Download Chapter 13 Answers

Chapter 13. Geometry of the Circle
C
13-1 Arcs and Angles (pages 541–542)
mArBr 5 60, m/A9O9B9 5 60. Therefore,
/AOB > /A9O9B9. Radii of the same circle are
congruent, so OA > OB and OrAr > OrBr . By
definition, nAOB and nA9O9B9 are isosceles
triangles with m/OAB 5 m/OBA 5 60 and
m/O9A9B9 5 m/O9B9A9 5 60. Then
/OAB > /OBA > /O9A9B9 > /O9B9A9
and nAOB , nA9O9B9 by AA,.
Writing About Mathematics
1. Yes. If two lines intersect at the center of a circle,
they form four central angles that are also
vertical angles. Since vertical angles are
congruent, the arcs intercepted by each pair of
vertical central angles are also congruent.
2. Yes. Since the degree measure of a circle is 360,
the four arcs each measure 90 degrees, and the
central angles that intercept the arcs each
g
13-2 Arcs and Chords (pages 551–552)
g
Writing About Mathematics
1. The apothem is perpendicular to the chord, so
the triangle formed is a right triangle. Also, the
apothem is the distance from the center to the
chord, so the length of one leg is 3 inches. The
hypotenuse of the triangle is the radius and
measures 5 inches. By the Pythagorean Theorem,
the length of the other leg is 4 inches, so a 3-4-5
right triangle is formed. However, the apothem
intersects the chord at its midpoint, so the length
of the chord is 8 inches.
2. No. If the arcs belong to two circles with different
radii, then the arcs are not congruent.
Developing Skills
3. 3 in.
4. 4.5 cm
5. 1.5 ft
6. 12 mm
7. !6 cm
8. 10 in.
9. 24 ft
10. 14 cm
11. 12.4 mm
12. 2!5 yd
13. radius 5 11, diameter 5 22
14. 5
15. 25
16. 30
17. 12!5
18. 12
19. 5!5
20. OB 5 30, DE 5 60
21. OB 5 30, OC 5 15!3
22. LM
Applying Skills
measure 90 degrees. Therefore, AC and BD are
perpendicular.
Developing Skills
3. 35
4. 48
5. 90
6. 140
7. 180
8. 60
9. 75
10. 100
11. 120
12. 170
13. 91
14. 89
15. 91
16. 138
17. 138
18. 133
19. 133
20. 227
21. 180
22. 180
23. 100
24. 110
25. 35
26. 115
27. 115
28. 210
29. 145
30. 145
31. 150
32. 215
Applying Skills
33. In a circle, central angles are congruent if their
intercepted arcs are congruent. Since AB > CD ,
/AOB > /COD. All radii of the same circle are
congruent, so OA > OB > OC > OD. Therefore,
nAOB > nCOD by SAS.
C C
34. /AOC and /BOD are vertical angles, so they
are congruent. Radii of the same circle are
congruent, so OA > OB > OC > OD. Then
nAOC > nBOD by SAS. Since corresponding
parts of congruent triangles are congruent,
AC > BD.
23. We are given (O > (Or and CD > AB > ArBr.
Since congruent circles are circles with congruent
radii, OA > OB > OC > OD > OrAr > OrBr,
and nCOD > nAOB > nA9O9B9 by SSS.
Corresponding parts of congruent triangles are
congruent, so /COD > /AOB > /A9O9B9.
C 5 90, mADC
C 5 270
35. mAC
36. Since AC and BD, the diagonals of ABCD, are
perpendicular, ABCD is a rhombus. /AOC and
/BOC are right angles since perpendicular lines
intersect to form right angles. Radii of the same
circle are congruent, so OA > OB > OC.
Therefore, nAOB and nBOC are isosceles right
triangles and m/ABO 5 m/CBO 5 45. Then
m/ABC 5 m/ABO 1 m/CBO 5 90. A
rhombus with a right angle is a square, so ABCD
is a square.
Hands-On Activity
In 1–2, results will vary.
3. By the definition of the degree measure of an
arc, since mAB 5 60, m/AOB 5 60 and since
24. We are given (O > (Or and CD > AB > ArBr .
Since congruent circles are circles with congruent
radii, OA > OB > OC > OD > OrAr > OrBr ,
and nCOD > nAOB > nA9O9B9 by SSS. Corresponding parts of congruent triangles are congruent, so /COD > /AOB > /A9O9B9. In a
circle or in congruent circles, if central angles are
congruent, then their intercepted arcs are
congruent. Therefore, CD > AB > ArBr .
C
C
333
C
C
16. a. 40
17. a. 54
b. 80
b. 48
c. 80
c. 50
d. 40
d. 55
e. 100
e. 50
18. a. 60
19. a. 120
b. 90
b. 120
c. 210
c. 120
d. 45
d. 60
e. 105
e. 60
f. 30
f. 60
Applying Skills
20. a. Since vertical angles are congruent,
/LPR > /SPM. If two inscribed angles of a
circle intercept the same arc, then they are
congruent. Therefore, /RLP > /MSP and
nLPR , nSPM by AA,.
25. We are given that diameter AOB intersects chord
CD at E and bisects CD at B. Then, CB > BD .
In a circle, congruent chords have congruent
central angles, so /COE > /DOE. All radii of a
circle are congruent, so OC > OD, and
OE > OE by the reflexive property of
congruence. Therefore, nOEC > nOED by
SAS. Corresponding parts of congruent triangles
are congruent, so /DEO > /CEO. If two lines
intersect to form congruent adjacent angles, then
they are perpendicular, so AOB ' CD. A
diameter perpendicular to a chord bisects the
chord, so E is the midpoint of CD and DE > CE.
