle is sometimes inside
the triangle. If the scalene triangle is obtuse or right, then
the circumcenter is outside or on the triangle, respectively.
However, if the scalene triangle is acute, then the
circumcenter is inside the triangle.
30. If the perpendicular bisector of one side of a triangle
intersects the opposite vertex, the triangle is always
isosceles. If the perpendicular bisector of one side of a
triangle intersects the opposite vertex, then it divides the
triangle into two congruent triangles. So, two sides of the
original triangle are congruent because corresponding parts
of congruent triangles are congruent.
31. The perpendicular bisectors of a triangle intersect at a point
that is sometimes equidistant from the midpoints of the
sides of the triangle. This only happens when the triangle is
equilateral.
32. The angle bisectors of a triangle intersect at a point that is
always equidistant from the sides of the triangle. This is the
Incenter Theorem (Thm. 6.6).
5.5 = −16 + b
21.5 = b
— is
The equation of the line perpendicular to AB
y = −4x + 21.5.
1
−2
3−5
—: m = —
= — = −—
Slope of AC
5
12 − 2
10
— is m = 5.
The slope of the line perpendicular to AC
2 + 12 5 + 3
14 8
—= —
, — = —, — = (7, 4)
midpoint of AC
2
2
2 2
y = mx + b
(
) ( )
⋅
4=5 7+b
4 = 35 + b
−31 = b
— is y = 5x − 31.
The equation of the line perpendicular to AC
1
−3
3−6
—: m = —
= — = −—
Slope of BC
2
12 − 6
6
— is m = 2.
The slope of the line perpendicular to BC
6 + 12 6 + 3
18 9
—= —
, — = —, — = (9, 4.5)
midpoint of BC
2
2
2 2
y = mx + b
(
) ( )
⋅
4.5 = 2 9 + b
4.5 = 18 + b
−13.5 = b
— is
The equation of the line perpendicular to BC
y = 2x − 13.5.
Find the intersection of y = −4x + 21.5 and y = 5x − 31.
−4x + 21.5 = 5x − 31
−9x + 21.5 = −31
−9x = −52.5
35
x=—
6
11
35
y = 5 — − 31 = −—
6
6
⋅
(
11
)
35
The point of intersection is —
, −—
6 . So, the coordinates of
6
the circumcenter of △ABC are
194
Geometry
Worked-Out Solutions
(
35
,
—
6
11
)
−—
6 .
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Chapter 6
34. Graph △DEF.
3
F(−2, −2)
−8
−4
x
D(−9, −5)
⋅(
−8
E(−5, −9)
−9 − (−5) −9 + 5 −4
—: m = —
= — = — = −1
Slope of DE
−5 − (−9) −5 + 9
4
— is m = 1.
The slope of the line perpendicular to DE
(
) (
)
⋅
)
−14 = 2b
−7 = b
— is y = −—3x −7.
The equation of the line perpendicular to EF
7
7
49
Find the intersection of y = x and y = −—3 x − —
.
3
7
49
x = −—x − —
3
3
3x = −7x − 49
−7 = 1 (−7) + b
−7 = −7 + b
0=b
— is y = x.
The equation of the line perpendicular to DE
−2 − (−5) −2 + 5 3
—: m = —
=—=—
Slope of DF
−2 − (−9) −2 + 9 7
— is m = −—7.
The slope of the line perpendicular to DF
3
11 7
−9 + (−2) −5 + (−2)
—
midpoint of DF = —, — = −—, −—
2
2
2
2
y = mx + b
7
7
11
−— = −— −— + b
2
3
2
7 77
−— = — + b
2
6
7
77
6 −— = 6 — + 6b
2
6
−21 = 77 + 6b
) (
(
⋅(
⋅(
)
( ) ()
−9 + (−5) −5 + (−9)
−14 −14
—= —
, — = —, —
midpoint of DE
2
2
2
2
= (−7, −7)
y = mx + b
)
) (
(
y
−4
⋅(
— is m = −—.
The slope of the line perpendicular to EF
7
7 11
−5 + (−2) −9 + (−2)
—
midpoint of EF = —, — = −—, −—
2
2
2
2
y = mx + b
3
11
7
−— = −— −— + b
2
7
2
11 21
−— = — + b
2
14
11 3
−— = — + b
2
2
11
3
2 −— = 2 — + 2b
2
2
−11 = 3 + 2b
)
)
10x = −49
35
x = −— = −4.9
10
y = −4.9
The point of intersection is (−4.9, −4.9). So, the coordinates
of the circumcenter of △DEF are (−4.9, −4.9).
35. 352 + (2x)2 = 372
1225 +
−2 − (−9) −2 + 9 7
—: m = —
,—=—
Slope of EF
−2 − (−5) −2 + 5 3
= 1369
= 144
x2 = 36
576 + 196x2 = 625
196x2 = 49
49
x2 = —
196
4x2
)
−98 = 6b
98
−— = b
6
49
−— = b
3
49
— is y = −—7x − —
The equation of the line perpendicular to DF
.
3
3
36. 242 + (14x)2 = 252
4x2
7
x=—
= —12
14
x=6
The value of x that will
make N the incenter is 6.
The value of x that will
make N the incenter is —12.
37. The circumcenter of any right triangle is located at the
midpoint of the hypotenuse of the triangle.
y
A(0, 2b)
MAB(0, b)
B(0, 0)
MAC (a, b)
MBC (a, 0)
C(2a, 0) x
Let A(0, 2b), B(0, 0), and C(2a, 0) represent the vertices of
a right triangle where ∠ B is the right angle. The midpoint
— is M—(0, b). The midpoint of BC
— is M—(a, 0). The
of AB
AB
— is M—(a, b). Because AB
— is BC
midpoint of AC
vertical, its
AC
perpendicular bisector is horizontal. So, the equation of the
horizontal line passing through M—(0, b) is y = b. Because
— is horizontal, its perpendicularABbisector is vertical. So,
BC
the equation of the vertical line passing through MBC
—(a, 0)
is x = a. The circumcenter of △ABC is the intersection of
perpendicular bisectors, y = b and x = a, which is (a, b).
—.
This point is also the midpoint of AC
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Geometry
Worked-Out Solutions
195
Chapter 6
38. Given
— bisects ∠ CAB, BD
— bisects ∠ CBA,
△ABC, AD
—
—
—
—
—
—
DE ⊥ AB , DF ⊥ BC , DG ⊥ CA.
42. a. The archaeologists need to locate the circumcenter of
the three stones because that will be the center of the
circle that contains all three stones. In order to locate the
circumcenter, the archaeologists need to find the point of
concurrency of the perpendicular bisectors of the sides of
the triangle formed by the three stones.
Prove The angle bisectors intersect at D, which is
—, BC
—, and CA
—.
equidistant from AB
C
G
A
F
b. This is a circumcenter problem. The approximate
D
coordinates of the point at which the archaeologists
should look for the fire pit are (7, 7).
E
B
10
— ⊥ AB
—, DF
— ⊥ BC
—, and DG
— ⊥ CA
—, ∠ DFB, ∠ DEB,
Because DE
∠ DEA, and ∠ DGA are congruent right angles. Also, by
definition of angle bisector, ∠ DBF ≅ ∠ DBE and
— ≅ DB
— and DA
— ≅ DA
— by
∠ DAE ≅ ∠ DAG. In addition, DB
the Reflexive Property of Congruence (Thm. 2.1). So,
△DFB ≅ △DEB and △DEA ≅ △DGA by the AAS
Congruence Theorem (Thm. 5.11). Next, because
corresponding parts of congruent triangles are congruent,
— ≅ DE
— and DG
— ≅ DE
—. By the Transitive Property of
DF
— ≅ DE
— ≅ DG
—. So, point D is
Congruence (Thm. 2.1), DF
—
—
—
equidistant from AB , BC , and CA . Because D is equidistant
— and CB
—, by the Converse of the Angle Bisector
from CA
Theorem (Thm. 6.4), point D is on the angle bisector of
∠ ACB. So, the angle bisectors intersect at point D.
39. The circumcenter is the point of intersection of the
it must be inside the triangle.
41. a. To determine the location of the pool so that it touches
the edges, construct two angle bisectors ⃗
RD and ⃗
QD .
—.
Construct a perpendicular bisector through point D to QR
Label the intersection E. With D as the center, construct a
—. If the incenter point were to move
circle with radius DE
in any direction, the circle contained within the triangle
would become smaller and not touch all three sides of the
triangle. So, the circle with the center at the incenter is the
largest circle that touches all three sides.
(
E
Q
)
R
D
P
A(2, 10)
8
B(13, 6)
D(7.09, 6.87)
6
4
2
C(6, 1)
2
4
6
8
10
12
x
43. B; by the Perpendicular Bisector Theorem
44. no; When you find the circumcenter of three of the points
and draw the circle that circumscribes those three points, it
does not pass through the fourth point. An example of one
circle is shown.
perpendicular bisectors of the sides of a triangle, and it is
equidistant from the vertices of the triangle. In contrast, the
incenter is the point of intersection of the angle bisectors of a
triangle, and it is equidistant from the sides of the triangle.
40. no; Because the incenter is the center of an inscribed circle,
y
A
B
E
C
D
45. yes; In an equilateral triangle, each perpendicular bisector
passes through the opposite vertex and divides the triangle
into two congruent triangles. So, it is also an angle bisector.
46. The incenter is at the center of the hub of the windmill where
the blades, acting as angle bisectors, connect to the hub.
47. a. All triangles have exactly three angle bisectors and
20
three perpendicular bisectors. Only three —
segments
3
( )
are needed to represent them on an equilateral triangle
because each perpendicular bisector also bisects the
opposite angle.
b. All six segments are needed to represent the three angle
bisectors and three perpendicular bisectors of a scalene
triangle because none of the perpendicular bisectors will
also bisect an angle.
b. Yes, and the radius would be decreased by 1 foot. You
would keep the center of the pool as the incenter of the
triangle, but you would make the radius of the pool at
least 1 foot shorter.
196
Geometry
Worked-Out Solutions
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Chapter 6
—=
midpoint of PN
48. Sample answer:
— —
⋅
⋅ ⋅
⋅
75 ft
36.87°
20 = 3b
60 ft
90°
— —
y = mx + b
1
11 = — 13 + b
3
1
3 11 = 3 — 13 + 3b
3
33 = 13 + 3b
53.13°
45 ft
( 10 +2 16, 20 2+ 2 ) = ( 262 , 222 ) = (13, 11)
20
3
20
— is y = —1x + —
The equation of the line perpendicular to PN
.
3
3
—=b
49. To determine the radius of the circle, the angle bisectors
would be used.
1
y
20
Find the intersection of y = −—2 x + 15 and y = —13x + —
.
3
E(4, 128)
10
8
6
4
⋅(
1
20
1
−—x + 15 = —x + —
2
3
3
1
1
20
−— x + 6 15 = 6 —x + 6 —
2
3
3
−3x + 90 = 2x + 40
)
⋅
⋅
⋅
−5x + 90 = 40
−5x = −50
2
D(0, 0)
6 C(8, 0) x
The radius of the circle is approximately 3 inches.
50. To determine the coordinates of the center of the circle and the
radius of the circle, use perpendicular bisectors.
Graph △LMN.
——
—
—
P(10, 20)
—
= √82 + 62 = √64 + 36 = √ 100 = 10
E(13, 11)
The radius of the circle is 10 units.
⋅ ⋅
⋅⋅
⋅⋅
⋅⋅
⋅⋅
⋅⋅
⋅
⋅ ⋅
1
AB AC
2
51. Total area of △BAC = —
F(6, 12)
8
−8
⋅
distance = √ (10 − 2)2 + (10 − 4)2
y
24
x = 10
1
y = −—x + 15
2
1
= −— 10 + 15 = −5 + 15 = 10
2
The coordinates of the center of the circle are (10, 10).
Distance between the center (10, 10) and T(2, 4):
8
T(2, 4)
16
24 x
N(16, 2)
20 − 4 16
—: m = —
=—=2
Slope of TP
10 − 2
8
— is m = −—1.
The slope of the line perpendicular to TP
2
10 + 2 20 + 4
12 24
—= —
, — = —, — = (6, 12)
midpoint of TP
2
2
2 2
y = mx + b
1
12 = −— 6 + b
2
12 = −3 + b
(
) (
)
⋅
15 = b
— is y = −—x + 15.
The equation of the line perpendicular to TP
2
18
20 − 2
—
Slope of PN : m = — = — = −3
10 − 16 −6
— is —1.
The slope of the line perpendicular to PN
3
1
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1
Area of △ADC = — x AC
2
1
Area of △BDC = — x BC
2
1
Area of △BDA = — x AB
2
1
1
1
1
— AB AC = — x AC + — x BC + — x AB
2
2
2
2
1
1
— AB AC = — x(AC + BC + AB)
2
2
1
1
2 — AB AC = 2 — x(AC + BC + AB)
2
2
AB AC = x(AC + BC + AB)
⋅ ⋅
⋅ ⋅
⋅ ⋅ ⋅
⋅
AB ⋅ AC
—— = x
(AC + BC + AB)
⋅⋅
—, BC
—, and
The expression for x in terms of the lengths of AB
AB AC
— is ——
.
AC
(AC + BC + AB)
⋅
Geometry
Worked-Out Solutions
197
Chapter 6
Maintaining Mathematical Proficiency
—
52. midpoint of AB =
(
58. 2x + 3y = 18
−3 + 3 5 + 5
0 10
—, — = —, — = (0, 5)
2
2
2 2
) ( )
———
3y = −2x + 18
2
18
y = −—3 x + —
3
—
—
AB = √( 3 − (−3) )2 + (5 − 5)2 = √ 62 + 0 2 = √36 = 6
—
( 2 +2 10 −1 2+ 7 ) ( 122 62 )
The slope of the new line is —32 .
53. midpoint of AB = —, — = —, — = (6, 3)
y = mx + b
———
AB = √( 10 − 2 )2 + ( 7 − (−1) )2
—
—
—
y = —32 x + b
—
= √82 + 82 = √64 + 64 = √ 128 = 8√ 2 ≈ 11.3
—
−5 + 4 1 + (−5)
2
2
1
−1 −4
= —, — = −—, −2
2
2 2
———
2
2
AB = √( 4 − (−5) ) + (−5 − 1)
(
(
) (
)
—
—
—
54. midpoint of AB = —, —
2
y = −—3 x + 6
⋅
−6 = —32 (−8) + b
)
−6 = −12 + b
6=b
The equation of the line passing through P(−8, −6) and
perpendicular to y = 2x + 1 is y = —32 x + 6.
= √ 92 + (−6)2 = √ 81 + 36 = √ 117 ≈ 10.8
—
12
( −72+ 5 5 +2 9 ) ( −22 142 )
8
55. midpoint of AB = —, — = —, — = (−1, 7)
AB = √( 5 − (−7) ) + (9 −
—
5)2
—
—
= √122 + 42 = √ 144 + 16 = √ 160 ≈ 12.6
56. The slope of the new
1
line is −—2.
10
y = mx + b
1
y = −—2x
1
8 = −—2
−12
y=
1
−2 x
+9
y = 2x + 1
2
9=b
The equation of the line
passing through P(2, 8)
and perpendicular to
1
y = 2x + 1 is y = −—2x + 9.
