Tangent Lines

1
Tangent Lines
GEOMETRY LESSON 12-1
(For help, go to the Skills Handbook, page 754
and Lesson 8-1.)
Find each product. 1. (p 3)2 2. (w
10)2 3. (m 2)2 Find the value of x. Leave
your answer in simplest radical
form. 4. 5. 6.
Check Skills Youll Need
12-1
2
Tangent Lines
GEOMETRY LESSON 12-1
Solutions
1. (p 3)2 (p 3)(p 3) p2 3p 3p 32
p2 6p 9 2. (w 10)2 (w 10)(w 10)
w2 10w 10w 102 w2 20w 100 3. (m 2)2
(m 2)(m 2) m2 2m 2m (2)2 m2 4m
4 4. Use the Pythagorean Theorem a2 b2 c2
(8)2 (16)2 x2 64 256 x2 x2
320 x 320 26 5 23 5 8
5 5. Use the Pythagorean Theorem a2 b2 c2
x2 (13)2 (17)2 x2 169 289 x2
289 169 120 x 120 22 2
3 5 2 2 3 5 2 30 6. Use the
Pythagorean Theorem a2 b2 c2 x2 (16)2
(20)2 x2 256 400 x2 400 256
144 x 144 122 12
12-1
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Tangent Lines
GEOMETRY LESSON 12-1
BA is tangent to C at point A. Find the value
of x.
90 22 x 180 Substitute.
112 x 180 Simplify.
x 68 Solve.
Quick Check
12-1
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Tangent Lines
GEOMETRY LESSON 12-1
Because opposite sides of a rectangle have the
same measure, DW 3 cm and OD 15 cm.
12-1
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Tangent Lines
GEOMETRY LESSON 12-1
Quick Check
OD2 PD2 OP2 Pythagorean Theorem
152 42 OP2 Substitute.
241 OP2 Simplify.
The distance between the centers of the pulleys
is about 15.5 cm.
12-1
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Tangent Lines
GEOMETRY LESSON 12-1
Quick Check
O has radius 5. Point P is outside O such
that PO 12, and point A is on O such that PA
13. Is PA tangent to O at A? Explain.
144 ? 194 Simplify.
12-1
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Tangent Lines
GEOMETRY LESSON 12-1
Theorem 11-3 states that two segments tangent to
a circle from a point outside the circle are
congruent.
OS OT because all radii of a circle are
congruent.
Two pairs of adjacent sides are congruent.
Quadrilateral QSOT is a kite if no opposite
sides are congruent or a rhombus if all sides
are congruent.
By theorems in Lessons 6-4 and 6-5, both the
diagonals of a rhombus and the diagonals of a
kite are perpendicular.
Quick Check
12-1
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Tangent Lines
GEOMETRY LESSON 12-1
p XY YZ ZW WX Definition of
perimeter p
XR RY YS SZ ZT TW WU UX
Segment Addition Postulate
11 8 8 6 6 7 7 11
Substitute.
64 Simplify.
The perimeter is 64 ft.
Quick Check
12-1
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Tangent Lines
GEOMETRY LESSON 12-1
PA and PB are tangent to C. Use the figure for
Exercises 13. 1. Find the value of x. 2. Find
the perimeter of quadrilateral PACB. 3. Find
CP. HJ is tangent to A and to B. Use the
figure for Exercises 4 and 5. 4. Find AB to the
nearest tenth. 5. What type of special
quadrilateral is AHJB? Explain how you know.
87.2
82 cm
29 cm
20.4 cm
12-1
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Chords and Arcs
GEOMETRY LESSON 12-2
(For help, go to Lesson 8-2.)
Find the value of each variable. Leave your
answer in simplest radical form. 1. 2. 3.

Check Skills Youll Need
12-2
11
Chords and Arcs
GEOMETRY LESSON 12-2
Solutions
12-2
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Chords and Arcs
GEOMETRY LESSON 12-2
Quick Check
12-2
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Chords and Arcs
GEOMETRY LESSON 12-2
QS QR RS Segment Addition Postulate
QS 7 7 Substitute.
QS 14 Simplify.
AB QS Chords that are equidistant from the
center of a circle are congruent.
AB 14 Substitute 14 for QS.
Quick Check
12-2
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Chords and Arcs
GEOMETRY LESSON 12-2
OP 2 PM 2 OM 2 Use the Pythagorean Theorem.
r 2 82 152 Substitute.
r 2 289 Simplify.
r 17 Find the square root of each side.
Quick Check
12-2
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Chords and Arcs
GEOMETRY LESSON 12-2
7.8 in.