C
C
C
26. 5 cm
27. In a circle, if the lengths of two chords are
unequal, then the shorter chord is farther from
the center. AB is farther from the center than
BC, so AB < BC. BC is farther from the center
than AC, so BC < AC. By the transitive property
of inequality, AB < BC < AC. If the lengths of
two sides of a triangle are unequal, then the
measures of the angles opposite these sides are
unequal and the larger angle lies opposite the
longer side. Therefore, /B is the largest angle of
nABC.
b. 8 cm
21. If mAB 5 100 and mBC 5 130, mCA 5 130.
Therefore, AB > CA. In a circle, congruent arcs
have congruent chords, so AB > CA and nABC
is isosceles.
C
C
C
C C
22. a. In a parallelogram, opposite angle are
congruent. Therefore, /ABC > /ADC or
m/ABC 5 m/ADC. The measure of an
inscribed angle of a circle is equal to one-half
the measure of its intercepted arc, so
13-3 Inscribed Angles and Their Measures
(pages 555–558)
C
C
m/ABC 5 12mABC and m/ADC 5 12mADC .
Writing About Mathematics
C
By the substitution postulate, 12mABC
1. Draw diameter AOB. Choose any point C on the
circle and draw CA and CB. Inscribed /ACB
intercepts a semicircle, so it is a right angle.
Triangle ACB is a right triangle with hypotenuse
AOB and legs CA and CB.
2. Yes. The measure of an inscribed angle of a circle
is equal to one-half the measure of its
intercepted arc. Therefore, m/ABC 5 25 and
m/PQR 5 25, so /ABC > /PQR.
Developing Skills
3. 44
4. 36
5. 85
6. 100
7. 140
8. 24
9. 90
10. 120
11. 190
12. 250
13. a. 160
14. a. 50
15. a. 34
b. 44
b. 100
b. 68
c. 56
c. 50
c. 84
d. 112
d. 80
d. 34
e. 200
e. 160
e. 42
C C C
C 5 mADC
C 5 180
b. mABC
5 12mADC or mABC 5 mADC .
c. By the reasoning in part a, m/BAD 5
m/BCD 5 90 and ABCD is equiangular. An
equiangular parallelogram is a rectangle, so
ABCD is a rectangle.
C
C
C
23. It is given that mDE 1 mEF 5 mFGD,
C C
C
C
C
so mDEF 5 mFGD. Since mDEF 1 mFGD
5 360, mDEF 5 180. The measure of an
inscribed angle of a circle is equal to one-half the
measure of its intercepted arc, so m/DEF 5 90
and nDEF is a right triangle.
24. It is given that AB > CD. Since AOC and BOD
are diameters, and /ABC and /BCD are
inscribed in semicircles. Therefore, /ABC and
/BCD are right angles and congruent. If two
inscribed angles of a circle intercept the same
arc, then they are congruent, so /BAC > /CDB.
Thus, nABC > nDCB by ASA.
334
25. In a circle, congruent chords have congruent arcs.
Since AB > CD, AB > CD. Vertical angles are
congruent, so /AEB > /DEC. If two inscribed
angles of a circle intercept the same arc, then
they are congruent, so /BAE > /CDE.
Therefore, nABC > nDCB by AAS.
30. In Exercise 26, we proved that a trapezoid
inscribed in a circle must be isosceles. Since
AB y DC, ABCD is a trapezoid and thus
isosceles. It follows that DA > BC. Two chords
are congruent if and only if their arcs are
congruent. Thus, DA > BC , so CA > CB 1
BA > DA 1 BA > BD . Similarly, CB y ED
and DBCE is an isosceles trapezoid. Again,
two chords are congruent if and only if
their arcs are congruent. Thus, EC > DB > CA .
Since EF y DC, DCFE is also an isosceles
trapezoid with ED > FC. Thus,
DF > DE 1 EF > FC 1 EF > EC , and
CA > DB > DF > EC .
C C
C
26. Let ABCD be inscribed in circle O with AB y CD.
Draw AC. If parallel lines are cut by a
transversal, then alternate interior angles are
congruent, so /ACD > /BAC or m/ACD 5
m/BAC. The measure of an inscribed angle of a
circle is equal to one-half the measure of its
C
C
m/BDC 5 12mBC . By the substitution postulate,
1
1
2 mAD 5 2 mBC or mAD 5 mBC . Therefore,
C
C C
C
C
AD > BC . In a circle, congruent arcs have
C
C
C
13-4 Tangents and Secants (pages 564–567)
Writing About Mathematics
1. No. Since l is tangent to circle O at A, l ' OA.
Since m is tangent to circle O at B, m ' OB.
Both OA and OB are segments of AOB. If two
lines are each perpendicular to the same line,
then they are parallel. Therefore, l || m and the
lines do not intersect.
2. A polygon inscribed in a circle intersects the
circle at each of its vertices, whereas a circle
inscribed in a polygon intersects each of its sides
at exactly one point.
Developing Skills
C
C or
m/ACD 5 12mAD and m/BDC 5 12mBC . By
C5
C
C
C
C.
mAD 5 mBC . Therefore, AD > BC
C > BC
C , mAD
C 5 mBC
C . The measure of
28. Since AD
1
2 mBC
an inscribed angle of a circle is equal to one-half
the measure of its intercepted arc, so
m/ACD 5 12mAD and m/BDC 5 12mBC .