2
4
6
8
57. The line y = −5 is horizontal. The equation of the line
10 x
passing through P(6, −3) and perpendicular to y = −5 is
x = 6. The slope of the perpendicular line is undefined so
the equation is vertical.
y
2
4
x=6
−2
−4
−6
198
x
12 x
8
−12
59. y + 3 = −4(x + 3)
4
8 = −1 + b
4
−4
−8
P(2, 8)
6
⋅2 + b
−8
P(−8, −6)
y
8
+b
2x + 3y = 18
3
y = 2x + 6
——
2
y
y + 3 = −4x − 12
y = −4x − 15
The slope of the new line is —14 .
y = mx + b
y
y = —14x + b
4
⋅
1 = —14 (−4) + b
1 = −1 + b
1
y = 4x + 2
2=b
The equation of the line
passing through P(−4, 1)
and perpendicular to
y + 3 = −4(x + 3)
is y = —14 x + 2.
P(−4, 1)
−8
−6
−4
−2
2 x
−2
−4
y + 3 = −4(x + 3)
P(6, −3)
y = −5
Geometry
Worked-Out Solutions
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Chapter 6
6.3 Explorations (p. 319)
2. a. Check students’ work.
1. a. Draw △ABC and plot the midpoints of each side and
b. Sample answer:
construct the medians. Sample answer:
7
7
6
6
B
5
5
F
A
4
B
G
0
C
0
4
F
0
1
3
A
1
E
2
G
2
D
2
1
D
3
3
0
E
4
5
6
7
8
9
3
the triangle.
c. Segment AD is divided into AG ≈ 4 and GD ≈ 2. So, the
ratio is about —12 . Segment BE is divided into BG ≈ 2 and
GE ≈ 1, so the ratio is about —12 . The ratio of the length of
the shorter segment to the length of the longer segment is
1 : 2 or —12.
AD ≈ 6, AG ≈ 4; The ratio is —23 .
5
C
6
7
8
9
c. The altitudes that connect a vertex and a point on the
opposite side are all perpendicular to that side. If the
triangle is acute, the altitudes meet inside of the triangle.
If the triangle is a right triangle, the legs of the right
triangle are the altitudes, and therefore, meet at a point on
the triangle. If the triangle is obtuse, the altitudes meet at
a point on the outside of the triangle.
3. The medians meet at a point inside the triangle that divides
2
—3 .
The ratio of the length of the longer segment to the length
of the whole median is 2 : 3 or —23. If the shorter segment of
the median is a and the longer segment is b, the shorter
(
)
segment is —12 of the longer segment a = —12 b . The median
3
equals a + b = —12 b + b = —12b + —22b = —2 b. Because the
3
median is —2 b, by multiplying by the reciprocal, the longer
each median into two segments whose lengths have the ratio
1 : 2. The altitudes meet at a point inside, on, or outside the
triangle depending on whether the triangle is acute, right, or
obtuse.
—
4. The two segments of RU have lengths of 1 inch and 2 inches.
6.3 Monitoring Progress (pp. 321–323)
⋅ 2100 = 700 ft
PC = ⋅ 2100 = 1400 ft
2.
BT = — ⋅ BC
1000 = — ⋅ BC
2 ⋅ 1000 = 2 ⋅ — ⋅ BC
1
1. PS = —3
2
—3
1
2
1
2
1
2
2000 = BC
So, BC is 2000 feet.
TC =
TC =
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4
The altitudes meet at the same point.
b. The medians of a triangle are concurrent at a point inside
segment, b, is —23 of the median.
2
−1
−1
BE ≈ 3, BG ≈ 2; The ratio is
1
1
—2
1
—2
⋅ BC
⋅ 2000
3.
⋅
⋅
2 ⋅ 800 = 2 ⋅ — ⋅ PA
PT = —12 PA
800 = —12 PA
1
2
1600 = PA
So, PA is 1600 feet.
TA = PT + PA
TA = 800 + 1600
TC = 1000
TA = 2400
So, TC is 1000 feet.
So, TA is 2400 feet.
Geometry
Worked-Out Solutions
199
Chapter 6
4.
y
6.
G(4, 9)
A(0, 3)
8
6
4
y
4
2
F(2, 5)
M(5, 5)
−2
P(4, 5)
2
B(0, −2)
H(6, 1)
2
4
x
6
4+6 9+1
10 10
— is —
, — = —, — = (5, 5).
The midpoint of GH
2
2
2 2
2 —
The centroid is — of FM .
3
(
) (
)
——
FM = √(5 − 2)2 + (5 − 5)2
= √32 = √ 9 = 3
⋅
The centroid is —23 3 = 2, so the centroid P is 2 units to
the right of F, which are the coordinates (4, 5). So, the
coordinates of the centroid of △FGH are (4, 5).
5.
L(−1, 4)
X(−3, 3)
P(−1, 2)
−5
5
y Y(1, 5)
y = mx + b
3=b
⋅
— is y = 6x + 3.
The equation of the line perpendicular to BC
−3 − 3 −6
—= —
= — = −1
slope AC
6−0
6
— is 1 and the line
The slope of the line perpendicular to AC
passes through vertex B(0, −2).
−2 = b
3 x
1
⋅
−2 = 1 0 + b
— is y = x − 2.
The equation of the line perpendicular to AC
Z(−1, −2)
Find the intersection of y = 6x + 3 and y = x − 2.
1 + (−1) 5 + (−2)
0 3
— is —
, — = —, — = (0, 1.5).
The midpoint of YZ
2
2
2 2
1
−1.5
1.5 − 3
—
The slope of XJ is — = — = −—.
2
0 − (−3)
3
y = mx + b
1
1.5 = −— 0 + b
2
1.5 = b
(
) ( )
⋅
3
b=—
2
— though J(0, 1.5) is y = −—1x + —3.
The equation of XJ
2
2
−3 + 1 3 + 5
−2 8
—
The midpoint of XY is —, — = —, — = (−1, 4).
2
2
2 2
6
4 − (−2)
—
The slope of ZL is — = — = undefined.
−1 − (−1) 0
— through J(−1, 4) is x = −1.
The equation of ZL
(
) (
)
The centroid has the coordinates of the intersection of
1
3
y = −—x + — and x = −1.
2
2
1
3
y = −—x + —
2
2
1
3
y = −—(−1) + —
2
2
1 3 4
y=—+—=—=2
2 2 2
So, the centroid has coordinates (−1, 2).
200
1
−3 − (−2)
—=—
= −—
slope of BC
6
6−0
— is 6 and the line
The slope of the line perpendicular to BC
passes through vertex A(0, 3).
y = mx + b
3
J(0, 1.5)
−3
C(6, −3)
3=6 0+b
—
—
6 x
2
Geometry
Worked-Out Solutions
6x + 3 = x − 2
5x + 3 = −2
5x = −5
x = −1
⋅
y = 6 (−1) + 3 = −6 + 3 = −3
So, the orthocenter is outside the triangle and the coordinates
are (−1, −3).
7. △JKL is a right triangle; therefore, the orthocenter is on the
triangle at the right angle (−3, 4).
K(−3, 4) y
L(5, 4)
3
−1
3
5 x
−3
J(−3, −4)
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Chapter 6
8. Proving △ABD ≅ △CBD by the SSS Congruence Theorem
(Thm. 5.8) at the beginning of the proof would be the same.
But then you would state that ∠ ABD ≅ ∠ CBD because
corresponding parts of congruent triangles are congruent.
— is also an angle bisector by definition.
This means that BD
⋅ CE
5 = — ⋅ CE
3 ⋅ 5 = 3 ⋅ — ⋅ CE
1
7. DE = —3
1
3
1
3
CD =
Vocabulary and Core Concept Check
1. The four types of triangular concurrencies are circumcenters,
incenters, centroids, and orthocenters. The circumcenters are
formed by the intersection of the perpendicular bisectors.
The incenters are formed by the intersection of the angle
bisectors. The centroids are formed by the intersection of the
medians. The orthocenters are formed by the intersection of
the altitudes.
2. The length of a segment from a vertex to the centroid is
CD =
Monitoring Progress and Modeling with Mathematics
PN =
PN =
4. PN =
⋅
PN =
6
PN = 6 units
QP =
QP =
1
—3
1
—3
QP = 3
⋅ QN
⋅9
QP = 3 units
2
5. PN = —3 QN
PN =
PN =
⋅ 30
2
—3
60
—
3
= 20
PN = 20 units
QP =
QP =
1
—3
1
—3
⋅
⋅ 30
QN
33 = CE
PN =
2
—3 QN
2
—3 21
42
= 14
—
3
⋅
PN = 14 units
QP =
QP =
1
—3
1
—3
QP = 7
⋅ QN
⋅ 21
2
PN =
PN =
⋅ 42
2
—3
84
—
3
= 28
PN = 28 units
QP =
QP =
1
—3
1
—3
⋅
⋅ 42
QN
QP = 10
QP = 14
QP = 10 units
QP = 14 units
⋅ CE
⋅ 15
⋅
⋅
CD = —23 CE
CD = —23 33
66
CD = —
= 22
3
CD = 10 units
CD = 22 units
⋅ CE
9 = — ⋅ CE
3 ⋅ 9 = 3 ⋅ — ⋅ CE
1
9. DE = —3
10.
1
3
1
3
27 = CE
45 = CE
CE = 27 units
CD =
CD =
CE = 45 units
⋅ CE
⋅ 27
2
—3
2
—3
54
—
3
⋅
⋅
3 ⋅ 15 = 3 ⋅ — ⋅ CE
DE = —13 CE
15 = —13 CE
1
3
⋅
⋅
CD = —23 CE
CD = —23 45
= 18
90
CD = —
= 30
3
CD = 18 units
CD = 30 units
CD =
—
—
midpoint of AC . Therefore, FC = 12 units.
11. Because G is a centroid, BF is a median and F is the
—
⋅ BF.
2
12. Because G is a centroid, BF is a median and BG is —3
BG =
QP = 7 units
6. PN = —3 QN
CE = 33 units
= 10
CD =
two-thirds the length of the median from that vertex.
3. PN =
1
3
15 = CE
2
—3
2
—3
30
—
3
⋅
⋅
3 ⋅ 11 = 3 ⋅ — ⋅ CE
DE = —13 CE
11 = —13 CE
CE = 15 units
6.3 Exercises (pp. 324–326)
2
—3 QN
2
—3 9
18
=
—
3
8.
6=
3
—2
⋅6 =
2
—3
2
—3
3
—2
⋅ BF
⋅ BF
⋅ — ⋅ BF
2
3
9 = BF
Therefore, BF = 9 units.
—
2
13. Because G is a centroid, AE is a median and AG = —3
AG =
AG =
2
—3
2
—3
⋅ AE
⋅ 15
⋅ AE.
AG = 10
Therefore, AG = 10 units.
—
1
14. Because G is a centroid, AE is a median and GE = —3
GE =
GE =
1
—3
1
—3
GE = 5
⋅ AE
⋅ 15
⋅ AE.
Therefore, GE = 5 units.
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Geometry
Worked-Out Solutions
201
Chapter 6
15.
8
6
4
16.
y
( )
(
F(6.5, 4)
A(2, 3)
2
G(−2, 7)
7
−3 , 5
C(5, 7)
11
5, 3
)
y
8
I(−4, 5)
F(1, 5)
4
E(−2.5, 4)
H(−6, 3)
D(5, 2)
2
B(8, 1)
4
2
6
−6
8 x
⋅
⋅ ⋅
⋅
) ( )
2 x
( −6 +2(−2), 3 +2 7 ) = ( −82 , 102 ) = (−4, 5).
— —
— —
0
5−5
— is —
= — = 0.
The slope of FI
1 − (−1) 2
— through I(−4, 5) is y = 5.
The equation of FI
2
4 = — 6.5 + b
9
2
9 4 = 9 — 6.5 + 9 b
9
36 = 2 6.5 + 9b
⋅
−2
— is
The midpoint of HG
5+8 7+1
13 8
— is —
, — = —, — = (6.5, 4).
The midpoint of CB
2
2
2 2
1
2
4−3
—
The slope of AF is — = — = —.
6.5 − 2 4.5 9
y = mx + b
(
−4
−6 + 1 3 + 5
−5 8
— is —
The midpoint of HF
, — = —, — = (−2.5, 4).
2
2
2 2
3
3
7−4
—
The slope of GE is —— = — = — = 6.
−2 − (−2.5) −2 + 2.5 0.5
y = mx + b
) (
(
⋅
36 = 13 + 9b
)
⋅
23 = 9b
4 = 6 (−2.5) + b
4 = −15 + b
19 = b
— through E(−2.5, 4) is y = 6x + 19.
The equation of GE
The centroid has the coordinates of the intersection of
y = 6x + 19 and y = 5.
5 = 6x + 19
−14 = 6x
7
14
x = −— = −—
6
3
7
The centroid has coordinates −—, 5 .
3
23
—=b
9
23
— through F(6.5, 4) is y = —2x + —
.
The equation of AF
9
9
2+8 3+1
10 4
— is —
, — = —, — = (5, 2).
The midpoint of AB
2
2
2 2
2 − 3 −1
—
The slope of CD is — = — = undefined.
5−5
0
(
) ( )
— through D(5, 2) is x = 5.
The equation of CD
(
The centroid has the coordinates of the intersection of
23
y = —29x + —
and x = 5.
9
17.
23
2
y = —x + —
9
9
2
23
y=— 5+—
9
9
10 23
y=—+—
9
9
33 11
y=—=—
9
3
y
)
S(5, 5)
4
⋅
2
E(8, 1)
(5, 1)
U(−1, 1)
−4
(
)
11
The centroid has coordinates 5, —
.
3
12 x
8
−2
D(5, −1)
T(11, −3)
16 2
5 + 11 5 + (−3)
— is —
, — = —, — = (8, 1).
The midpoint of ST
2
2 2
2
0
1−1
—
The slope of UE is — = — = 0.
8 − (−1) 9
— through E(8, 1) is y = 1.
The equation of UE
(
) ( )
— is
The midpoint of UT
( −1 +2 11, 1 + 2(−3) ) = ( 102 , −22 ) = (5, −1).
— —
— —
−1 − 5 −3
— is —
= — = undefined.
The slope of SD
5−5
0
—
The equation of SD through D(5, −1) is x = 5.
The centroid has the coordinates of the intersection of x = 5
and y = 1, which is (5, 1).
202
Geometry
Worked-Out Solutions
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Chapter 6
18.
4
3
19.
y X(1, 4)
L(0, 5) y
4
D(4, 3)
y=1
Z(2, 3)
2
(103, 3)
2
Y(7, 2)
E(4.5, 2.5)
1
M(3, 1)
−2
4
N(8, 1)
8 x
6
−2
2
4
y = 2x − 5
8 x
6
−4
8 6
1+7 4+2
— is —
, — = —, — = (4, 3).
The midpoint of XY
2
2 2
2
3−3 0
— is —
= — = 0.
The slope of ZD
4−2 2
— through D(4, 3) is y = 3.
The equation of ZD
(
) ( )
9 5
2+7 3+2
— is —
, — = —, — = (4.5, 2.5).
The midpoint of ZY
2
2 2
2
15
3
1.5
4 − 2.5
—
The slope of XE is — = — = −— = −—.