22 cm
6.9 cm
12-2
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Inscribed Angles
GEOMETRY LESSON 12-3
(For help, go to Lesson 10-6.)
Identify the following in P at the
right. 1. a semicircle 2. a minor arc 3. a
major arc 4. a central angle Find the measure
of each arc in P. 5. ST 6. STQ 7.
RST 8. TQ
Check Skills Youll Need
12-3
17
Inscribed Angles
GEOMETRY LESSON 12-3
Solutions
12-3
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Inscribed Angles
GEOMETRY LESSON 12-3
Solutions (continued)
7. The measure of an arc is the measure of its
central angle. m SPQ 180 and m RPQ
145. By the Angle Addition Postulate, m SPR
m RPQ m SPQ, so m SPR 145 180
and m SPR 180 145 35. By the Angle
Addition Postulate, m RPT m RPS m
SPT 35 86 121. Thus, mRST 121. 8. The
measure of an arc is the measure of its central
angle. m SPQ 180 and m SPT 86. By the
Angle Addition Postulate, m TPQ m SPT m
SPQ, so m TPQ 86 180 and m TPQ
180 86 94. Thus, mTQ 94.
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Inscribed Angles
GEOMETRY LESSON 12-3
x 75 Simplify.
12-3
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Inscribed Angles
GEOMETRY LESSON 12-3
y 95 Simplify.
Quick Check
12-3
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Inscribed Angles
GEOMETRY LESSON 12-3
By Corollary 2 to the Inscribed Angle Theorem, an
angle inscribed in a semicircle is a right
angle, so a 90.
Therefore, the angle whose intercepted arc has
measure b must have measure 180 90 32, or
58.
Because the inscribed angle has half the measure
of the intercepted arc, the intercepted arc has
twice the measure of the inscribed angle, so b
2(58) 116.
Quick Check
12-3
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Inscribed Angles
GEOMETRY LESSON 12-3
12-3
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Inscribed Angles
GEOMETRY LESSON 12-3
Quick Check
12-3
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Inscribed Angles
GEOMETRY LESSON 12-3
In the diagram below, O circumscribes
quadrilateral ABCD and is inscribed in
quadrilateral XYZW. 1. Find the measure of
each inscribed angle. 2. Find m DCZ. 3. Are
XAB and XBA congruent? Explain. 4. Find
the angle measures in quadrilateral XYZW.
45
Yes each is formed by a tangent and a chord, and
they intercept the same arc.
5. Does a diagonal of quadrilateral ABCD
intersect the center of the circle? Explain how
you can tell.
12-3
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Angle Measures and Segment Lengths
GEOMETRY LESSON 12-4
(For help, go to Lessons 12-1 and 12-3.)
Check Skills Youll Need
12-4
26
Angle Measures and Segment Lengths
GEOMETRY LESSON 12-4
Solutions
12-4
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Angle Measures and Segment Lengths
GEOMETRY LESSON 12-4
Solutions (continued)
7. By Theorem 11-3, two tangents tangent to a
circle from a point outside the circle are
congruent, so DF EF 2. 8. Draw CF. Since FE
is tangent to the circle, FE CE . By def. of
, CEF is a right angle. By def. of right
angle, m CEF 90, so CEF is a right
triangle. From Exercise 6, CE 4. Also, EF
2. Use the Pythagorean Theorem a2 b2 c2
CE 2 EF 2 CF 2 42 22 CF 2
16 4 CF 2 20 CF 2 CF 20 2 5
4.5 9. CEFD is a quadrilateral, so the sum
of its angles is 360. From Exercise 8, m ACE
90. Similarly, m CDF 90. Also, m DCE
57. So, m EFD m FEC m DCE m
CDF 360 m EFD 90 57 90 360 m
EFD 237 360 m EFD 360 237 123
12-4
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Angle Measures and Segment Lengths
GEOMETRY LESSON 12-4
Find the value of the variable.
a.
x 88 Simplify.
12-4
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Angle Measures and Segment Lengths
GEOMETRY LESSON 12-4
(continued)
Quick Check
76 x Multiply each side by 2.
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Angle Measures and Segment Lengths
GEOMETRY LESSON 12-4
12-4
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Angle Measures and Segment Lengths
GEOMETRY LESSON 12-4
72 180 x Distributive Property
x 72 180 Solve for x.
x 108
Quick Check
A 108 arc will be in the advertising agencys
photo.
12-4
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Angle Measures and Segment Lengths
GEOMETRY LESSON 12-4
5 x 3 7 Along a line, the product of the
lengths of two segments from a point to a
circle is constant (Theorem 12-12 (1)).