Halves of equal quantities are equal, so m/ACD
5 m/BDC and /ACD > /BDC. If two lines
are cut by a transversal so that the opposite
interior angles formed are congruent, then they
are parallel. Therefore, AB y CD.
C
C
C
27. We are given that AB y CD. Draw AC. If parallel
lines are cut by a transversal, then alternate
interior angles are congruent, so
/ACD > /BAC or m/ACD 5 m/BAC. The
measure of an inscribed angle of a circle is equal
to one-half the measure of its intercepted arc, so
the substitution
C
31. In a circle, two chords are congruent if and only
if their arcs are congruent. Since AB is not
congruent to CD , AB is not congruent to CD.
Since opposite sides of ABCD are not congruent,
ABCD is not a parallelogram.
congruent chords, so AD > BC and ABCD is an
isosceles trapezoid.
postulate, 12mAD
C
C C C C C C
C C C C
C
intercepted arc, so m/ACD 5 12mAD and
C
C C
C C C
C
3. AB 5 10, BC 5 15, CA 5 15; since BC 5 CA,
BC > CA so nABC is isosceles.
4. AB 5 40, BC 5 50, CA 5 30; since 302 1 402 5
502, nABC is a right triangle.
5. a. 25
6. a. 7
7. a. 24
b. 20
b. 14
b. 16
c. 40
c. 10
c. 36
8. a. 7
9. a. 41
10. a. 12
b. !13
b. 40
b. 12
c. !13
c. 40
c. 5
d. 13
11. a. 3
12. a. 4
d. 8
b. 6
b. 3
e. 11
c. 8
c. 3
f. 12
d. 10
29. In a circle, all radii are congruent, so
OA > OB > OC > OD. Vertical angles are
congruent, so /AOB > /COD. Then
nAOB > nCOD by SAS. Corresponding parts
of congruent triangles are congruent, so
/BAC > /DCA. If two lines are cut by a
transversal so that the opposite interior angles
formed are congruent, then they are parallel.
Therefore, AB y CD.
335
13. a. !3 cm
b. 1.73 cm
14. a. 80
b. 70
c. 50
e. 80
f. 10
g. 20
i. 110
j. 130
k. 360
Applying Skills
g
perpendicular to a radius at a point of
tangency, so AC ' OA and m/OAC 5 90.
The sum of the measures of the angles in a
triangle is 180, so m/AOC 5 45. Since
/OAC > /AOC, AC > AO.
d. 160
h. 100
g
15. PQ and PR are tangent to circle O at Q and R,
respectively. Tangent segments drawn to a circle
from an external point are congruent, so
PQ > PR. PO > PO by the reflexive property of
congruence. Radii of a circle are congruent, so
OQ > OR. Therefore, nOPQ > nOPR by SSS.
Corresponding parts of congruent triangles are
congruent, so /OPQ > /OPR. By definition,
b. Since nAOC is a 45-45-degree right triangle,
OC 5 "2OA.
c. From part a, m/ACB 5 90 and AC > AO.
Radii of a circle are congruent, so OA > OB.
By the same reasoning in part a, BC > BO. By
the transitive property of congruence,
AO > OB > BC > AC. Since AOBC is
equilateral and has a right angle, it is a square.
h
PO bisects /RPQ.
g
g
19. Isosceles nABC with vertex A is circumscribed
about circle O, and D, E, and F are the points of
tangency of AB, BC, and AC, respectively. By the
isosceles triangle theorem, /DBE > /FCE. If
two tangents are drawn to a circle from an
external point, then the line segment from the
center of the circle to the external point bisects
the angle formed by the tangents. Therefore,
m/OBE 5 12m/DBE and m/OCE 5 12m/FCE,
so /OBE > /OCE. Since BC is tangent to circle
O at E, BC ' OE, forming congruent right
angles /BEO and /CEO. OE > OE by the
reflexive property of congruence, so nBEO >
nCEO by AAS. Corresponding parts of
congruent triangles are congruent, so BE > CE
and E is the midpoint of BC.
16. PQ and PR are tangent to circle O at Q and R,
respectively. Tangent segments drawn to a circle
from an external point are congruent, so
PQ > PR. PO > PO by the reflexive property of
congruence. Radii of a circle are congruent, so
OQ > OR. Therefore, nOPQ > nOPR by SSS.
Corresponding parts of congruent triangles are
congruent, so /QOP > /ROP. By definition,
h
OP bisects /QOR.
17. a. Tangent segments drawn to a circle from an
external point are congruent, so PQ > PR.
By the isosceles triangle theorem,
/PQR > /PRQ.
b. From part a, PQ > PR and /PQR > /PRQ.
SP > SP by the reflexive property of
congruence, so nQPS > nRPS by SAS.
Corresponding parts of congruent triangles
are congruent, so QS > RS and QS 5 RS.
Also, /QSP > /RSP. If two lines intersect to
form congruent adjacent angles, then they are
perpendicular. Therefore, OP ' QR.
20. a. If a line is tangent to a circle, then it is
perpendicular to a radius at a point on the
g
g
circle. Therefore, AB ' OA and AB ' OrB. If
two lines are each perpendicular to the same
line, then they are parallel, so OA y OrB. We
are given that OA < O9B. Since one pair of
opposite sides of OABO9 is not both
congruent and parallel, OABO9 is not a
c. The length of the altitude to the hypotenuse of
a right triangle is the mean proportional
between the lengths of the segments of the
hypotenuse. Thus, if x 5 OS, then x4 5 10 42 x or
2x2110x 5 16. Solving for x gives x 5 2 or 8.