35
7
1 − 4.5 −3.5
y = mx + b
3 9
2.5 = −— — + b
7 2
27
5
— = −— + b
14
2
27
5
14 — = 14 −— + 14b
14
2
35 = −27 + 14b
(
) ( )
⋅
⋅
⋅(
)
62 = 14b
y = mx + b
y = 2x + b
⋅
1=2 3+b
1=6+b
−5 = b
— containing point
The equation of the line perpendicular to LN
(3, 1) is y = 2x − 5.
y = 2x − 5
31
— through E(4.5, 2.5) is y = −—3x + —
.
The equation of XE
7
7
The centroid has the coordinates of the intersection of
3
31
y = −—x + — and y = 3.
7
7
3
31
y = −—x + —
7
7
3
31
3 = −—x + —
7
7
3
31
7 3 = 7 −— x + 7 —
7
7
21 = −3x + 31
⋅(
1−1 0
— is —
= — = 0.
The slope of the line containing MN
8−3 5
— is undefined and
The slope of the line perpendicular to MN
passes through (0, 5). Therefore, the equation of that line
is x = 0.
1
4
5−1
— is —
= — = −—.
The slope of the line containing LN
2
0 − 8 −8
— is 2.
The slope of the line perpendicular to LN
The orthocenter is the intersection of x = 0 and y = 2x − 5.
62 31
b=—=—
14
7
⋅
(0, −5)
)
⋅(
)
−10x =−3x
⋅
y=2 0−5
y = −5
The orthocenter of △LMN is located on the outside and the
coordinates are (0, −5).
20. △XYZ is a right triangle; therefore, the orthocenter is on the
triangle at the intersection of the legs.
y
Z(−3, 6)
6
4
Y(5, 2)
X(−3, 2)
10
—=x
3
10
The centroid has coordinates —, 3 .
3
( )
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−2
2
4
x
The coordinates of the orthocenter are (−3, 2).
Geometry
Worked-Out Solutions
203
Chapter 6
21.
C(−1, 3)
(−1, 2)
−2
6
4
(0, 73)
B(1, 0)
A(−4, 0)
−6
22.
y
x = −1
4
y
V(0, 4)
T(−2, 1)
U(2, 1)
2 x
−2
2
8
y = 3x + 3
0
0−0
— is —
= — = 0.
The slope of the line containing AB
1 − (−4) 5
— is undefined and
The slope of the line perpendicular to AB
passes through (−1, 3). Therefore, the equation of that line is
x = −1.
3
3
3−0
— is —
= — = −—.
The slope of the line containing CB
2
−1 − 1 −2
— is —2.
The slope of the line perpendicular to CB
3
y = mx + b
⋅
2
0 = — (−4) + b
3
8
0 = −— + b
3
8
—=b
3
— containing
The equation of the line perpendicular to CB
8
2
point (−4, 0) is y = —x + —.
3
3
The orthocenter is the intersection of x = −1 and
8
2
y = —x + —.
3
3
8
2
y = —x + —
3
3
2
8
y = — (−1) + —
3
3
2 8
y = −— + —
3 3
6
y=—=2
3
The orthocenter of △ABC is inside the triangle with
coordinates (−1, 2).
⋅
2
2
4 x
7
y = 3x + 3
0
1−1
— is —
= — = 0.
The slope of the line containing TU
2 − (−2) 4
— is undefined and
The slope of the line perpendicular to TU
passes through (0, 4). Therefore, the equation of that line
is x = 0.
3
3
4−1
— is —
= — = −—.
The slope of the line containing VU
2
0 − 2 −2
— is —2.
The slope of the line perpendicular to VU
3
y = mx + b
⋅
2
1 = — (−2) + b
3
4
1 = −— + b
3
4
3 1 = 3 −— + 3 b
3
3 = −4 + 3b
⋅(
⋅
)
⋅
7 = 3b
7
b=—
3
— containing
The equation of the line perpendicular to VU
7
2
point (−2, 1) is y = —x + —.
3
3
7
2
The orthocenter is the intersection of x = 0 and y = —x + —.
3
3
7
2
y = —x + —
3
3
2
7
y=— 0+—
3
3
7
y=—
3
The orthocenter of △TUV is inside the triangle with
7
coordinates 0, — .
3
⋅
( )
23. Construct the medians of an isosceles right triangle by
finding the midpoint of each side and connecting the
midpoint and the vertex opposite that midpoint. Where
the medians intersect is the location of the centroid. The
orthocenter is on the triangle at the right angle.
centroid
orthocenter
204
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Chapter 6
24. Construct the medians of an obtuse scalene triangle by
finding the midpoint of each side and connecting the
midpoint and the vertex opposite that midpoint. Where the
medians intersect is the location of the centroid. Construct
the altitudes by drawing a perpendicular segment from each
vertex to the opposite side or to the line that contains the
opposite side. The orthocenter is located outside the triangle
where the altitudes intersect.
Sample answer:
—
—
1
27. The length of DE should be —3 of the length of AE because it
is the shorter segment from the centroid to the side.
DE = —13 AE
DE = —13 (18)
DE = 6
—
—
1
1
28. The length of DE is —2 of the length of AD because DE = —3 AE
and AD = —23 AE.
DE = —12 AD
centroid
DE = —12 (24)
DE = 12
orthocenter
29. Given
Prove
25. Construct the medians of a right scalene triangle by finding
— is an angle bisector of ∠ ABC.
Isosceles △ABC, AD
— is a median.
BD
B
the midpoint of each side and connecting the midpoint
and the vertex opposite that midpoint. Where the medians
intersect is the location of the centroid. The orthocenter is on
the triangle at the right angle.
Sample answer:
A
centroid
orthocenter
26. Construct the medians of an acute isosceles triangle by finding
the midpoint of each side and connecting the midpoint and the
vertex opposite that midpoint. Where the medians intersect is
the location of the centroid. Construct the altitudes by drawing
a perpendicular segment from each vertex to the opposite side.
The orthocenter is where the altitudes intersect.
Sample answer:
centroid
D
C
STATEMENTS
REASONS
1. △ABC is an
isosceles triangle.
1. Given
2. —
AB ≅ —
BC
3. —
AD is an angle bisector
of ∠ ABC.
3. Given
4. △ABD ≅ ∠ CBD
4. Definition of angle bisector
5.
5. Reflexive Property of
Congruence (Thm. 2.1)
—
BD ≅ —
BD
6. △ABD ≅ △CBD
6. SAS Congruence
Theorem (Thm. 5.5)
7. —
AD ≅ —
CD
orthocenter
8. D is the midpoint of —
AC .
9.
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2. Definition of isosceles
triangle
—
BD is a median.
7. Corresponding parts of
congruent triangles are
congruent.
8. Definition of midpoint
9. Definition of median
Geometry
Worked-Out Solutions
205
Chapter 6
30. Given
Prove
Isosceles △ABC,
— is an altitude to AC
—.
BD
35. The centroid and orthocenter are sometimes the same point.
B
The centroid and the orthocenter are not the same point
unless the triangle is equilateral.
— is a perpendicular bisector.
BD
36. The centroid is always formed by the intersection of the
three medians. This is the definition of a centroid.
A
STATEMENTS
1. △ABC is an
isosceles triangle.
2. —
AB ≅ —
BC
3. —
BD is an
altitude to —
AC .
D
C
1. Given
2. Definition of isosceles
triangle
5. ∠ ADB and ∠ CDB
are right angles.
5. Definition of perpendicular
8.
—
AD ≅ —
CD
9. D is the midpoint
of —
AC .
BD is a
10. —
perpendicular
bisector.
38. All are segments that pass through the vertex of a triangle.
3. Given
4. Definition of altitude
7. △ABD ≅ △CBD
their point of intersection can fall either inside, on, or outside
of the triangle. However, the altitude does not necessarily
bisect the side, but the perpendicular bisector does. Also, the
perpendicular bisector does not necessarily pass through the
opposite vertex, but the altitude does.
REASONS
4. —
BD ⊥ —
AC
6. —
BD ≅ —
BD
37. Both segments are perpendicular to a side of a triangle, and
6. Reflexive Property of
Congruence (Thm. 2.1)
7. HL Congruence Theorem
(Thm. 5.9)
8. Corresponding parts of
congruent triangles are
congruent.
A median connects a vertex with the midpoint of the opposite
side. An altitude is perpendicular to the opposite side. An
angle bisector bisects the angle through which it passes.
The medians of a triangle intersect at a single point, and the
same is true for the altitudes and angle bisectors of a triangle.
Medians and angle bisectors always lie inside the triangle, but
altitudes may be inside, on, or outside of the triangle.
1
39. Area = —2 bh
The area of the triangle in solid red is
1
—2
⋅ — ⋅ 3 = — = 6.75 square inches.
9
2
27
4
The special segment of the triangle used was the altitude.
——
—
40. K is the centroid and DH , EJ , and FG are medians.
9. Definition of midpoint
a. EJ = 3KJ
F
b. DK = 2KH
10. Definition of perpendicular
bisector
3
c. FG = —2 FK
32. The orthocenter is sometimes outside the triangle. An
orthocenter can be inside, on, or outside the triangle
depending on whether the triangle is acute, right, or obtuse.
33. A median is sometimes the same line segment as a
perpendicular bisector. A median is the same line segment
as the perpendicular bisector if the triangle is equilateral
or if the segment is connecting the vertex angle to the base
of an isosceles triangle. Otherwise, the median and the
perpendicular bisectors are not the same segment.
34. An altitude is sometimes the same line segment as an angle
K
H
D
31. The centroid is never on the triangle. Because medians are
always inside a triangle, and the centroid is the point of
concurrency of the medians, it will always be inside the
triangle.
J
1
d. KG = —3 FG
G
41.
BD =
4x + 5 =
2
—3 BF
2
—3 9x
42.
⋅
2x − 8 =
4x + 5 = 6x
3)
2x − 8 = x + 1
5 = 2x
5
—2
GD =
E
1
—3 GC
1
—3 (3x +
x=9
=x
43. AD = 2DE
44.
DF = —12 BD
5x = 2(3x − 2)
4x − 1 = —12 (6x + 4)
5x = 6x − 4
4x − 1 = 3x + 2
−x = −4
x=3
x=4
bisector. An altitude is the same line segment as the angle
bisector if the triangle is equilateral or if the segment is
connecting the vertex angle to the base of an isosceles
triangle. Otherwise, the altitude and the angle bisector are
not the same segment.
206
Geometry
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Chapter 6
45.
51. Given
y
median
8
3
y2 = 4 x + 5
Prove
— is a median.
△ABC is an equilateral triangle. BD
— is angle bisector,
BD
6
4
A
median
(0, 2)
STATEMENTS
−8
−6
−4
2
median
B
perpendicular bisector,
and altitude.
4
x
6
−2
y1 = 3x − 4
−4
y3 = −2 x − 4
y1 = 3x − 4: Graph the x-intercept
y-intercept (0, −4).
1. Given
2. —
AB ≅ —
AC ≅ —
BC
4.
(
5.
)
20
y2 = —34x + 5: Graph the x-intercept −—
3 , 0 and the
y-intercept (0, 5).
(
)
2. Definition of equilateral
triangle
3. —
BD is a median.
( 0 ) and the
4
—3 ,
C
REASONS
1. △ABC is an
equilateral triangle.
3
D
3. Given
—
AD ≅ —
CD
—
BD ≅ —
BD
4. Definition of median
5. Reflexive Property of
Congruence (Thm. 2.1)
y3 = −—2x − 4: Graph the x-intercept −—3 , 0 and the
y-intercept (0, −4).
6. △ABD ≅ △CBD
6. SSS Congruence
Theorem (Thm. 5.8)
The points of intersection that form the triangle are (4, 8),
(−4, 2), and (0, −4). The equation of the median from y1
to the opposite vertex is y = 2. The equation of the median
from y2 to the opposite vertex is x = 0. The coordinates of
the centroid are (0, 2).
7. ∠ADB ≅ ∠CDB,
∠ABD ≅ ∠CBD
7. Corresponding parts of
congruent triangles are
congruent.
8. ∠ADB and ∠CDB
form a linear pair and
are supplementary.
8. Linear Pair Postulate
9. ∠ADB and ∠CDB
are right angles.
9. If two angles are congruent
and supplementary then
they are right angles.
3
8
46. right triangle; The orthocenter of a right triangle is the vertex
of the right angle.
1
1
47. PE = —3 AE, PE = —2 AP, PE = AE − AP
—
—
b. KN is an altitude. It contains the orthocenter.
48. a. KM is a median. It contains the centroid.
c. The area of △JKM is
1
—2
⋅⋅
1
—2
9
—2 h,
⋅9 ⋅h =
9
—2 h
10. —
BD ⊥ —
AC
10. Definition of perpendicular
AD ≅ —
CD
11. —
11. Corresponding parts of
congruent triangles are
congruent.
and the area of
△KLM is 9 h =
which indicates that the two
areas are equal. Yes, triangles formed by the median will
always have the same area because they will have the
same base length and height.
49. yes; If the triangle is equilateral, then the perpendicular
bisectors, angle bisectors, medians, and altitudes will all be
the same three segments.
50. centroid; Because the triangles formed by the median of any
triangle will always be congruent, the mass of the triangle on
either side of the median is the same. So, the centroid is the
point that has an equal distribution of mass on all sides.
12. D is the midpoint
BC .
of —
AD is a perpendicular 13. Definition of
13. —
bisector.
perpendicular bisector
AD is an altitude.
14. —
14. Definition of altitude
52. The orthocenter, circumcenter, and the centroid are all inside
the triangle. They are all three distinct points. The three
concurrent points are all collinear.
y
A
4
orthocenter
3
circumcenter
2
1
B
centroid 1
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12. Definition of midpoint
C
2
3
4
5
x
Geometry
Worked-Out Solutions
207
Chapter 6
53. Sample answer: The circle passes through nine significant
points of the triangle. They are the midpoints of the sides, the
midpoints between each vertex and the orthocenter, and the
points of intersection between the sides and the altitudes.
y
A
9
7
K
I
J
5
D
4
1.
E
3. Vertical Angles
Congruence Theorem
(Thm. 2.6)
4. △LPM ≅ △RPN,
△MQL ≅ △SQN
4. SAS Congruence
Theorem (Thm. 5.5)
L
1
1
54. Given
Prove
2
3
C
4
5
6
x
— and MQ
— are medians of scalene △LMN. Point R
LP
— ≅ PR
—. Point S is on ⃗
⃗
is on LP such that LP
MQ such
— ≅ QS
—.
that MQ
— ≅ NR
—
a. NS
— and NR
— are both parallel to LM
—.
b. NS
c. R, N, and S are collinear.
R
M
6. —
NS ≅ —
NR
H
G
2. Definition of median
3. ∠ LPM ≅ ∠ RPN,
∠ MQL ≅ ∠ SQN
5. —
NR ≅ —
LM , —
NS ≅ —
LM
F
B
1. Given
medians of scalene
△LMN; —
LP ≅ —
PR ,
—
—
MQ ≅ QS
3
2
REASONS
—
LP and —
MQ are
NP ≅ —
MP , —
LQ ≅ —
NQ
2. —
8
6
a. STATEMENTS
5. Corresponding parts of
congruent triangles are
congruent.