5x 21 Solve for x.
x 4.2
8(y 8) 152 Along a line, the product of the
lengths of two segments from a point to a
circle is constant (Theorem 12-12 (3)).
8y 64 225 Solve for y.
8y 161
Quick Check
y 20.125
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Angle Measures and Segment Lengths
GEOMETRY LESSON 12-4
Quick Check
Because the radius is 125 ft, the diameter is 2
125 250 ft.
The length of the other segment along the
diameter is 250 ft 50 ft, or 200 ft.
x x 50 200 Along a line, the product of the
lengths of the two segments from a point to a
circle is constant (Theorem 12-12 (1)).
x2 10,000 Solve for x.
x 100
The shortest distance from point A to point B is
200 ft.
12-4
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Angle Measures and Segment Lengths
GEOMETRY LESSON 12-4
a 60 b 28
82
15.5
24
22
12-4
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Circles in the Coordinate Plane
GEOMETRY LESSON 12-5
(For help, go to Lesson 1-8.)
Find the length of each segment to the nearest
tenth. 1. 2. 3.
Check Skills Youll Need
12-5
36
Circles in the Coordinate Plane
GEOMETRY LESSON 12-5
1. The endpoints are (2, 1) and (3, 4). The
distance between them is d (x2 x1)2
(y2 y1)2 (3 ( 2))2 (4 1)2
52 32 25 9 34 5.8 2. The
endpoints are (6, 4) and (4, 4). The distance
between them is d (x2 x1)2 (y2
y1)2 (4 ( 6))2 ( 4 4)2
102 ( 8)2 100 64 164 12.8
3. The endpoints are (3, 3) and (0, 2). The
distance between them is d (x2 x1)2
(y2 y1)2 (0 ( 3))2 (2 ( 3))2
32 52 9 25 34 5.8
Solutions
12-5
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Circles in the Coordinate Plane
GEOMETRY LESSON 12-5
(x h)2 (y k)2 r2 Standard form
(x 8)2 y2 5 Simplify.
Quick Check
12-5
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Circles in the Coordinate Plane
GEOMETRY LESSON 12-5
First find the radius.
Then find the standard equation of the circle
with center (5, 8) and radius 29.
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Circles in the Coordinate Plane
GEOMETRY LESSON 12-5
(x h)2 (y k)2 r2 Standard form
(x 5)2 (y 8)2 292 Substitute (5, 8) for
(h, k) and 29 for r.
(x 5)2 (y 8)2 841 Simplify.
Quick Check
12-5
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Circles in the Coordinate Plane
GEOMETRY LESSON 12-5
Quick Check
12-5
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Circles in the Coordinate Plane
GEOMETRY LESSON 12-5
The center of a circular range is at (6, 12),
and the radius is 80.
(x h)2 (y k)2 r2 Use standard
form.
(x 6)2 y (12)2 802 Substitute.
(x 6)2 (y 12)2 6400 This is an
equation for the tower.
Quick Check
12-5
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Circles in the Coordinate Plane
GEOMETRY LESSON 12-5
1. Find the center and radius of the circle with
equation (x 1)2 (y 1)2 9. Then graph
the circle. 2. A cellular phone tower with a
range of 25 units is located on a coordinate
grid at (10, 35). Write an equation that
describes its position and range. Write the
standard equation of each circle. 3. center (0,
6) radius 11 4. center (3, 2) diameter
18 5. center (9, 5) passing through (7, 1)
(x 10)2 (y 35)2 625
x2 (y 6)2 11
(x 3)2 (y 2)2 81
(x 9)2 (y 5)2 20
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Locus A Set of Points
GEOMETRY LESSON 12-6
(For help, go to Lessons 1-7 and 3-8.)
Sketch each of the following. 1. the
perpendicular bisector of CD 2. EFG bisected
by FH 3. line k parallel to line m and
perpendicular to line w, all in plane N
Check Skills Youll Need
12-6
44
Locus A Set of Points
GEOMETRY LESSON 12-6
Solutions 1. 2. 3.
12-6
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Locus A Set of Points
GEOMETRY LESSON 12-6
Quick Check
12-6
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Locus A Set of Points
GEOMETRY LESSON 12-6
The set of points in a plane that are 6 cm from
point P is a circle with center P and radius 6
cm.
The set of points in a plane that are 8 cm from
point Q is a circle with center Q and radius 8
cm.
Quick Check
12-6
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Locus A Set of Points
GEOMETRY LESSON 12-6
You can use the floor to imagine this.
Quick Check
12-6
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Locus A Set of Points
GEOMETRY LESSON 12-6
the set of points in a coordinate plane 5 units
from the point (2, 8)
12-6