Since OS < SP, OS 5 2 and SP 5 8.
18. a. We are given perpendicular tangents
AC and BC. Since AC ' BC, m/ACB 5 90.
If two tangents are drawn to a circle from an
external point, then the line segment from the
center of the circle to the external point
bisects the angle formed by the tangents.
Therefore, m/ACO 5 45. A tangent line is
g
parallelogram. Therefore, OOr cannot be
g
parallel to AB.
g
g
b. From part a, AB ' OA and AB ' OrB.
Therefore, /OAC and /O9BC are right
angles and congruent. By the reflexive
property of congruence, /C > /C. Therefore,
nOAC , nO9BC by AA,.
c. AC 5 8, AB 5 4, OC 5 2!17, O9C 5 3!17,
OO9 5 "17
336
2. Yes. Vertical angles are congruent, and each
central angle is equal to the measure of its
intercepted arc. One-half of the sum of two equal
intercepted arcs is the intercepted arc.
Developing Skills
3. 35
4. 30
5. 30
6. 166
7. 170
8. 60
9. 50
10. 45
11. 50
12. 130
13. 130
14. 180
15. 20
16. 100
17. 80
18. 30
19. 60
21. a. We are given that OA 5 O9B, so OA > OrB.
If a line is tangent to a circle, then it is
perpendicular to a radius at that point on the
g
g
circle. Therefore, AB ' OA and AB ' OrB,
forming congruent right angles /OAC and
/O9BC. Vertical angles are congruent, so
/OCA > /O9CB. nOCA > nO9CB by AAS.
Corresponding parts of congruent triangles
are congruent, so OC > OrC and OC 5 O9C.
b. From part a, nOCA > nO9CB.
Corresponding parts of congruent triangles
are congruent, so AC > BC and AC 5 BC.
Hands-On Activity
a. Results will vary.
b. (1) By definition, the center, P, is the
intersection of the bisectors of the angles of a
regular polygon. If a point lies on the
bisector of an angle, then it is equidistant
from the sides of the angle. Thus, the center P
is equidistant from the sides of the regular
polygon. The distance from a point to a line
is equal to the length of the perpendicular
from the foot. Thus, the apothems are all
congruent.
C
C
28. a. 150
d. 30
Applying Skills
b. 150
e. 75
20.
21.
24.
27.
mRQ 5 135, mRSQ 5 225
55
22. 140
23. 35
140
25. 70
26. 220
a. 125
b. 55
c. 55
d. 90
e. 90
f. 125
g. m/AED 5 12 (55 1 125) 5 90. Therefore,
BD ' AC. A line through the center of a
circle that is perpendicular to a chord bisects
the chord, so BD bisects AC.
c. 30
f. 105
g
29. a. We are given that AB y CP. If two parallel
lines are cut by a transversal, then the
alternate interior angles formed are
congruent. Therefore, /OCP > /OEA and
/OPC > /OAE, and nOPC , nOAE by
AA,.
(2) Each circle constructed in part a used one of
the apothems as the radius. Since the
apothems are all congruent, every apothem
is also a radius of the circle, so the foot of
each apothem is on the circle.
(3) By construction, each side of the regular
polygon is perpendicular to one of the
apothems at the foot of the apothem. If a line
is perpendicular to a radius at a point on a
circle, then it is tangent to the circle. Since
the foot of each apothem is on the circle and
each apothem is a radius of the circle, the
sides of the regular polygon are tangent to
the circle.
C
C
C
C
C
b. mAD 5 75, mCF 5 45, mFB 5 60,
mBD 5 75, mAC 5 105, m/P 5 30
30. Vertical central angles are congruent and
intercept congruent arcs. Therefore, /FOE >
C
C
/GOD and mEF 5 mDG . We are given that
C C
C 5 mFB
C and mDGB
C
C 2 mEF
C 2 mDG
mEFB
C . Thus, by substitution, mFB
C 5 mEFB
C2
5 mGB
C
C
C
C
mEF 5 mDGB 2 mDG 5 mDG . The measure
mEFB 5 mDGB . By the partition postulate,
2
c. r2 5 a2 1 A 2s B
13-5 Angles Formed By Tangents, Chords,
and Secants (pages 572–574)
of an angle formed by a tangent and a secant is
Writing About Mathematics
cepted arcs, so m/A 5 21 (mDGB 2 mFB )
and m/C 5 12 (mEFB 2 mGB ) . By substitu-
equal to one-half the difference of the inter-
C
g
1. Yes. Let AB be tangent to the circle O at A. Then
g
g
AB ' OA and m/AOB 5 90. Let AC be a
secant that intersects circle O at A and C. Then
m/OAB 5 m/OAC 1 m/CAB. Therefore,
m/OAC Þ 90 and the radius OA is not
tion, m/A 5
C
C C
C) 5
C 2 mFB
1
2 (mDGB
C ) 5 m/C, so /A > /C.
C 2 mGB
1
2 (mEFB
Therefore, OA > OC and nAOC is isosceles.
g
perpendicular to the secant AC.
337
C
12.
15.
18.
19.
20.
21.
22.
23.
31. Let D be a point on major ADB . Since
C 5 120 and mADB
C 5 240.
m/AOB 5 120, mAB
The measure of an angle formed by two tangents
intersecting outside the circle is equal to one-half
the difference of the measures of the intercepted
arcs, so m/P 5 12 (240 2 120) 5 60. Tangent
segments drawn to a circle from an external
point are congruent, so /PAB > /PBA. Since
the sum of the measures of the angles of a
triangle is 180, m/PAB 5 m/PBA 5 60.