6. Transitive Property of
Congruence (Thm. 2.1)
b. It was shown in part (a) that △LPM ≅ △RPN and
△MQL ≅ △SQN. So, ∠ LMP ≅ ∠ RNP and
∠ MLQ ≅ ∠ SNQ because corresponding parts of
— LM
— and
congruent triangles are congruent. Then, NS
—
—
NR LM by the Alternate Interior Angles Converse (Thm. 3.6).
— —
—
LM , they would have to be parallel to each other by the
c. Because NS and NR are both parallel to the same segment,
Transitive Property of Parallel Lines (Thm. 3.9). However,
because they intersect at point N, they cannot be parallel.
So, they must be collinear.
Maintaining Mathematical Proficiency
—
−3 1
3−6
−1 − 5 −6 2
−6
−6
1
3−9
—: ——
=—=—=—
Slope of CD
−16 − (−4) −16 + 4 −12 2
— and CD
— are equal, so AB
— CD
—.
The slopes of AB
55. Slope of AB : — = — = —
P
N
Q
L
S
—
1
4−6
−2
−2
4
5 − (−3) 5 + 3
8
3
1
−7 − (−10) −7 + 10
—
Slope of CD : —— = — = — = —
−2 − (−14) −2 + 14 12 4
56. Slope of AB : — = — = — = −—
— and CD
— are not equal, so AB
— is not parallel
The slopes of AB
—
to CD .
—
5
2 − (−3) 2 + 3
5−6
−1
−1
2+4
6
2 − (−4)
—: —
= — = — = −6
Slope of CD
−5 − (−4) −5 + 4 −1
57. Slope of AB : — = — = — = −5
— and CD
— are not equal, so AB
— is not parallel
The slopes of AB
—
to CD .
—
−4
2−6
−7 − (−5) −7 + 5
−5 − 1 −6
—: —
=—=2
Slope of CD
4−7
−3
−4
−2
58. Slope of AB : — = — = — = 2
— and CD
— are equal, so AB
— is parallel to CD
—.
The slopes of AB
208
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Chapter 6
6.1– 6.3 What Did You Learn? (p. 327)
4. Graph △ABC.
1. Sample answer: yes; You realize that if you construct the
—, you have created two isosceles triangles that
segment AC
share vertices A and B, and a third triangle that shares the
same vertices. Then you look back at the Perpendicular
Bisector Theorem (Thm. 6.1) and its converse to see that points
D and E would have to be on the perpendicular bisector of
—. Then, in the same way, in order for B to also be on the same
AC
— and CB
— would have to be congruent.
perpendicular bisector, AB
2. Sample answer: These can be constructed with a compass
and a straightedge, or they can be constructed with geometry
software; If you construct them with geometry software, you
create a triangle that fits the description first. Then use the
software to draw the perpendicular bisectors of each side.
Next, label the point where these three lines meet. Finally,
draw a circle with its center at this point of intersection
that passes through one vertex of the triangle. It will
automatically pass through the other two vertices.
3. Sample answer: Right triangles are the only kind of
triangles that have one of the points of intersection on the
vertex of the triangle; While all segment types can be inside
the triangle, only the perpendicular bisectors and altitudes
can be on or outside the triangle.
6.1–6.3 Quiz (p. 328)
— — VW ⊥ ⃖⃗
1. Because SW ≅ UW and ⃖⃗
SU, point V is on the
—.
perpendicular bisector of SU
SV = UV
Perpendicular Bisector Theorem (Thm. 6.1)
2x + 11 = 8x − 1
−6x + 11 = −1
−6x = −12
x=2
⋅
UV = 8x − 1 = 8 2 − 1 = 16 − 1 = 15
—
—
2. ⃗
SQ is an angle bisector of ∠ PSR, PQ ⊥ ⃗
SP , and RQ ⊥ ⃗
SR .
PQ = RQ
Angle Bisector Theorem (Thm. 6.3)
6x = 3x + 9
3x = 9
x=3
⋅
PQ = 6x = 6 3 = 18
3. Because J is equidistant from ⃗
GH and ⃗
GK, ⃗
GJ bisects
∠ HGK by the Angle Bisector Theorem (Thm. 6.3).
D(−2, −1)
A(−4, 2)
y
2
−6
2 x
(−4, −1)
−2
C(0, −4)
B(−4, −4)
(−2, −4)
−4 + (−4) 2 + (−4)
—= —
,—
midpoint of AB
2
2
−8 −2
= —, — = (−4, −1)
2 2
−4 + 0 2 + (−4)
—
midpoint of AC = —, —
2
2
−4 −2
= —, — = (−2, −1)
2 2
— through its
The equation of the perpendicular bisector of AB
midpoint (−4, −1) is y = −1.
3
6
2 − (−4)
—= —
= — = −—
slope of AC
2
−4 − 0
−4
— is —2.
The slope of the perpendicular line to AC
3
(
(
(
(
)
)
)
)
y = mx + b
2
−1 = — (−2) + b
3
4
−1 = −— + b
3
4
3 (−1) = 3 −— + 3b
3
−3 = −4 + 3b
⋅
⋅
⋅(
)
1 = 3b
1
—=b
3
— through its
The equation of the perpendicular bisector of AC
2
1
midpoint (−2, −1) is y = —3x + —3 .
Find the point of intersection of y = −1 and y = —23x + —13.
1
2
y = —x + —
3
3
2
1
−1 = —x + —
3
3
−3 = 2x + 1
−4 = 2x
x = −2
The coordinates of the circumcenter of △ABC are (−2, −1).
m∠ HGJ = m∠ JGK
5x − 4 = 4x + 3
x−4=3
x=7
⋅
m∠ JGK = 4x + 3 = 4 7 + 3 = 28 + 3 = 31°
31° + m∠ GJK + 90° = 180°
m∠ GJK + 121° = 180°
Triangle Sum Theorem
(Thm. 5.1)
m∠ GJK = 59°
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Geometry
Worked-Out Solutions
209
Chapter 6
5. Graph △DEF.
16
9.
K(5, 6)
H(3, 2)
D(5, 2)
4
E(7, 9)
8
4
J(−1, 2)
G(9, 7)
F(11, 5)
(7, 5)
4 D(3, 5)
−2
(
— —
)(
)
(
⋅
7 = −5 + b
12 = b
— through the
The equation of the line perpendicular to DE
midpoint (5, 7) is y = −x + 12. The intersection of x = 7
and y = −x + 12 is y = −7 + 12 = 5. The coordinates of
the circumcenter of △DFE are (7, 5).
NU = NV
7.
−3x + 6 = −5x
−2x + 1 = −9
6 = −2x
−2x = −10
−3 = x
10.
y
−8
NU = NV = NT
NQ = NR = NS
⋅
= −3 (−3) + 6
=2 5+1
=9+6
= 11
= 15
⋅
NZ = NY
4x − 10 = 3x − 1
−6
−4
−2
N(−4, −2)
H(−4, −4)
D(−6, −4)
= −2x + 6
= 2x + 1
M(−8, −6)
2 x
−2
⋅
NZ = NY = NW = 4x − 10 = 4 9 − 10 = 36 − 10 = 26
P(0, −4)
F(−4, −5)
−8
— is
The midpoint D of MN
−4 + (−8) −2 + (−6)
−12 −8
—, — = —, — = (−6, −4).
2
2
2
2
−4 − (−4) 0
— is —
= — = 0.
The slope of DP
0 − (−6)
6
— through D(−6, −4) is y = −4.
The equation of DP
(
x − 10 = −1
x=9
) ( )
⋅
7 = −1 5 + b
x=5
— —
0
2−2
— is —
= — = 0.
The slope of JD
5 − (−1) 6
— through D(5, 2) is y = 2.
The equation of JD
— is
The midpoint F of JL
5 + (−1) −2 + 2
4 0
—, — = —, — = (2, 0).
2
2
2 2
6−0 6
— is —
= — = 2.
The slope of KF
5−2 3
y = mx + b
0=2 2+b
0=4+b
−4 = b
— through F(2, 0) is y = 2x − 4.
The equation of KF
The centroid has the coordinates of the intersection of y = 2
and y = 2x − 4.
2 = 2x − 4
6 = 2x
3=x
The coordinates of the centroid are (3, 2).
y = mx + b
2x + 1 = 4x − 9
L(5, −2)
( 5 +2 5, 6 + 2(−2) ) = ( 102 , 42 ) = (5, 2).
3 + 7 5 + 9 10 14
—= —
, — , —, — = (5, 7)
midpoint of DE
2
2
2 2
9−5 4
—
slope of DE = — = — = 1
7−3 4
— is −1.
The slope of the line perpendicular to DE
NQ = NR
F(2, 0)
— is
The midpoint D of KL
16 x
8
x
6
4
−2
0
5−5
—=—
=—=0
slope of DF
11 − 3 8
— is undefined, so the
The slope of the line perpendicular to DF
—
equation of the line perpendicular to DF is x = 7.
8.
y
y
H(5, 7)
6.
6
) (
)
— is
The midpoint F of MP
−8 −10
0 + (−8) −4 + (−6)
2
2
2
2
−2 − (−5) −2 + 5 3
—
The slope of NF is — = — = — = undefined.
−4 − (−4) −4 + 4 0
— through F(−4, −5) is x = −4.
The equation of NF
(
) (
)
—, — = —, — = (−4, −5).
The centroid has the coordinates of the intersection of y = −4
and x = −4. So, the coordinates of the centroid are (−4, −4).
210
Geometry
Worked-Out Solutions
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Chapter 6
11.
13. a. The point of concurrency at the center of the circle is the
y
T(−2, 5)
—, CG
—, and AG
— are angle bisectors.
incenter. BG
6
V(2, 5)
b. Sample answer: △BGF ≅ △BGE by HL Congruence
Theorem (Thm. 5.9).
1
y = −2 x + 4
c. Because △BGF ≅ △BGE and corresponding parts of
congruent triangles are congruent, BE = BF = 3 centimeters.
So, AE = 10 − 3 = 7 centimeters. Then you can use the
Pythagorean Theorem (Thm. 9.1) for △AEG to find EG,
which is the radius of the wheel.
U(0, 1)
−4
−2
4 x
2
0
5−5
— is —
= — = 0.
The slope of the line containing TV
2 − (−2) 4
— is undefined and
The slope of the line perpendicular to TV
passes through U(0, 1). Therefore, the equation of that line
is x = 0.
AG2 = GE2 + EA2
82 = GE2 + 72
64 = GE2 + 49
15 = GE2
5−1 4
— is —
= — = 2.
The slope of the line containing UV
2−0
—
GE = √15 ≈ 3.9
2
— is −—1.
The slope of the line perpendicular to UV
2
y = mx + b
1
y = −—x + b
2
1
5 = −—(−2) + b
2
5=1+b
4=b
— containing
The equation of the line perpendicular to UV
1
point T(−2, 5) is y = −—x + 4.
2
1
The orthocenter is the intersection of x = 0 and y = −—x + 4.
2
1
y = −—x + 4
2
1
y = −— 0 + 4
2
y=4
The radius of the wheel is about 3.9 centimeters.
14. a. The point of concurrency used was the centroid.
b. The circumcenter should have been used because the
Circumcenter Theorem (Thm. 6.5) can be used to find a
point equidistant from the three points (the three cities).
6.4 Explorations (p. 329)
1. a. Check students’ work.
— —
—
—
—
which indicates that DE AC , and the length of DE
1—
is — AC .
b. Sample answer: The slopes of DE and AC are equal
2
3 − 4.5 −1.5
—=—
= — ≈ −0.43
slope of DE
5 − 1.5
3.5
——
—
length of DE = √ (1.5 − 5)2 + (4.5 − 3)2 ≈ 3.8
−3
1−4
—= —
= — ≈ −0.43
slope of AC
5 − (−2)
7
— = √——
length of AC
( 5 − (−2) )2 + (1 − 4)2 ≈ 7.6
⋅
The coordinates of the orthocenter are located on the inside
of the triangle and are (0, 4).
12.
4
y
c. The segment connecting the midpoints of two sides of a
triangle is parallel to the third side and is half as long as
that side.
Z(7, 4)
2. a. Check students’ work.
b. Sample answer: The triangle formed by the midsegments
2
4
6
of a triangle, △EFD, is similar to the original triangle,
△ABC. The side lengths of △EFD are —12 the side lengths
of △ABC.
x
−2
X(−1, −4)
Y(7, −4)
△XYZ is a right triangle. Therefore, the orthocenter is on the
triangle at the intersection of the legs, which is (7,−4).
3. The midsegment that is formed by joining the midpoints of
two sides is parallel to the third side and is —12 the length of the
third side.
4.
UV = —12 RT
⋅
T
12 = —12RT
⋅
V
2 12 = 2 —12 RT
12
24 = RT
In △RST, if UV = 12,
then RT = 24.
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R
U
Geometry
Worked-Out Solutions
S
211
Chapter 6
( 0 +2 2p 0 +2 0 ) ( 2p2 )
6.4 Monitoring Progress (pp. 330–332)
—
2 − (−1) 2 + 1
2−0
2
0 − (−6) 6 3
—
Slope of AC : — = — = —
5−1
4 2
3. The midpoint F is —, — = —, 0 = ( p, 0).
3
2
1. Slope of DE : — = — = —
— equals the slope of AC
—, DE
— AC
—
Because the slope of DE
because parallel lines have equal slopes.
——
=√
+
—
——
—
= √ 4q2 + 4r2
—
—
= √4 + 9 = √13
AC = √(5 − 1)2 + ( 0 − (−6) )2
OB = 2FE
1
FE = —OB
2
—
= √(4)2 + (6)2
—
= √16 + 36
—
⋅
—
—
= √ 4(q2 + r2) = 2√ q2 + r2
——
= √52
—
FE = √( p + q − p)2 + (r − 0)2 = √q2 + r2
OB = √(2q− 0)2 + (2r − 0)2
—
(3)2
— and OB
— are equal, FE
— OB
—.
Because the slopes of FE
——
DE = √(2 − 0)2 + ( 2 − (−1) )2
(2)2
r
r−0
—: —
=—
Slope of FE
p+q−p q
r
2r − 0
—: —
=—
Slope of OB
2q − 0 q
—
4. The third midsegment is UW . If UW = 81, then
ST = 2 81 = 162; therefore, VS = —12 162 = 81 inches.
⋅
—
= √4 13 = 2√ 13
—
AC = 2√ 13
⋅
S
AC = 2DE
—
R
2. The midpoint of AC is
(
5 + 1 0 + (−6)
2
2
) ( 26 −62 )
— is −1 + 5, 4 + 0 = 4, 4 = (2, 2).
The midpoint of BC
( 2 2 ) (2 2) —
—
—, — = —, — = (3, −3).
— —
— —
The midpoint F of AC is (3, −3) and the midpoint E of BC
is (2, 2).
5
2 − (−3)
—: —
= — = −5
Slope of EF
2−3
−1
−10
−6 − 4
—: —
= — = −5
Slope of AB
1 − (−1)
2
— AB
— because the slopes are equal.
EF
———
AB = √( 1 − (−1) )2 + (−6 − 4)2
——
=√
(2)2
+
(−10)2
—
= √4 26 = 2√ 26
——
2)2
+ (−3 −
——
2)2
= √(1)2 + (−5)2
—
6. From Peach Street to Plum Street is 2.25 miles; from
Plum Street to Cherry Street is 1.4 miles; from Cherry Street
to Pear Street is 1.3 miles; from Pear Street to Peach Street
is —12 1.4 is 0.7 mile; from Pear Street back home is
(⋅ )
( — ⋅ 2.25 ) is 1.125 miles. The total distance is
1
2
2.25 + 1.4 + 1.3 + 0.7 + 1.125 = 6.775 miles.