Therefore, nABP is equiangular. If a triangle is
equiangular, then it is equilateral, so nABP is
equilateral.
27
13. 20 and 6
16
16. 6
8
AC 5 24, AB 5 6, BC 5 18
AC 5 20, AB 5 5, BC 5 15
AC 5 16, AB 5 4, BC 5 12
AC 5 25, AB 5 9, BC 5 16
20
24. 12
26. AB 5
27.
29.
31.
,
3 $21
5 AC
5
8
28. 8!2
AD 5 3, AE 5 6
14. 14 and 2
17. 4
25. 30
15 $21
5
30. AD 5 7, AE 5 12
15
2
13-7 Circles in the Coordinate Plane
(pages 584–587)
g
32. We are given secant ABC intersecting circle O at
A and B and chord BD. Assume that m/CBD 5
Writing About Mathematics
1. Yes. Consider a circle with center (h, k) and
radius r. Two points on the circle are (h 2 r, k)
and (h 1 r, k). The chord connecting these points
goes through the center, so it is a diameter. These
points have the same y-coordinate, so this
diameter is horizontal. Two other points on the
circle are (h, k 1 r) and (h, k 2 r). The chord
connecting these points goes through the center,
so it is a diameter. These points have the same
y-coordinate, so this diameter is vertical.
2. Yes. Dividing each side by 3 gives x2 1 y2 5 4.
This is the equation of a circle centered at the
origin with a radius of 2.
Developing Skills
3. x2 1 y2 5 9
4. (x 2 1) 2 1 (y 2 3) 2 5 9
5. (x 1 2) 2 1 y2 5 36
6. (x 2 4) 2 1 (y 1 2) 2 5 100
7. (x 2 6) 2 1 y2 5 81
8. (x 1 3) 2 1 (y 1 3) 2 5 4
9. x2 1 y2 5 4
10. x2 1 y2 5 16
11. (x 2 2) 2 1 (y 2 9) 2 5 16
12. (x 1 1) 2 1 (y 2 3) 2 5 16
13. x2 1 (y 2 12) 2 5 25
14. (x 1 6) 2 1 (y 2 1) 2 5 9
15. (x 2 1) 2 1 (y 2 6.5) 2 5 76.25
16. (x 2 10) 2 1 (y 1 1) 2 5 73
17. (x 1 3) 2 1 (y 2 2) 2 5 4
18. (x 2 4) 2 1 (y 2 3) 2 5 16
19. (x 2 5) 2 1 (y 1 5) 2 5 9
20. (x 1 4) 2 1 (y 1 4) 2 5 25
21. (x 1 1) 2 1 y2 5 1
22. x2 1 (y 1 2) 2 5 36
C
C
1
2 mBD . /ABD is an inscribed angle, so m/ABD
5 12mAD . Since /ABD and /CBD form a linear
pair, m/CBD 1 m/ABD 5 180. By the
C
C
C 1 mAD
C 5 360. However, mBD
C1
or mBD
C
C
C
C 5 360.
mAD 2 360 since mBD 1 mAD 1 mAB
C.
Our assumption is false and m/CBD Þ 12mBD
substitution postulate, 12mBD 1 12mAD 5 180
13-6 Measures of Tangent Segments,
Chords, and Secant Segments
(pages 579–581)
Writing About Mathematics
1. Yes. If two chords intersect, the product of the
measures of the segments of one chord is equal
to the product of the segments of the other. Since
AB 5 24, and M is the midpoint of AB, AM 5
BM 5 24. Because 12 3 12 5 144, any chord of
circle O that intersects AB at M is separated by
M into two segments such that the product of the
lengths is also 144.
2. No. If two secant segments are drawn to a circle
from an external point, then the product of the
lengths of one secant segment and its external
segment is equal to the product of the lengths of
the other secant segment and its external
segment. Since (AP)(BP) 5 (CP)(DP) and
AP > CP, BP must be less than CP.
Developing Skills
3. 8
4. 6
5. 10
6. 7
7. 8
8. 28
9. 4
10. 8
11. 4
338
23.
28.
y
1 O
1
y
x
(2, 25)
(1, 1)
1
x
O 1
24.
y
(24, 4)
O
25.
1
29. (x 2 2) 2 1 (y 2 3) 2 5 169
30. Yes. The equation of the circle is x2 1 y2 5 32.
Substituting (4, 4) into the equation gives 42 1 42
5 32. Therefore, (4, 4) is a point on the circle.
31. Yes. It is equivalent to (x 1 2) 2 1 (y 2 1) 2 5 25,
which is the equation for a circle centered at
(22, 1) with a radius of 5.
Applying Skills
32. a. y 5 2x 1 2
b. x 5 0
1
c. y 5 3x 1 2
d. (0, 2)
x
1
y
(232 , 1)
x
O
21 1
26.
y
O1
21
27.
e. The perpendicular bisector of a side is
equidistant from the endpoints of each side, so
the circumcenter, which is the intersection of
the three perpendicular bisectors, is
equidistant from the vertices of nABC.
f. Yes. Let (0, 2) be the center of the circle and
A(2, 6) be a point on the circle. Then the radius
is !20 and the equation of the circle is
x2 1 (y 2 2)2 5 20. Substituting B(24, 0)
and into the equation gives 242 1 (0 2 2) 2
5 16 1 4 5 20, so B is on the circle.
Substituting C(4, 0) into the equation gives 42
1 (0 2 2)2 5 16 1 4 5 20, so C is on the circle.
33. a. Yes. The x-coordinate of the incenter is 3.
6(3) 1 8y 5 10
x
(
5 , 23
2 4
)
y
18 1 8y 5 10
1 O
(0, 0) 1
y 5 21
x
The incenter is at (3, 21).
b. The incenter is on each of the angle bisectors,
and a point on the bisector of an angle is
equidistant from the sides of the angle.
c. Yes. The equation of PR is y 5 2, so (3, 2) is on
PR. Since the incenter of the triangle is
339
12. a. (5, 5)
b. Tangent
13. a. (2, 2)
b. Tangent
14. a. (0, 24) and (2, 22)
b. Secant
15. x 5 3
16. y 5 24
17. y 5 x 2 4
18. y 5 2x 2 10
Applying Skills
19. a. y 5 3x 2 5
b. x 5 0
c. P(0, 25)
d. PB 5 0 and PE 5 0, so
(PA)(PB) 5 (PD)(PE) 5 0.
20. a. y 5 x 1 2
b. y 5 10
c. P(8, 10)
d. PA 5 2!2, PB 5 16 !2, PD 5 8
?
(PA)(PB) 5
(PD) 2
(2 !2)(16 !2) 5 82
64 5 64 ✔
21. a. y 5 2x 1 6
b. y 5 x 2 6
c. (6, 0)
d. PA 5 PB 5 18
22. The line and the circle intersect at a single point:
(18, 24).
23. a. Both points are solutions to the equation
(x 2 2) 2 1 (y 2 3) 2 5 25.
b. 4
24. a. The equation of AB is y 5 234x 1 23 . The
equation of BC is x 5 2. The equation of AC is
equidistant from sides of the circle, PQ and
QR also intersect the circle at one point.
d. (x 2 3) 2 1 (y 1 1) 2 5 9
34. The equation of the perpendicular bisector of
AB is y 5 22x 2 4 and the equation of the
perpendicular bisector of BC is y 5 21.
21 5 22x 2 4
2x 5 23
x 5 232
The circumcenter of nABC is (21.5, 21), which
is also the center of the circle. The radius equals
the distance between (21.5, 21) and
A(21, 3) 5 "(21 1 1.5) 2 1 (3 1 1) 2 5 !16.25.
The equation of the circle is
(x 1 1.5) 2 1 (y 1 1) 2 5 16.25.
35. a.
d.
36. a.
b.
c.
d.
12 !3 ft
b. (s, s !3)
c. A s, s !3
3 B
6 ft
e. 12 ft
(x 2 13)2 1 (y 2 13)2 5 169
(8, 1) and (8, 25); (18, 1) and (18, 25)
(8, 13) or (18, 13)
1,800 ft and 800 ft
y 5 10. AB intersects the circle at (2325, 515) .
13-8 Tangents and Secants in the
Coordinate Plane (pages 592–593)
BC intersects the circle at (2, 7). AC intersects
the circle at (21, 10).
b. The sides of the triangle each intersect the
circle at exactly one point.
c. Yes. Each side of the triangle is tangent to the
circle.
Writing About Mathematics
1. Yes. The equation of the circle is
(x 2 r)2 1 (y 2 k)2 5 r2. The equation of the
y-axis is x 5 0. Solving for y gives:
(0 2 r) 2 1 (y 2 k) 2 5 r2
r2 1 (y 2 k) 2 5 r2
Review Exercises (pages 598–599)
(y 2 k) 2 5 0
1. a. 100
2. a. 100
3. a. 40
b. 50
b. 110
b. 130
c. 80
c. 75
c. 65
d. 90
d. 30
d. 50
e. 90
e. 100
e. 95
4. PB 5 6, BC 5 18
5. 10
6. AE 5 3, EC 5 8
7. (1)
8. (3)
9. 60
10. 15 cm
11. AB 5 4, AC 5 12
12. mAB 5 mAC 5 100, mBC 5 160
13. Let ABCD be inscribed in circle O with AB y CD.
Draw AC. If parallel lines are cut by a
transversal, then alternate interior angles
are congruent, so /ACD > /BAC or
m/ACD 5 m/BAC. The measure of an
inscribed angle of a circle is equal to one-half
the measure of its intercepted arc, so
m/ACD 5 12mAD and m/BDC 5 12mBC . By
y5k
Thus, the y-axis intersects the circle in the point
(0, k), so it is tangent to the circle.
g
2. No. Since the slope of AB is not the negative
reciprocal of the slope of CA, it is not perpendicular to CA at the point of intersection.
g
Therefore, AB is not tangent at A.
Developing Skills
3. a. (0, 6)
b. Tangent
4. a. (6, 8) and (8, 6)
b. Secant
5. a. (3, 4) and (4, 3)
b. Secant
6. a. (1, 3) and (21, 23)
b. Secant
7. a. (0, 23) and (3, 0)
b. Secant
8. a. (2, 2) and (22, 22)
b. Secant
9. a. (4, 3) and (23, 24)
b. Secant
10. a. (2, 4) and (4, 2)
b. Secant
11. a. (23, 3)
b. Tangent
C
C
C
C
340
C
C
C
C C
the substitution postulate, 12mAD 5 12mBC or
mAD 5 mBC . Therefore, AD > BC . In a
circle, congruent arcs have congruent chords, so
AD > BC and ABCD is an isosceles trapezoid.