This route was less than that taken in Example 5.
6.4 Exercises (pp. 333–334)
—
—
2. If DE is the midsegment opposite AC in △ABC, then
— AC
— and DE
— = —1 AC
— by the Triangle Midsegment
DE
2
Theorem (Thm. 6.8).
—
= √1 + 25 = √ 26
—
AB = 2√26
AB = 2EF
1
EF = —AB
2
212
—
— —
5. In Example 4, DF is a midsegment; therefore, DF AB and
1—
—
DF = —2 AB .
midpoints of two sides of a triangle.
—
EF = √(3 −
T
1. The midsegment of a triangle is a segment that connects the
—
= √104
W
Vocabulary and Core Concept Check
—
= √4 + 100
⋅
V
U
1
DE = —AC
2
Geometry
Worked-Out Solutions
Monitoring Progress and Modeling with Mathematics
3. The coordinates are D(−4, −2), E(−2, 0), and F(−1, −4).
Copyright © Big Ideas Learning, LLC
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Chapter 6
—
2
0 − (−2)
−2 − (−4) 2
−2 − (−6) −2 + 6 4
—: —
=—=—=1
Slope of CB
1 − (−3)
4
4
4. Slope of DE : — = — = 1
— equals the slope of CB
—, DE
— CB
—.
Because the slope of DE
———
1
—
—
= √4 + 4
—
⋅
—
—
= √8 = √4 2 = 2√ 2
———
CB = √( 1 − (−3) )2 + ( −2 − (−6) )2
—
= √(4)2 + (4)2
—
= √16 + 16
x = —12 (26)
5 = —12 (AB)
x = 13
x = 10
⋅
x=8
— —
11. JK YZ
— —
12. JL XZ
— —
13. XY KL
14. JY ≅ JX ≅ KL
— — —
= √16 2 = 4√ 2
—
—
1
1
Because 2√ 2 = —( 4√ 2 ), DE = —CB.
2
2
AB = —12(GL)
—
3y − 5 = 2y + 1
y=6
x=2
HB = AC
⋅
HB = 3y − 5
⋅
= 4 + 24 = 28
=3 6−5
AB = —12 (28) = 14
= 18 − 5 = 13
CB = —12(GA)
2(4z − 3) = 2 —12 (7z − 1)
——
8z − 6 = 7z − 1
= √(2)2 + (−8)2
—
z − 6 = −1
—
z=5
= √4 17 = 2√ 17
—
y−5=1
⋅
———
⋅
3x + 8 = x + 12
4z − 3 = —12(7z − 1)
AC = √( −3 − (−5) )2 + (−6 − 2)2
—
3y − 5 = —12 (4y + 2)
19.
= √1 + 16 = √17
= √4 + 64
3x + 8 = —12 (2x + 24)
GL = 2 2 + 24
———
—
AC = —12 (HJ)
18.
2x = 4
— equals the slope of AC
—, EF
— AC
—.
Because the slope of EF
——
— — —
16. JK ≅ YL ≅ LZ
2x + 8 = 12
−4
−4
— −4 − 0
5. Slope of EF : — = — = — = −4
−1 − (−2) −1 + 2
1
−8
−8
−6 − 2
—
Slope of AC : — = — = — = −4
−3 − (−5) −3 + 5
2
= √(1)2 + (−4)2
— — —
15. JL ≅ XK ≅ KZ
—
EF = √( −1 − (−2) )2 + (−4 − 0)2
10. BE = EC
6=x
17.
—
8. DE = —2 AB
9. AE = EC
DE = √( −2 − (−4) )2 + ( 0 − (−2) )2
= √(2)2 + (2)2
1
7. DE = —2 BC
—
1
1
Because √ 17 = —( 2√ 17 ), EF = —AC.
2
2
—
6. Slope of DF :
2
−4 − (−2) −4 + 2
— = — = −—
3
−1 − (−4) −1 + 4
2
−4
−4
−2 − 2
—
Slope of AB : — = — = — = −—
3
1 − (−5) 1 + 5
6
— equals the slope of AB
—, DF
— AB
—.
Because the slope of DF
GA = CB
⋅
CB = 4z − 3 = 4 5 − 3 = 20 − 3 = 17
GA = 17
—
—
—
20. DE is not parallel to BC . So, DE is not a midsegment. So,
according to the contrapositive of the Triangle Midsegment
— does not connect the midpoints of
Theorem (Thm. 6.8), DE
—
—
AC and AB .
———
DF = √( −1 − (−4) )2 + ( −4 − (−2) )2
—
——
—
= √(3)2 + (−2)2 = √ 9 + 4 = √ 13
———
AB = √( 1 − (−5) )2 + (−2 − 2)2
——
= √(6)2 + (−4)2
—
= √36 + 16
—
⋅
—
= √52 = √4 13
—
= 2√13
—
21. The distance between first base and second base is 90 feet.
Because the shortstop is halfway between second and third
bases, and the pitcher is halfway between first and third
bases, using the Triangle Midsegment Theorem, the distance
between the shortstop and the pitcher is —12 90 = 45. So, the
distance between the shortstop and the pitcher is 45 feet.
⋅
—
Because √ 13 = —12( 2√13 ), DF = —12AB.
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Geometry
Worked-Out Solutions
213
Chapter 6
—.
F is the midpoint of OC
— BC
— and DF = —1(BC)
Prove DF
2
24. Sample answer:
22. Given
y
The crossbars on the
ends of the swing set
are midsegments.
B(2q, 2r)
D(q, r)
O(0, 0)
E
C(2p, 0) x
F
0 + 2p
2p
—= —
, 0 = —, 0 = F(p, 0)
midpoint of OC
2
2
r
—=—
slope of DF
q−p
2r
r
2r − 0
—
slope of BC = — = — = —
2q − 2p 2(q − p) q − p
(
25. a. The perimeter of the shaded triangle is
) ( )
8 + 8 + 8 = 24 units.
16
8
— equals the slope of BC
—, DF
— BC
—.
Because the slope of DF
——
8
16
——
DF = √(q − p)2 + (r − 0)2 = √ (q − p)2 + r2
16
8
——
BC = √(2q − 2p)2 + (2r − 0)2
——
= √( 2(q − p) )2 + 4r2
——
= √4(q − p)2 + 4r2
b. The perimeter of all the shaded triangles is
——
⋅
= √4(q − p)2 + r2
24 + 3(3 4) = 24 + 36 = 60 units.
——
= 2√(q − p)2 + r2
Because BC = 2DF, DF =
—
1
—2 BC.
4
—
4
23. An eighth segment, FG , would connect the midpoints of DL
—.
and EN
16
4
4
8
8
4
4
8
DE = —12(XY + LN)
[
= —12 —12(LN) + LN
16
4
]
4
16
4
= —14LN + —12LN
= —34LN
FG = —34LN + LN = —78 LN
Because you are finding quarter segments and eighth segments,
use 8p, 8q, and 8r. So, L(0, 0), M(8q, 8r), and N(8p, 0).
Find the coordinates of X, Y, D, E, F, and G.
—, the coordinates are X(4q, 4r).
Because X is the midpoint of LM
— , the coordinates are
Because Y is the midpoint of MN
Y(4q + 4p, 4r).
—, the coordinates are D(2q, 2r).
Because D is the midpoint of XL
—, the coordinates are
Because E is the midpoint of YN
E(2q + 6p, 2r).
—, the coordinates are F(q, r).
Because F is the midpoint of DL
—, the coordinates are
Because G is the midpoint of EN
G(q + 7p, r).
— has a
The y-coordinates of D and E are the same, so DE
slope of 0. The y-coordinates of F and G are also the same,
— has a slope of 0. LN
— is on the x-axis, so its slope is 0.
so FG
— LN
— FG
—.
Because their slopes are the same, DE
Use the Ruler Postulate (Post. 1.1) to find DE, FG, and LN.
DE = 6p, FG = 7p, LN = 8p
Because 6p = —34(8p), DE = —34 LN. Because 7p = —78(8p),
FG = —78LN.
214
Geometry
Worked-Out Solutions
c. The perimeter of all the shaded triangles is
⋅
⋅
24 + 3 12 + 9 6 = 24 + 36 + 54 = 114 units.
16
6
12
6
24
6
16
6
6
12
12
6
6
6
16
6
26. Two sides of the red triangle have a length of 4 tile widths. A
yellow segment connects the midpoints, where there are 2 tile
lengths on either side. The third red side has a length of
4 tile diagonals, and the other two yellow segments meet at the
midpoint, where there are 2 tile diagonals on either side.
Copyright © Big Ideas Learning, LLC
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Chapter 6
—
—
—) with midpoint R
the length of QR . Construct a line (EF
—. Construct a line (DE
—)
parallel to and twice the length of QP
—
with midpoint Q parallel to and twice the length of PR . The vertices
27. Construct a line (DF ) with midpoint P parallel to and twice
of the original large triangle are D(−1, 2), E(9, 8), and F(5, 0).
10
2. a. Check students’ work. Using the sample in the text:
AB ≈ 3.61
AC ≈ 5.10
BC = 5
b. Check students’ work. Using the sample in the text:
y
BC = 5 < 8.71 = AC + AB
E
8
AC = 5.10 < 8.61 = AB + BC
AB = 3.61 < 10.10 = BC + AC
6
Q
c. Sample answer:
4
R
D
P
−2
F
2
4
8
10 x
−2
A(x, y)
B(x, y)
C(x, y)
AB
AC + BC
AC
A(5, 1)
B(7, 4)
C(2, 4)
3.61
9.24
4.24
A(2, 4)
B(4, −2)
C(7, 6)
6.32
13.93
5.39
A(1, 0)
B(7, 0)
C(1, 7)
6
16.22
7
A(1, 0)
B(7, 0)
C(5, 1)
6
6.36
4.12
AB + BC
BC
AB + AC
8.61
5
7.85
14.86
8.54
11.71
15.22
9.22
13
8.24
2.24
10.12
Maintaining Mathematical Proficiency
28. Sample answer: A counterexample to show that the
difference of two numbers is not always less than the greater
of the two numbers is 6 and −2: 6 − (−2) = 8, which is not
less than 6, so the original conjecture is false.
29. Sample answer: An isosceles triangle has at least two sides
that are congruent. An isoseles triangle whose sides are 5
centimeters, 5 centimeters, and 3 centimeters is not equilateral.
6.5 Explorations (p. 335)
3. The largest angle is opposite the longest side, and the
1. a. Check students’ work. Using the sample in the text:
AC ≈ 6.08, AB ≈ 4.47, BC ≈ 3.61, m∠ A ≈ 36.03°,
m∠ B ≈ 97.13°, m∠ C ≈ 46.85°
b. Check students’ work. Using the sample in the text,
BC < AB < AC and m∠ A < m∠ C < m∠ B. The shortest
side is opposite the smallest angle, and the longest side is
opposite the largest angle.
c. Sample answer:
A(x, y)
B(x, y)
The length of each side is less than the sum of the other two.
smallest angle is opposite the shortest side; The sum of any
two side lengths is greater than the third side length.
4. no; The sum 3 + 4 is not greater than 10, and it is not
possible to form a triangle when the sum of the lengths
of the two sides is less than the length of the third side.
6.5 Monitoring Progress (pp. 336–339)
1. Given
C(x, y)
AB
AC
BC
A(5, 1)
B(7, 4)
C(2, 4)
3.61 4.24
A(2, 4)
B(4, −2)
C(7, 6)
6.32 5.39 8.54
A(1, 0)
B(7, 0)
C(1, 7)
m∠ A
m∠ B
m∠ C
78.69°
56.31°
45°
93.37°
38.99°
47.64°
90°
49.4°
40.6°
6
7
5
9.22
If one side of a triangle is longer than another side, then
the angle opposite the longer side is larger than the angle
opposite the shorter side. Similarly, if one angle of a triangle
is larger than another angle, then the side opposite the larger
angle is longer than the side opposite the smaller angle.
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All rights reserved.
Prove
△ABC is a scalene triangle.
△ABC does not have two congruent angles.
Assume temporarily that △ABC is a scalene triangle with
∠ A ≅ ∠ B. By the Converse of Base Angles Theorem
(Thm. 5.7), if ∠ A ≅ ∠ B, then the opposite sides are
— ≅ AC
—. A scalene triangle cannot have two
congruent: BC
congruent sides. So, this contradicts the given information.
So, the assumption that △ABC is a scalene triangle with two
congruent angles must be false, which proves that a scalene
triangle cannot have two congruent angles.
——
—
PQ . So, by the Triangle Longer Side Theorem, the angles
2. The sides of △PQR from smallest to largest are PR , RQ , and
from smallest to largest are ∠Q, ∠P, and ∠R.
3. The angles of △RST from smallest to largest are ∠R, ∠T,
and ∠S. So, by the Triangle Larger Angle Theorem, the sides
—, RS
—, and RT
—.
from shortest to longest are ST
Geometry
Worked-Out Solutions
215
Chapter 6
4. Let x represent the length of the third side. By the Triangle
Inequality Theorem: x + 12 > 20 and 12 + 20 > x.
x + 12 > 20
and
12 + 20 > x
x>8
32 > x, or x < 32
The length of the third side must be greater than 8 inches and
less than 32 inches.
5. 4 + 9 > 10 → 13 > 10
—
—
arc with radius AB , then draw an arc with B as the center
— to intersect the first arc on both sides of AB
—.
and radius AB
10. Draw a segment AB . Using point A as the center, draw an
Construct a segment from one arc intersection toward the
—. Label the arc-segment intersection as C
other but stop at AB
— as G. △BGC is a right scalene
and the intersection with AB
triangle.
Yes
4 + 10 > 9 → 14 > 9
Yes
9 + 10 > 4 → 19 > 4
Yes
C
yes; The sum of any two side lengths of a triangle is greater
than the length of the third side.
6. no; The sum 8 + 9 = 17 is not greater than 18.
A
G
B
7. no; The sum 5 + 7 = 12 is not greater than 12.
6.5 Exercises (pp. 340–342)
Vocabulary and Core Concept Check
1. In an indirect proof, rather than proving a statement directly,
you show that when the statement is false, it leads to a
contradiction.
2. The longest side of a triangle is opposite the largest angle
and the shortest side is opposite the smallest angle.
Monitoring Progress and Modeling with Mathematics
3. Assume temporarily that WV = 7 inches.
4. Assume temporarily that xy is even.
5. Assume temporarily that ∠ B is a right angle.
—
6. Assume temporarily that JM is not a median.
7. A and C; The angles of an equilateral triangle are always 60°.
So, an equilateral triangle cannot be a right triangle.
8. B and C; If both ∠ X and ∠ Y have measures less than 30°,
then their sum is less than 60°. Therefore, the sum of their
measures cannot be 62°.
9. To construct a scalene triangle, draw a segment and label it
—. Ensuring that AB
—, BC
—, and AC
— are all different lengths,
AC
— and an arc with
draw an arc with center A and radius AB
—
center C with radius CB . Where the two arcs intersect place
point B.
B
A
C
—, is the
The largest angle is ∠ ABC and the opposite side, AC
longest side. The smallest angle is ∠ ACB and the opposite
—, is the shortest side.
side, AB
216
Geometry
Worked-Out Solutions
The largest angle is ∠CGB because it is the right angle and
—, is the longest side. The smallest angle
the opposite side, CB
—, is the shortest side.
is ∠GCB and the opposite side, GB
——
11. The sides of △RST from smallest to largest are RT , TS , and
—. So, by the Triangle Longer Side Theorem, the angles
RS
from smallest to largest are ∠ S, ∠ R, and ∠ T.
——
—
JK . So, by the Triangle Longer Side Theorem, the angles
12. The sides of △JKL from smallest to largest are KL , JL , and
from smallest to largest are ∠ J, ∠ K, and ∠ L.
13. The angles of △ABC from smallest to largest are ∠ C, ∠ A,
and ∠ B. So, by the Triangle Larger Angle Theorem, the
—, BC
—, and AC
—.
sides from shortest to longest are AB
14. The angles of △XYZ from smallest to largest are ∠ Z, ∠ X,
and ∠ Y. So, by the Triangle Larger Angle Theorem, the
—, ZY
—, and ZX
—.
sides from shortest to longest are XY
15. m∠ M = 180° − (127° + 29°) = 24°
The angles of △MNP from smallest to largest are ∠ M, ∠ P,
and ∠ N. So, by the Triangle Larger Angle Theorem, the
—, MN
—, and MP
—.
sides from shortest to longest are PN
16. m∠ D = 180° − (90° + 33°) = 57°
The angles of △DFG from smallest to largest are ∠ G, ∠ D,
and ∠ F. So, by the Triangle Larger Angle Theorem, the
—, GF
—, and GD
—.
sides from shortest to longest are DF
17. x + 5 > 12
x>7
5 + 12 > x
17 > x or x < 17
The possible lengths of the third side are greater than
7 inches and less than 17 inches.
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Chapter 6
28. Assume temporarily that the second group has 15 or more
18. x + 12 > 18
x>6
12 + 18 > x
30 > x or x < 30
The possible lengths of the third side are greater than
6 feet and less than 30 feet.
students. Because the first group has 15 students, the total
number of students in the class would be 30 students or
more. Because the class has fewer than 30 students, the
assumption must be false, and the second group must have
fewer than 15 students.
29. C; m∠ U = 180° − (84° + 48°) = 180° − 132° = 48°,
which indicates that △UTV is isosceles. By the Triangle
Longer Side Theorem, UV > TV.
19. x + 24 > 40
x > 16
30. C and D; m∠ R = 180° − (65° + 56°) = 180° − 121° = 59°;
24 + 40 > x
64 > x or x < 64
The possible lengths of the third side are greater than
16 inches and less than 64 inches.
20. x + 25 > 25
By the Triangle Inequality Theorem, the order of the
angles from smallest to largest is ∠ T, ∠ R, and ∠ S.
The order of the sides from shortest to longest is
RS < ST < RT ⇒ 8 < ST < RT. ST could possibly be
9 or 10, but not 7 or 8.
31. Given
x>0
An odd number
Prove An odd number is not divisible by 4.
25 + 25 > x
50 > x or x < 50
The possible lengths of the third side are greater than
0 meters and less than 50 meters.
21. 6 + 7 = 13 → 13 > 11
Yes
7 + 11 = 18 → 18 > 6
Yes
11 + 6 = 17 → 17 > 7
Yes
Assume temporarily that an odd number is divisible by 4.
Let the odd number be represented by 2y + 1 where y is a
positive integer. Then there must be a positive integer x such
that 4x = 2y + 1. However, when you divide each side of the
equation by 4, you get x = —12 y + —14 , which is not an integer.
So, the assumption must be false, and an odd number is not
divisible by 4.
32. Given
yes; The sum of any two side lengths of a triangle is greater
than the length of the third side.
22. no; The sum 3 + 6 = 9 is not greater than 9.
23. no; The sum 28 + 17 = 45 is not greater than 46.
24. 35 + 120 = 155 → 155 > 125
Yes
120 + 125 = 255 → 255 > 35
Yes
125 + 35 = 160 → 160 > 120
Yes
yes; The sum of any two side lengths of a triangle is greater
than the length of the third side.
25. An angle that is not obtuse could be acute or right. Assume
temporarily that ∠ A is not obtuse.
—
26. Because 30° < 60° < 90° and 1 < √ 3 < 2, the longest side,
which is 2 units long, should be across from the largest
angle, which is the right angle.
Prove
△QRS, m∠ Q + m∠ R = 90°
m∠ S = 90°
Assume temporarily that in △QRS, m∠ Q + m∠ R = 90°
and m∠ S ≠ 90°. By the Triangle Sum Theorem (Thm. 5.1),
m∠ Q + m∠ R + m∠ S = 180°. Using the Substitution
Property of Equality, 90° + m∠ S = 180°. So, m∠ S = 90°
by the Subtraction Property of Equality, but this contradicts
the given information. So, the assumption must be false,
which proves that in △QRS, if m∠ Q + m∠ R = 90°,
then m∠ S = 90°.
33. The right angle of a right triangle must always be the largest
angle because the other two will have a sum of 90°. So,
according to the Triangle Longer Angle Theorem (Thm. 6.10),
because the right angle is larger than either of the other
angles, the side opposite the right angle, which is the
hypotenuse, will always have to be longer than either of
the legs.
34. yes; If the sum of the lengths of the two shortest sides is
greater than the length of the longest side, then the other
two inequalities will also be true.
60°
1
2
——
30°
√3
27. Assume temporarily that your client committed the crime.
Then your client had to be in Los Angeles, California, at the
time of the crime. Security footage shows that your client
was in New York at the time of the crime. Therefore, the
assumption must be false, and the client must be innocent.
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Geometry
Worked-Out Solutions
217
Chapter 6
35. a. The width of the river must be greater than 35 yards and
—,
less than 50 yards. In △BCA, the width of the river, BA
—
must be less than the length of CA , which is 50 yards,
— is less than
because the measure of the angle opposite BA
—
the measure of the angle opposite CA , which must be 50°. In
—, must be greater than the
△BDA, the width of the river, BA
—
length of DA , which is 35 yards, because the measure of
— is greater than the measure of the
the angle opposite BA
—
angle opposite DA , which must be 40°.
40.
3x + 21 > 5x − 9
−2x > −30
x < 15
JK + JL > KL
x + 11 + 5x − 9 > 2x + 10
6x + 2 > 2x + 10
b. You could measure from distances that are closer together.
In order to do this, you would have to use angle measures
that are closer to 45°.
4x > 8
x>2
36. a. By the side length requirements for a triangle,
KL + JL > JK
x < 489 + 565 = 1054 kilometers and
x > 565 − 489 = 76 kilometers.
2x + 10 + 5x − 9 > x + 11
7x + 1 > x + 11
b. Because ∠2 is the smallest angle, the distance between
6x > 10
Granite Peak and Fort Peck Lake must be the shortest side
of the triangle. So, the second inequality becomes
x < 489 kilometers.
37. ∠WXY, ∠ Z, ∠ YXZ, ∠WYX and ∠ XYZ, ∠W;
In △WXY, because WY < WX < YX, by the Triangle Longer
Side Theorem (Thm. 6.9), m∠WXY < m∠WYX < m∠W.
Similarly, in △XYZ, because XY < YZ < XZ, by the Triangle
Longer Side Theorem (Thm. 6.9), m∠ Z < m∠ YXZ < m∠ XYZ.
Because m∠WYX = m∠ XYZ and ∠W is the only angle
greater than either of them, you know that ∠W is the largest
angle. Because △WXY has the largest angle and one of
the congruent angles, the remaining angle, ∠WXY, is the
smallest.
38.
m∠ D + m∠ E + m∠ F = 180°
(x + 25)° + (2x − 4)° + 63° = 180°
3x + 84 = 180
3x = 96
x = 32
m∠D = 32 + 25 = 57°
⋅
m∠E = 2 32 − 4 = 60°
m∠F = 63°
The order of the angles from least to greatest is
m∠ D < m∠ E < m∠ F. The order of the sides from
least to greatest is EF < DF < DE.
39. By the Exterior Angle Theorem (Thm. 5.2),
m∠1 = m∠ A + m∠B. Then by the Subtraction Property of
Equality, m∠1 − m∠B = m∠ A. If you assume temporarily
that m∠1 ≤ m∠B, then m∠ A ≤ 0. Because the measure
of any angle in a triangle must be a positive number, the
assumption must be false. So, m∠1 > m∠B. Similarly, by the
Subtraction Property of Equality, m∠1 − m∠ A = m∠B. If
you assume temporarily that m∠1 ≤ m∠ A, then m∠B ≤ 0.
Because the measure of any angle in a triangle must be a
positive number, the assumption must be false. So,
m∠1 > m∠ A.
218
Geometry
Worked-Out Solutions
JK + KL > JL
x + 11 + 2x + 10 > 5x − 9
10
x>—
= —53 ≈ 1.667
6
The possible values for x are x > 2 and x < 15.
41.
UV + VT > TU
3x − 1 + 2x + 3 > 6x − 11
5x + 2 > 6x − 11
−x > −13
x < 13
TU + TV > UV
6x − 11 + 2x + 3 > 3x − 1
8x − 8 > 3x − 1
5x > 7
x > —75 = 1—25
UV + TU > TV
3x − 1 + 6x − 11 > 2x + 3
9x − 12 > 2x + 3
7x > 15
15
x>—
= 2—17
7
The possible values for x are x > 2—17 and x < 13.
42. The shortest route is along Washington Avenue. By the
Triangle Inequality Theorem (Thm. 6.11), the length of
Washington Avenue must be shorter than the sum of the
lengths of Eighth Street and View Street, as well as the sum
of the lengths of Hill Street and Seventh Street.
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Chapter 6
BC > AB, BD = BA
43. Given
m∠ BAC > m∠ C
Prove
B
A
1
2
3
Prove
D
2 3
C
D
1
C
A
— is longer than or the same length as each of
Assume BC
— and AC
—. Then AB + BC > AC and
the other sides, AB
AC + BC > AB. The proof for AB + AC > BC follows.
STATEMENTS
1. △ABC
REASONS
1. Given
2. Extend —
AC to D so
that —
AB ≅ —
AD .
44. x + x >ℓ
2x >ℓ
x > —12ℓ
Because the sum of the lengths of the legs must be greater
than the length of the base, the length of a leg must be
greater than —12ℓ.
45. no; If one side is 13 inches and the perimeter is 24 inches
(2 feet), then the other two sides would total 11 inches, but
they must have a sum greater than 13 inches for a triangle
to exist.
46. As an example, if the 24-centimeter string is divided into
10 centimeters, 10 centimeters, and 4 centimeters, the
triangle is an acute isosceles triangle.
10 cm
△ABC
AB + BC > AC, AC + BC > AB, and
AB + AC > BC
B
It is given that BC > AB and BD = BA. By the Base Angles
Theorem (Thm. 5.6), m∠ 1 = m∠ 2. By the Angle Addition
Postulate (Post. 1.4), m∠ BAC = m∠ 1 + m∠ 3. So,
m∠ BAC > m∠ 1. Substituting m∠ 2 for m∠ 1 produces
m∠ BAC > m∠ 2. By the Exterior Angle Theorem (Thm. 5.2),
m∠ 2 = m∠ 3 + m∠ C. So, m∠ 2 > m∠ C. Finally, because
m∠ BAC > m∠ 2 and m∠ 2 > m∠ C, you can conclude that
m∠ BAC > m∠ C.
10 cm
2. Ruler Postulate
(Post. 1.1)
3. AB = AD
3. Definition of segment
congruence
4. AD + AC = DC
4. Segment Addition
Postulate (Post. 1.2)
5. ∠ 1 ≅ ∠ 2
5. Base Angles Theorem
(Thm. 5.6)
6. m∠ 1 = m∠ 2
6. Definition of angle
congruence
7. m∠ DBC > m∠ 2
7. Protractor Postulate
(Post. 1.3)
8. m∠ DBC > m∠ 1
8. Substitution Property
9. DC > BC
9. Triangle Larger Angle
Theorem (Thm. 6.10)
10. AD + AC > BC
10. Substitution Property
11. AB + AC > BC
11. Substitution Property
48. The perimeter of △HGF must be greater than 4 and less
4 cm
For a right scalene triangle, the sides could be 6 centimeters,
8 centimeters, and 10 centimeters.
8 cm
47. Given
than 24; Because of the Triangle Inequality Theorem
(Thm. 6.11), FG must be greater than 2 and less than 8,
GH must be greater than 1 and less than 7, and FH must
be greater than 1 and less than 9. So, the perimeter must be
greater than 2 + 1 + 1 = 4 and less than 8 + 7 + 9 = 24.
10 cm
6 cm
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Geometry
Worked-Out Solutions
219
Chapter 6
—
49. Assume temporarily that another segment, PA , where A is on
plane M, is the shortest segment from P to plane M. By definition
— ⊥ plane M.
of the distance between a point and a plane, PA
This contradicts the given statement because there cannot
be two different segments that share an endpoint and are
both perpendicular to the same plane. So, the assumption is
false, and because no other segment exists that is the shortest
— that is the shortest
segment from P to plane M, it must be PC
segment from P to plane M.
3. Because the sides of the hinge do not change in length, the
angle of the hinge can model the included angle and the
distance between the opposite ends of the hinge can model
the third side. When the hinge is open wider, the angle is
larger and the ends of the hinge are farther apart. If the hinge
is open less, the ends are closer together.
6.6 Monitoring Progress (pp. 345–346)
1.
Maintaining Mathematical Proficiency
—
—
50. The included angle between AE and BE is ∠ AEB.
m∠ QPR > m∠ QPS
— —
51. The included angle between AC and DC is ∠ ACD.
—
—
—
—
PR = PS
— ≅ PQ
—
PQ
2.
53. The included angle between CE and BE is ∠ BEC.
Given
RQ > SQ
Hinge Theorem (Thm. 6.12)
PR = PS
Given
— ≅ PQ
—
PQ
6.6 Explorations (p. 343)
RQ < SQ
1. a. Check students’ work.
m∠ RPQ < m∠ SPQ
b. Check students’ work.
c. Check students’ work.
— —
d. AC ≅ DC , because all points on a circle are equidistant
— ≅ BC
— by the Reflexive Property of
from the center; BC
—
e. As drawn, the length of AB is 3.6 units and the length of
— is 2.7 units, so AB > DB; m∠ ACB = 90° and
DB
m∠ DCB = 61°, so m∠ ACB > m∠ DCB; yes; the results
are as expected because the triangle with the longer third
side has the larger angle opposite the third side.
f. Sample answer:
Reflexive Property of Congruence
(Thm. 2.1)
Given
Converse of the Hinge Theorem
(Thm. 6.13)
∠ SPQ is the larger angle.
3. Assume temporarily that the third side of the first triangle
with the larger included angle is not longer than the third
side of the second triangle with the smaller included angle.
This means the third side of the first triangle is equal to or
shorter than the third side of the second triangle.