C
C
(PAB)(PA) 5 (PCD)(PD). By the division
postulate, PA 5 PD. Then, by the subtraction
postulate, PAB 2 PA 5 PCD 2 PD, so
AB 5 CD and AB > CD. Two chords are
equidistant from the center of a circle if and only
if the chords are congruent. Therefore, AB and
CD are equidistant from the center of the circle.
14. a. Since AB y CD, alternate interior angles
/B and /C are congruent so m/B 5 m/C.
m/B 5 12mAC and m/C 5 12mBD . By the
C
C
C 5 mBD
C . If two
substitution postulate, mAC
18. 12 cm
inscribed angles of a circle intercept the same
arc, then they are congruent. Therefore,
/A > /C and /D > /B. By the transitive
property of congruence, /A > /B and
/C > /D. If two angles of a triangle are
congruent, then the sides opposite these
angles are congruent, so AE > BE and
CE > DE. Therefore, nABE and nCDE are
isosceles.
C
Exploration (pages 599–600)
a. Place the point of your compass on a point P on a
g
line AB. With any convenient radius, draw arcs
h
C
b. From part a, mAC 5 mBD . Therefore,
AC > BD .
C C
b. The interior angles of a regular hexagon each
measure 120°. The angle bisectors of the angles
of the polygon intersect in the center of the
polygon and form congruent equilateral
triangles. Use the construction from part a to
construct equilateral nABP. Then, construct
equilateral triangle nBCP. Next, construct
equilateral triangle nCDP. Continue this process
until you have constructed equilateral nFAP.
Since the equilateral triangles all share two sides,
they are all congruent. Thus, the sides of the
resulting hexagon, ABCDEF, are all congruent.
Each interior angle of the hexagon is the union
of two angles of two adjacent equilateral
triangles. Since the angles of an equilateral all
measure 60°, each interior angle of the hexagon
measures 60 1 60 5 120°. Therefore, the hexagon
formed is a regular hexagon.
c. From part a, /A > /C and /D > /B.
Therefore, nABE , nCDE by AA,.
15. We are given that AB > BC > CD > DA.
Therefore, mAB 5 mBC 5 mCD 5 mDA 5 90.
The measure of an inscribed angle of a circle is
equal to one-half the measure of its intercepted
arc, so m/A 5 12 (mBC 1 mCD ) 5 12 (180) 5 90.
Therefore, m/A is a right angle. In a circle, if two
arcs are congruent, then their chords are
congruent. Therefore, AB > BC > CD > DA
and ABCD is a rhombus. A rhombus with a right
angle is a square, so ABCD is a square.
C
C C C
C
C
C
h
that intersect PA at C and PB at D. With C and
D as centers and the compass open to a radius
equal to CD, draw arcs that intersect at a point E.
Connect C, D, and E to form nCDE. Congruent
radii were used to draw CD, DE, and CE, so
CD > DE > CE and nCDE is equilateral.
C
C
C
16. Let ABC be a triangle with right angle /B and
M the midpoint of AC. Then, MA > MC. Let
nABC be inscribed in a circle. Since a right angle
C
is inscribed in a semicircle, ABC is a semicircle
and AC is a diameter. Since MA > MC, M is the
center of the circle. All radii of the same circle
are congruent, so MA > MC > MB and MA 5
MC 5 MB, and the midpoint of the hypotenuse
is equidistant from the vertices of the triangle.
c. The diagonals of a square bisect each other, are
perpendicular, and are congruent. Let O be the
center of the circle. Draw a diameter AOB of the
circle. Construct the perpendicular bisector of
the diameter. Let C and D be the points where
the perpendicular bisector intersects the circle.
Since AO and OB are radii, AO > OB, and the
perpendicular bisector DC passes through O. But
OD and OC are also radii. Therefore, AB and CD
bisect each other, are congruent, and are
perpendicular, so ACBD is a square.
17. Since PAB > PCD, PAB 5 PCD. If two secant
segments are drawn to a circle from an external
point, then the product of the lengths of one
secant segment and its external segment is equal
to the product of the lengths of the other secant
segment and its external segment. Therefore,
341
d. From the construction in part c, AC , CB , BD ,
and DA are all congruent since their central
angles are all right angles. Thus, the arcs formed
by bisecting these arcs are all congruent (since
halves of congruent arcs are congruent). Two
chords are congruent if and only if their arcs are
congruent. Therefore, the chords formed by the
arcs are all congruent, so the sides of the polygon
formed are all congruent. Draw segments from
the vertices of the polygon to the center of the
circle. These segments are all congruent because
they are radii of the same circle. Thus, congruent
isosceles triangles are formed by SSS. The base
angles of these triangles are congruent, so it
follows that the interior angles of the octagon
are all congruent. Therefore, the octagon
formed is a regular octagon.
Part III
13. Statements
f. Construct six congruent and adjacent equilateral
triangles with the center of the circle as a vertex
and using two radii as sides.
14.
CC C
C
Reasons
g
g
g
g
1. EF ' AB and
1. Given.
EF ' CD
2. /AEF and /CEF
are right angles.
3. /AEF > /CEF
4. EF > EF
5. EA > EC
6. nAEF > nCEF
7. FA > FC
2. Definition of
perpendicular lines.
3. Right angles are
congruent.
4. Reflexive property
of congruence.
5. Given.
6. SAS.
7. Corresponding
parts of congruent
triangles are
congruent.
x2 1 (x 2 7) 2 5 (x 1 2) 2
x2 1 x2 2 14x 1 49 5 x2 1 4x 1 4
x2 2 18x 1 45 5 0
Cumulative Review (pages 600–603)
(x 2 15)(x 2 3) 5 0
Part I
1. 3
2. 1
3. 4
4. 3
5. 1
6. 3
7. 3
8. 2
9. 1
10. 3
Part II
11. a. (2, 2)
b. (22, 0)
c. Yes. The slope of DE is 12 and the slope of AC
x 5 15 u x 5 3
x cannot be 3 since 3 2 7 5 24.