4. Group A: 135°
Group B: 150°
D
AC
BC
m∠ ACB
m∠ BCD
1. (4.75, 2.03)
2
3
3.61 2.68
90°
61.13°
2. (4.94, 2.5)
2
3
3.61 3.16
90°
75.6°
3. (5, 3)
2
3
3.61 3.61
90°
90°
4. (4.94, 3.5)
2
3
3.61
90°
104.45°
5. (3.85, 4.81)
2
3
3.61 4.89
90°
154.93°
AB
BD
4
g. If two sides of one triangle are congruent to two sides
of another triangle, and the included angle of the first is
larger than the included angle of the second, then the third
side of the first is longer than the third side of the second.
2. If the included angle of one is larger than the included angle
of the other, then the third side of the first is longer than the
third side of the second. If the included angles are congruent,
then you already know that the triangles are congruent by
the SAS Congruence Theorem (Thm. 5.5). Therefore, the
third sides are congruent because corresponding parts of
congruent triangles are congruent.
220
Reflexive Property of Congruence
(Thm. 2.1)
— is the longer segment.
RQ
52. The included angle between AD and DC is ∠ ADC.
Congruence (Thm. 2.1).
Given
Geometry
Worked-Out Solutions
Group C: 180° − 40° = 140°
Because 135° < 140° < 150°, Group C is closer to the camp
than Group B, but not as close as Group A.
6.6 Exercises (pp. 347–348)
Vocabulary and Core Concept Check
1. Theorem 6.12 refers to two angles with two pairs of sides
that have the same measure, just like two hinges whose sides
are the same length. Then the angle whose measure is greater
is opposite a longer side, just like the ends of a hinge are
farther apart when the hinge is open wider.
— —— —
2. In △ABC and △DEF, AB ≅ DE , BC ≅ EF , and AC < DF.
So, m∠ E > m∠ B by the Converse of the Hinge Theorem
(Theorem 6.13).
Monitoring Progress and Modeling with Mathematics
3. m∠ 1 > m∠ 2; By the Converse of the Hinge Theorem
(Thm. 6.13), because ∠ 1 is the included angle in the triangle
with the longer third side, its measure is greater than that
of ∠ 2.
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Chapter 6
13. Your flight:
4. m∠ 1 < m∠ 2; By the Converse of the Hinge Theorem
(Thm. 6.13), because ∠ 1 is the included angle in the triangle
with the shorter third side, its measure is less than that of ∠ 2.
50 miles
5. m∠ 1 = m∠ 2; The triangles are congruent by the SSS
160°
H
20°
Congruence Theorem (Thm. 5.8). So, ∠ 1 ≅ ∠ 2 because
corresponding parts of congruent triangles are congruent.
100 miles
Friend’s flight:
6. m∠ 1 > m∠ 2; By the Converse of the Hinge Theorem
(Thm. 6.12), because ∠ 1 is the included angle in the triangle
with the longer third side, its measure is greater than that of ∠2.
—
7. AD > CD; By the Hinge Theorem (Thm. 6.12), because AD
is the third side of the triangle with the larger included angle,
—.
it is longer than CD
—
50 miles
30°
150°
100 miles
Because 160° >150°, the distance you flew is a greater distance
than your friend flew by the Hinge Theorem (Thm. 6.12).
14. Your flight:
8. MN < LK; By the Hinge Theorem (Thm. 6.12), because MN
is the third side of the triangle with the smaller included
—.
angle, it is shorter than LK
—
9. TR < UR; By the Hinge Theorem (Thm. 6.12), because TR is
the third side of the triangle with the smaller included angle,
—.
it is shorter than UR
—
210 miles
10. AC > DC; By the Hinge Theorem (Thm. 6.12), because AC is
the third side of the triangle with the larger included angle, it
—.
is longer than DC
11. Given
Prove
— ≅ YZ
—, m∠ WYZ > m∠ WYX
XY
W
WZ > WX
110°
Z
X
STATEMENTS
REASONS
1.
1. Given
—
XY ≅ —
YZ
2. —
WY ≅ —
WY
70°
80 miles
Y
Friend’s flight:
2. Reflexive Property of
Congruence (Thm. 2.1)
210 miles
3. m∠ WYZ > m∠ WYX 3. Given
4. WZ > WX
4. Hinge Theorem (Thm. 6.12)
50°
12. Given
— ≅ DA
—, DC < AB
BC
Prove
m∠ BCA > m∠ DAC
80 miles
D
STATEMENTS
BC ≅ —
DA
1. —
AC ≅ —
AC
2. —
130°
B
A
C
REASONS
1. Given
2. Reflexive Property of
Congruence (Thm. 2.1)
3. DC < AB
3. Given
4. m∠ BCA > m∠ DAC
4. Converse of the Hinge
Theorem (Thm. 6.13)
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Because 130° > 110°, the distance your friend flew is a
greater distance than the distance you flew by the Hinge
Theorem (Thm. 6.12).
15. The measure of the included angle in △PSQ is greater than
the measure of the included angle in △SQR; By the Hinge
Theorem (Thm. 6.12), PQ > SR.
Geometry
Worked-Out Solutions
221
Chapter 6
16.
24. The sum of the measures of the angles of a triangle in
8>6
spherical geometry must be greater than 180°; The area of
πr2
spherical △ABC = —(m∠ A + m∠B + m∠C − 180°),
180°
where r is the radius of the sphere.
LM > MJ
m∠ LKM > m∠ JKM
25° > m∠ JKM
25° > 15° → Yes
Maintaining Mathematical Proficiency
25° > 22° → Yes
25. x° + 115° + 27° = 180°
25° > 25° → No
x + 142 = 180
25° > 35° → No
17. Given EF > ED and GD = GF, then m∠ EGF > m∠ DGE by
the Hinge Theorem (Thm. 6.12).
18. The angle bisector of ∠ FEG will also pass through incenter H.
180°
Then, m∠ HEG + m∠ HFG + m∠ HGF = — = 90°,
2
because they are each half of the measure of an angle of
a triangle. By subtracting m∠ HEG from each side, you
can conclude that m∠ HFG + m∠ HGF < 90°. Also,
m∠ FHG + m∠ HFG + m∠ HGF = 180° by the Triangle
Sum Theorem (Thm. 5.1). So, m∠ FHG > 90°, which means
that m∠ FHG > m∠ HFG and m∠ FHG > m∠ HGF. So,
FG > FH and FG > HG.
— —— —
19. Because NR is a median, PR ≅ QR . NR ≅ NR by the
Reflexive Property of Congruence (Thm. 2.1). So, by
the Converse of the Hinge Theorem (Thm. 6.13),
m∠ NRQ > m∠ NRP. Because ∠ NRQ and ∠ NRP form a
linear pair, they are supplementary. So, ∠ NRQ must be
obtuse and ∠ NRP must be acute.
20. 180° − (27° + 102°) = 180° − 129° = 51°
110° > 51°
2x = 144
x = 38
A, B; The two possible measures for ∠ JKM are 15° and 22°.
—
26. 2x° + 36° = 180°
x = 72
27. 3x° = 180°
28. x° = 44°+ 64°
x = 60
x = 108
6.4–6.6 What Did You Learn? (p. 349)
1. Let n be the stage, then the side length of the new triangles in
each stage is 24 − n. So, the perimeter of each new triangle is
(3 24 − n ). The number of new triangles is given by ( 3n − 1).
So, to find the perimeter of all the shaded triangles in each
stage, start with the total from the previous stage and add
( 3 24 − n )( 3n − 1 ). The perimeter of the new triangles in
stage 4 will be ( 3 24 − 4 )( 34 − 1 ) = 81. The total perimeter
of the new triangles and old triangles is 81 + 114 = 195 units.
⋅
⋅
⋅
2. x + 5 > 12, x + 12 > 5, 5 + 12 > x; Because the length of
the third side has to be a positive value, the inequality
x + 12 > 5 will always be true. So, you do not have to
consider this inequality in determining the possible values
of x. Solve the other two inequalities to find that the length
of the third side must be greater than 7 and less than 19.
3. If △ABC is an acute triangle, then m∠ BAC < m∠ BDC and
the orthocenter D is inside the triangle.
B
3x + 2 > x + 3
2x > 1
x > —12
D
21. By the Exterior Angle Theorem, m∠ ABD = m∠ BDC + m∠ C.
So, m∠ ABD > m∠ BDC.
AD > BC
4x − 3 > 2x
2x > 3
x > —32
22. By the Converse of the Hinge Theorem (Thm. 6.13), because
A
C
If △ABC is a right triangle, then m∠ BAC = m∠ BDC
because the orthocenter D is on vertex A, where ∠ A is the
right angle.
B
28 ft > 22 ft → AD > AB, then m∠ ACD > m∠ ACB.
23. △ABC is an obtuse triangle; If the altitudes intersect inside
the triangle, then m∠ BAC will always be less than m∠ BDC
—. However,
because they both intercept the same segment, CD
because m∠ BAC > m∠ BDC, ∠ A must be obtuse, and the
altitudes must intersect outside of the triangle.
222
Geometry
Worked-Out Solutions
A
D
C
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Chapter 6
If △ABC is an obtuse triangle, then m∠ BAC > m∠ BDC and
the orthocenter D is outside the triangle, where ∠ A is the
obtuse angle.
5. Graph △LMN.
4
X(−2, 1)
B
y
D(4, 3)
2
−2
A
C
−4
x
2
6
Y(2, −3)
Z(6, −3)
−3 − (−3) −3 + 3 0
—: m = —
=—=—=0
Slope of YZ
6−2
4
4
— is undefined.
The slope of the line perpendicular to YZ
2 + 6 −3 + (−3)
—= —
,—
midpoint of YZ
2
2
8 −6
= —, — = (4, −3)
2 2
(
(
D
Chapter 6 Review (pp. 350–352)
—.
1. DC = 20; Point B is equidistant from A and C, and ⃖⃗
BD ⊥ AC
So, by the Converse of the Perpendicular Bisector Theorem
(Thm. 6.2), DC = AD = 20.
—
—
⃗. So,
2. RS = 23; ∠ PQS ≅ ∠ RQS, SR ⊥ ⃗
QR, and SP ⊥ QP
by the Angle Bisector Theorem (Thm. 6.3), SR = SP. This
means that 6x + 5 = 9x − 4, and the solution is x = 3. So,
RS = 9(3) − 4 = 23.
3. m∠ JFH = 47°; Point J is equidistant from ⃗
FG and ⃗
FH. So,
by the Converse of the Angle Bisector Theorem (Thm. 6.4),
m∠ JFH = m∠ JFG = 47°.
4. Graph △TUV.
2
x = −3
−6
−4
y=3
)
— is x = 4.
The equation of the line perpendicular to YZ
−4
−4
−3 − 1
—: m = —
= — = — = −1
Slope of XY
2 − (−2) 2 + 2
4
— is m = 1.
The slope of the line perpendicular to XY
−2 + 2 1 + (−3)
—= —
,—
midpoint of XY
2
2
0 −2
= —, — = (0, −1)
2 2
y = mx + b
(
(
)
)
⋅
−1 = 1 0 + b
−1 = b
— through (0, −1)
The equation of the line perpendicular to XY
is y = x − 1.
Intersection of x = 4 and y = x − 1:
y
y=4−1=3
So, the coordinates of the circumcenter of △XYZ are (4, 3).
U(0, −1) x
−2
)
6. By the Incenter Theorem (Thm. 6.6), x = 5.
−2
−4
T(−6, −5)
−6
V(0, −5)
−1 + (−5)
0 −6
— = —0, —
midpoint of UV
= —, — = (0, −3)
2
2
2 2
−6 + 0 −5 + (−5)
−6 −10
—= —
, — = —, —
midpoint of TV
2
2
2
2
(
(
) (
)
) (
)
= (−3, −5)
— through
The equation of the perpendicular bisector of UV
its midpoint (0, −3) is y = −3, and the equation of the
— through its midpoint
perpendicular bisector of TV
(−3, −5) is x = −3. The point of intersection of the two
perpendicular bisectors is (−3, −3). So, the coordinates of
the circumcenter of △TVU are (−3, −3).
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Geometry
Worked-Out Solutions
223
Chapter 6
7.
8.
y
A(−10, 3)
F(−6, 3)
B(−4, 5)
y
2
6
−2
6
4
x
8
E(2, −2) G(5, −2) F(8, −2)
4
D(−4, 3)
−10
−8
−4
2
E(−7, 2)
−6
−4
−2
x
−8
−4 + (−4) 5 + 1
—= —
,—
midpoint of BC
2
2
−8 6
= —, — = (−4, 3)
2 2
0
3−3
—
The slope of AD is —— = — = 0.
−4 − (−10) 6
(
(
C(4, −4)
H(2, −5)
−6
C(−4, 1)
)
)
— through (−4, 3) is y = 3.
The equation of AD
−10 + (−4) 3 + 1
— = ——
,—
midpoint of AC
2
2
(
)
−14 4
, = (−7, 2)
=(
2 2)
— —
3
3
5−2
— is —
= — = — = 1.
The slope of BE
−4 − (−7) −4 + 7 3
y = mx + b
⋅
2 = 1 (−7) + b
2 = −7 + b
9=b
— through (−7, 2) is y = x + 9.
The equation of BE
The centroid has the coordinates of the intersection of y = 3
and y = x + 9.
y=x+9
3=x+9
−6 = x
So, the coordinates of the centroid are (−6, 3).
D(2, −8)
2 + 8 −2 + (−2)
—= —
,—
midpoint of EF
2
2
10 −4
= —, — = (5, −2)
2 2
−2 − (−8) −2 + 8 6
—
The slope of DG is — = — = — = 2.
5−2
3
3
(
(
)
)
y = mx + b
⋅
−2 = 2 5 + b
−2 = 10 + b
−12 = b
— through (5, −2) is y = 2x − 12.
The equation of DG
2 + 2 −2 + (−8)
—= —
,—
midpoint of ED
2
2
4 −10
= —, — = (2, −5)
2 2
−2 − (−5) −2 + 5 3 1
—
The slope of FH is — = — = — = —.
8−2
6
6 2
(
(
)
)
y = mx + b
⋅
1
−5 = — 2 + b
2
−5 = 1 + b
−6 = b
— through (2, −5) is y = —1x − 6.
The equation of FH
2
The centroid has the coordinates of the intersection of
y = 2x − 12 and y = —12 x − 6.
1
2x − 12 = —x − 6
2
1
2x = —x + 6
2
1
2 2x = 2 —x + 2 6
2
4x = x + 12
⋅
⋅
⋅
3x = 12
x=4
⋅
1
y = — 4 − 6 = 2 − 6 = −4
2
So, the coordinates of the centroid are (4, −4).
224
Geometry
Worked-Out Solutions
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Chapter 6
9.
2
5
y= x+4
y
6
10.
y
K(−8, 5)
M(0, 5)
4
G(1, 6)
H(5, 6)
O(3, 5.2)
L(−6, 3)
y=x+5
2
J(3, 1)
x=3
4
6
x
6−6 0
—=—
=—=0
slope of GH
5−1 4
— through J(3, 1)
The slope of the line perpendicular to GH
is x = 3.
5
6−1
—=—
= −—
slope of GJ
2
1−3
— is —2.
The slope of the line perpendicular to GJ
5
y = mx + b
2
6=— 5+b
5
6=2+b
⋅
4=b
— that passes
The equation of the line perpendicular to GJ
2
through H(5, 6) is y = —x + 4.