The lengths of the sides of the triangle are 15, 8,
and 17.
Part IV
15. a. (25, 2); ry=x (5, 2) 5 (2, 5), R90 (2, 5) 5 (25, 2)
b. ry-axis
c. No. Rotation is a direct isometry and line
reflection is an opposite isometry.
is 12 .
d. Yes. By the reflexive property of congruence,
/B > /B. If two parallel lines are cut by a
transversal, then the corresponding angles are
congruent. Since DE y AC, /CAD > /EDB
and nABC , nDBE by AA,.
12. Statements
1. ABCD with
AC > BD, and
AC and BD bisect each other.
2. ABCD is a
parallelogram.
3. ABCD is a
rectangle.
16. a. We are given that AD : DC 5 1 : 3 and BE :
EC 5 1 : 3. Let AD 5 x, so DC 5 3x. Let BE
5 y, so EC 5 3y. AC 5 AD 1 DC 5 4x and
AC
4
5 4x
BC 5 BE 1 EC 5 4y. DC
3x 5 3 and
BC
EC
4y
5 3y 5 43 . Therefore, AC : DC 5 BC : EC.
b. From part a, AC : DC 5 BC : EC. By the
reflexive property of congruence, /C > /C.
Therefore, nABC , nDEC by SAS,.
Reasons
1. Given.
2. If the diagonals of a
quadrilateral bisect each
other, then the quadrilateral is a parallelogram.
3. If the diagonals of a
parallelogram are congruent, then the parallelogram is a rectangle.
342
Chapter 14. Locus and Construction
h
14-1 Constructing Parallel Lines
(pages 607–609)
radius, mark off point Y on BD. Draw XY.
ABYX is a square.
Writing About Mathematics
1. The distance between two parallel lines is
defined as the length of the perpendicular from
any point on one line to the other line. Therefore,
since CE is the perpendicular with length 2PQ,
g
c. Use Construction 1 to construct a segment AB
congruent to the segment with length a. Use
Construction 2 to construct an angle congruent
h
to /A on AB . On the side of this angle not
containing point B, use Construction 1 to
construct a segment AC congruent to the
segment with length b. Use Construction 7 to
construct a line parallel to AB through point
C. On this line, use Construction 1 to construct
CD congruent to the segment with length a
such that CD is on the same side of AC. Draw
BD. ABDC is a parallelogram.
g
every point on CD is at a distance 2PQ from AB.
2. If two coplanar lines are each perpendicular to
the same line, then they are parallel. Thus,
g
g
g
g
GH y AB and GH y CD. The distance between two
parallel lines is defined as the length of the
perpendicular from any point on one line to the
other line. Thus, the distance from any point on
g
g
g
GH to AB and CD is 12GH.
Developing Skills
3. a. Use Construction 7.
b. Mark any point Q on l. Use Construction 6 to
construct the perpendicular to m through Q.
Let R be the intersection of this perpendicular
and line m. Then use Construction 3 to
construct the perpendicular bisector of QR.
c. Construct a line parallel to l (and above l) that
5.
h
is the same distance away as m. Extend QR to
a point S such that QR 5 RS. Then use
Construction 7 to construct a line parallel to l
through S.
d. Yes. If two of three lines in the same plane are
each parallel to the third line, then they are
parallel to each other.
4. a. Use Construction 1 to construct a segment AB
congruent to the segment with length b. Use
g
6.
g
Construction 5 to construct AC and BD
perpendicular to AB with C and D on the
same side of AB. Set the compass radius to a.
7.
h
With A as the center, mark off point X on AC .
With B as the center and using the same
h
radius, mark off point Y on BD. Draw XY.
ABYX is a rectangle.
b. Use Construction 1 to construct a segment
AB congruent to the segment with length a.
g
8.
g
Use Construction 5 to construct AC and BD
perpendicular to AB with C and D on the
same side of AB. Set the compass radius to a.
h
With A as the center, mark off point X on AC .
With B as the center and using the same
343
d. Repeat part c, but construct AC to be
congruent to the segment with length a.
ABDC is a rhombus.
a. Construct the perpendicular bisector of AB at
C. Construct the perpendicular bisector of AC
at D and of CB at E. Then D, C, and E divide
AB into four congruent parts.
b. Set the compass radius to AB. Draw a circle.
c. Set the compass radius to AC. Draw a circle.
a. Use Construction 7.
b. Use Construction 3 to find the midpoint M of
AC. Draw BM. Then BM is the median to AC.
c. Use Construction 3 to find the midpoint N of
BC. Draw AN. Then AN is the median to BC.
d. Yes. Any two medians of a triangle intersect in
the same point, and two points determine a
line. Thus, the line drawn from C to the
median of AB is the same as that drawn from
C to P.
a. Use Construction 3.
b. Use Construction 3.
c. Use Construction 3 to find the midpoint M of
AB.
d. P and M. Any two perpendicular bisectors of
the sides of a triangle intersect in the same
point. Thus, the line containing P and M is the
perpendicular bisector of AB.
a. Use Construction 6 to construct a line
perpendicular to AC through B intersecting
AC at D. BD is the altitude to AC.
b. Use Construction 6 to construct a line
perpendicular to BC through A intersecting
BC at E. AE is the altitude to BC.