5
2
The orthocenter is the intersection of x = 3 and y = — x + 4.
5
2
y=— 3+4
5
6 20
y=—+—
5
5
26
y=—
5
The orthocenter of △GHJ is inside the triangle with
26
coordinates 3, — .
5
⋅
( )
2
x = −6
−4
O(−6, −1)
x
0
5−5
—=—
=—=0
slope of KM
0 − (−8) 8
— through
The equation of the line perpendicular to KM
L(−3, 3) is x = −6.
2
2
5−3
—=—
= — = — = −1
slope of KL
−8 − (−6) −8 + 6 −2
— is 1.
The slope of the line perpendicular to KL
y = mx + b
⋅
5=1 0+b
5=b
y=x+5
— that passes
The equation of the line perpendicular to KL
through M(0, 5) is y = x + 5.
The orthocenter is the intersection of x = −6 and y = x + 5.
y=x+5
y = −6 + 5
y = −1
The orthocenter of △KLM is outside the triangle with
coordinates (−6, −1).
—
( −6 +2 (−6) 8 +2 4 )
−12 12
,
=(
= (−6, 6)
2 2)
— = −6 + 0, 8 + 4 = −6, 12 = (−3, 6)
midpoint of AC
( 2 2 ) ( 2 2)
— = −6 + 0, 4 + 4 = −6, 8 = (−3, 4)
midpoint of BC
( 2 2 ) ( 2 2)
11. midpoint of AB = —, —
— —
— —
— —
— —
— —
The coordinates of the midsegments of △ABC are (−6, 6),
(−3, 6), and (−3, 4).
—
( −32+ 3 1 +2 5 ) ( 02 62 )
— = −3 + 1, 1 + (−5)
midpoint of DF
( 2 2 )
−2 −4
=( ,
= (−1, −2)
2 2 )
— = 3 + 1, 5 + (−5) = 4, 0 = (2, 0)
midpoint of EF
( 2 2 ) (2 2)
12. midpoint of DE = —, — = —, — = (0, 3)
— —
— —
— —
— —
The coordinates of the midsegments of △DEF are (0, 3),
(−1, −2), and (2, 0).
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Geometry
Worked-Out Solutions
225
Chapter 6
13. x + 4 > 8
4.
WY = WZ
Angle Bisector Theorem (Thm. 6.3)
6x + 2 = 9x − 13
x>4
4+8>x
−3x = −15
x=5
12 > x, or x < 12
⋅
WY = 6x + 2 = 6 5 + 2 = 32
The possible lengths for the third side of the triangle with
sides 4 inches and 8 inches are 4 in. < x < 12 in.
5. By the Incenter Theorem (Thm. 6.6), the incenter point is
equidistant to each side of the triangle. Because WC = 20,
BW = 20.
14. x + 6 > 9
x>3
6+9>x
6. AB > CB; By the Hinge Theorem (Thm. 6.12)
15 > x, or x < 15
The possible lengths for the third side of the triangle with
sides 6 meters and 9 meters are 3 m < x < 15 m.
7. m∠ 1 < m∠ 2; By the Converse of the Hinge Theorem
(Thm. 6.13)
8. m∠ MNP < m∠ NPM; By the Triangle Larger Angle
15. x + 11 > 18
Theorem (Thm. 6.10)
x>7
9.
11 + 18 > x
29 > x, or x < 29
The possible lengths for the third side of the triangle with
sides 11 feet and 18 feet are 7 ft < x < 29 ft.
16. Given
Prove
y
C(0, 6)
△XYZ, XY = 4, and XZ = 8
YZ > 4
Assume temporarily that YZ > 4. Then it follows that either
YZ < 4 or YZ = 4. If YZ < 4, then XY + YZ < XZ because
4 + YZ < 8 when YZ < 4. If YZ = 4, then XY + YZ = XZ
because 4 + 4 = 8. Both conclusions contradict the Triangle
Inequality Theorem (Thm. 6.11), which says that
XY + YZ > XZ. So, the temporary assumption that YZ > 4
cannot be true. This proves that in △XYZ, if XY = 4 and
XZ = 8, then YZ > 4.
17. Given that m∠ QRT > m∠SRT, by the Hinge Theorem
(Thm. 6.12), QT > ST.
18. Given that QT > ST, by the Converse of the Hinge Theorem
(Thm. 6.13), m∠ QRT > m∠SRT.
y=2
4
x=2
circumcenter
E(2, 2)
D(1.3, 0.7)
centroid
−2
6 x
4
A(0, −2)
orthocenter
B(4, −2)
Circumcenter:
0 + 4 −2 + (−2)
—= —
,—
midpoint of AB
2
2
4 −4
= —, — = (2, −2)
2 2
(
(
)
)
— through (2, −2) is x = 2.
The line perpendicular to AB
6 + (−2)
−4
— = —0, —
midpoint of AC
= 0, — = (0, 2)
2
2
2
(
) (
)
— through (0, 2) is y = 2.
The line perpendicular to AC
The intersection of x = 2 and y = 2 is (2, 2).
The coordinates of the circumcenter are (2, 2).
Orthocenter:
Chapter 6 Test (p. 353)
1. By the Triangle Midsegment Theorem (Thm. 6.8),
x = —12 12 = 6.
⋅
2. By the definition of midpoint, x = 9.
3.
RS = ST
Perpendicular Bisector
Theorem (Thm. 6.1)
3x + 8 = 7x − 4
−4x = −12
x=3
⋅
ST = 7 3 − 4 = 21 − 4 = 17
226
Geometry
Worked-Out Solutions
6 − (−2) 8
— is —
The slope of AC
= — = undefined.
0−0
0
— through
The equation of the line perpendicular to AC
B(4, −2) is y = −2.
−2 − (−2) 0
— is —
The slope of AB
= — = 0.
4−0
4
— through C(0, 6)
The equation of the line perpendicular to AB
is x = 0.
The intersection of x = 0 and y = −2 is (0, −2).
The coordinates of the orthocenter are (0, −2).
Centroid:
The slope of the line that contains B(4, −2) and the
4
2 − (−2)
—, (0, 2), is —
= — = −1.
midpoint of AC
0−4
−4
The equation of the line that contains B(4, −2) and the
—, (0, 2), is y = −x + 2.
midpoint of AC
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Chapter 6
The slope of the line that contains C(0, 6) and the midpoint
−2 − 6 −8
—, (2, −2), is —
= — = −4.
of AB
2−0
2
The equation of the line that contains C(0, 6) and the
—, (2, −2), is y = −4x + 6.
midpoint of AB
The intersection of y = x + 2 and y = −4x + 6 is:
16 > x
from the beach to Main Street, it must be perpendicular to
Main Street, and you ended up at the midpoint between
your house and the movie theater. So, the trail must be the
perpendicular bisector of the portion of Main Street between
your house and the movie theater. By the Perpendicular
Bisector Theorem (Thm. 6.1), the beach must be the same
distance from your house and the movie theater. So, Pine
Avenue is the same length as the 9-mile portion of Hill Street
between your house and the beach.
( )
— ≅ QR
— ≅ PR
—
PQ
R
∠R ≅ ∠P ≅ ∠Q
Prove
7+9 > x
14. 9 miles; Because the path represents the shortest distance
4 2
The coordinates of the centroid are —, — .
3 3
10. Given
x>2
The possible lengths of Pine Avenue are
2 miles < x < 16 miles.
−x + 2 = −4x + 6
3x = 4
4
x=—
3
4 6 2
4
y = −— + 2 = −— + — = —
3
3 3 3
13. 7 + x > 9
15. To determine the placement of the market, construct the
P
Q
Assume temporarily that △PQR is equilateral and
equiangular. Then it follows that m∠ P ≠ m∠ Q,
m∠ Q ≠ m∠ R, or m∠ P ≠ m∠ R. By the contrapositive of
the Base Angles Theorem (Thm. 5.6), if m∠ P ≠ m∠ Q,
then PR ≠ QR, if m∠ Q ≠ m∠ R, then QP ≠ RP, and
if m∠ P ≠ m∠ R, then PQ ≠ RQ. All three conclusions
contradict the fact that △PQR is equilateral. So, the
temporary conclusion must be false. This proves that if
△PQR is equilateral, it must also be equiangular.
11. By the Triangle Midsegment Theorem (Thm. 6.8),
GH = —12 FD. By the markings EG = GD. By the Segment
So, the area of △GEH = —12bh
(
)(
your
house
market
1. The definitions that are needed to prove the Converse of the
Perpendicular Bisector Theorem (Thm. 6.2) are the definition
of perpendicular bisector and the definition of segment
congruence.
2. Given
)
= —12 —12ED —12 FD
= —18 (ED)(FD).
Note that the area of △DEF = —12bh = —12 (ED)(FD).
So, the area of △GEH = —18(ED)(FD) = —14 —12 (ED)(FD)
[
12.
beach
movie
theater
Chapter 6 Standards Assessment (pp. 354–355)
Addition Postulate (Post. 1.2), EG + GD = ED. So, when
you substitute EG for GD, you get EG + EG = ED,
or 2(EG) = ED, which means that EG = —12 ED.
= —12 (EG)(GH)
perpendicular bisectors of each side. Where they intersect is
the location of the market.
Prove
— is the perpendicular bisector of DF
—.
YG
△DEY ≅ △FEY
Y
] = —A.
1
4
D
E
F
hiker 2
40°
1.8 miles
hiker 1
4 miles
140°
128°
Visitor
Center
4 miles
1.8 miles
X
G
Z
52°
Hiker one: 180° − 40° = 140°
Hiker two: 180° − 52° = 128°
Because 140° > 128°, the first hiker is farthest from the
visitor center, because the longer side is opposite the
larger angle.
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Geometry
Worked-Out Solutions
227
Chapter 6
STATEMENTS
—
1. YG is the perpendicular
—
bisector of DF .
2.
—
DE ≅ —
EF , —
YG ⊥ —
DF
REASONS
5. Given
1. Given
Prove
2. Definition of
perpendicular bisector
3. Definition of
perpendicular lines
4. ∠ DEY ≅ ∠ FEY
4. Right Angles Congruence
Theorem (Thm. 2.3)
5.
5. Reflexive Property of
Congruence (Thm. 2.1)
6. △DEY ≅ △FEY
6. SAS Congruence
Theorem (Thm. 5.5)
3. B;
1+5 5+2
6 7
—= —
, — = —, — = (3, 3.5)
midpoint of MN
2
2
2 2
—
The slope of the equation that contains the midpoint of MN
through L(3, 8):
8 − 3.5 4.5
slope = — = — = undefined
3−3
0
The equation of the line is x = 3.
(
) ( )
3+5 8+2
8 10
—= —
, — = —, — = (4, 5)
midpoint of LN
2
2
2 2
—
The slope of the equation that contains the midpoint of LN
through M(1, 5):
(
) ( )
5−5 0
slope = — = — = 0
4−1 3
The equation of the line is y = 5.
The intersection of x = 3 and y = 5 is (3, 5). So, the
coordinates of the centroid are (3, 5).
4. In Step 1, the constructed line connects two points that
are each equidistant from both A and B. So, it is the
—, and therefore every point
perpendicular bisector of AB
on the line is equidistant from A and B. In Step 2, the
constructed line connects two points that are each equidistant
—,
from both B and C. So, it is the perpendicular bisector of BC
and therefore every point on the line is equidistant from
B and C. So, the point where these two lines intersect is
equidistant from all three points. So, the circle with this point
as the center that passes through one of the points will also
pass through the other two.
228
∠B ≅ ∠C
B
3. ∠ DEY and ∠ FEY are
right angles.
—
YE ≅ —
YE
— ≅ AC
—
AB
Geometry
Worked-Out Solutions
A
D
C
STATEMENTS
REASONS
1. Draw ⃖⃗
AD, the angle
bisector of ∠ CAB.
1. Construction of angle
bisector
2. ∠ CAD ≅ ∠ BAD
2. Definition of angle bisector
3. —
AB ≅ —
AC
3. Given
4. —
DA ≅ —
DA
4. Reflexive Property of
Congruence (Thm. 2.1)
5. △ADB ≅ △ADC
5. SAS Congruence Theorem
(Thm. 5.5)
6. ∠ B ≅ ∠ C
6. Corresponding parts of
congruent triangles are
congruent.
—
6. a. Let T be the midpoint of QR . The coordinates of T are
0 14
−3 + 3 8 + 6
2
2 2
2
—. The coordinates of U are
Let U be the midpoint of SR
4 8
3+1 6+2
—, — = —, — = (2, 4).
2
2 2
2
—. The coordinates of V are
Let V be the midpoint of QS
−2 10
−3 + 1 8 + 2
—, — = —, — = (−1, 5).
2
2 2
2
The coordinates of the midsegments of the triangle are
T(0, 7), U(2, 4), and V(−1, 5).
(
(
(
) ( )
—, — = —, — = (0, 7).
) ( )
) (
)
y
Q
8
T
R
V
U
2
−4
−2
S
2
4
x
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Chapter 6
—
2
7−5
0 − (−1) 1
6−2 4
—=—
=—=2
slope of SR
3−1 2
— RS
—.
Because the slopes are the same, TV
b. slope of TV = — = — = 2
1
−1
4−5
—=—
= — = −—
slope of VU
3
2 − (−1)
3
1
−2
6−8
—
slope of QR = — = — = −—
3
3 − (−3)
6
——
Because the slopes are the same, VU
QR.
3
7
−
4
— = — = −—
slope of TU
2
0−2
3
6
8−2
—
slope of QS = — = — = −—
2
−3 − 1 −4
——
Because the slopes are the same, TU
QS.
——
—
—
TV = √ (0 − (−1))2 + (7 − 5)2 = √ 1 + 4 = √5
——
—
SR = √(3 − 1)2 + (6 − 2)2 = √4 + 16
—
⋅
—
—
= √20 = √ 4 5 = 2√ 5
—
—
1
1
Because √ 5 = —( 2√ 5 ), TV = — SR.
2
2
——
—
—
VU = √(2 − (−1))2 + (4 − 5)2 = √ 9 + 1 = √10
——
—
QR = √(−3 − 3)2 + (8 − 6)2 = √ 36 + 4
—
⋅
—
—
= √40 = √ 4 10 = 2√ 10
—
—
1
1
Because √ 10 = —( 2√ 10 ), VU = — QR.
2
2
——
—
—
TU = √(0 − 2)2 + (7 − 4)2 = √4 + 9 = √13
——
—
QS = √(−3 − 1)2 + (8 − 2)2 = √ 16 + 36
—
⋅
—
—
= √52 = √ 4 13 = 2√ 13
—
—
1
1
Because √ 13 = —( 2√ 13 ), TU = —QS.
2
2
7. yes; Because translations are rigid motions, the image will be
congruent to the original triangle after the translation. Then,
because dilations are similarity motions, the image after the
dilation will be similar to the previous image and the original
triangle.
1 − (−1)
4−1
8. The slope of ⃖⃗
BD is — = —.
2
3
2 − (−2) 4 2
The slope of ⃖⃗
B′D′ is — = — = —.
8−2
6 3
Because the slope of ⃖⃗
BD is equal to the slope of ⃖⃗
B′D′,
⃖⃗ ⃖⃗
BD
B′D′.
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Geometry
Worked-Out Solutions
229