featurelesson geometry lesson main pa and pb are tangent to c. use the figure for exercises 1–3....

FeatureLesson

GeometryGeometry

LessonMain

PA and PB are tangent to C. Use the figure for Exercises 1–3.

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

Lesson 12-1

Tangent LinesTangent Lines

.

..

Trapezoid; the tangent line forms right angles at vertices H and J, so HA || JB. Because HA = JB, AHJB is not a parallelogram but a trapezoid./

Lesson Quiz

12-2

FeatureLesson

GeometryGeometry

LessonMain

(For help, go to Lesson 8-2.)

Lesson 12-2

Find the value of each variable. Leave your answer in simplest radical form.

1. 2. 3.

Chords and ArcsChords and Arcs

Check Skills You’ll Need

Check Skills You’ll Need

12-2

5.5 2 5 28

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-2

Chords and ArcsChords and Arcs

Notes

12-2

A segment whose endpoints are on a circle is called a chord.

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-2

Chords and ArcsChords and Arcs

Notes

12-2

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-2

Chords and ArcsChords and Arcs

Notes

12-2

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-2

Chords and ArcsChords and Arcs

Notes

12-2

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-2

Chords and ArcsChords and Arcs

Notes

12-2

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-2

Chords and ArcsChords and Arcs

Notes

12-2

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-2

Chords and ArcsChords and Arcs

Notes

12-2

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-2

Chords and ArcsChords and Arcs

Notes

12-2

FeatureLesson

GeometryGeometry

LessonMain

AOX BOX by the definition of an angle bisector.

Lesson 12-2

Chords and ArcsChords and Arcs

In the diagram, radius OX bisects AOB. What can you conclude?

AX BX because congruent central angles have congruent chords.

AX BX because congruent chords have congruent arcs.

Quick Check

Additional Examples

12-2

Using Theorem 12-4

FeatureLesson

GeometryGeometry

LessonMain

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.

Find AB.

Lesson 12-2

Chords and ArcsChords and Arcs

Quick Check

Additional Examples

12-2

Using Theorem 12-5

FeatureLesson

GeometryGeometry

LessonMain

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.

PM = PQ A diameter that is perpendicular to a chord bisects the chord.

12

PM = (16) = 8 Substitute.12

The radius of O is 17 in..

Draw a diagram to represent the situation. The distance from the center of O to PQ is measured along a perpendicular line.

.

P and Q are points on O. The distance from O to PQ is

15 in., and PQ = 16 in. Find the radius of O.

..

Lesson 12-2

Chords and ArcsChords and Arcs

Quick Check

Additional Examples

12-2

Using Diameters and Chords

FeatureLesson

GeometryGeometry

LessonMain

FeatureLesson

GeometryGeometry

LessonMain

For Exercises 1–5, use the diagram of L.

1. If YM and ZN are congruent chords, what can you conclude?

2. If YM and ZN are congruent chords, explain why you cannot conclude that LV = LC.

3. Suppose that YM has length 12 in., and its distance from point L is 5 in. Find the radius of L to the nearest tenth.

For Exercises 4 and 5, suppose that LV YM, YV = 11 cm, and L has a diameter of 26 cm.

4. Find YM.

5. Find LV to the nearest tenth.

.

.

.

YM ZN; YLM ZLN

6.9 cm

22 cm

7.8 in.

Lesson 12-2

Chords and ArcsChords and Arcs

You do not know whether LV and LC are perpendicular to the chords.

Lesson Quiz

12-2

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-2

Chords and ArcsChords and Arcs

Solutions

2 2

5 2 2

11 2 2

11 2

11 2

Check Skills You’ll Need

1. The triangle is a 45°-45°-90° triangle, so each leg is the length of the

hypotenuse divided by 2 : = • = , or 5.5 2

2. The triangle is a 45°-45°-90° triangle, so each leg is the length of the

hypotenuse divided by 2: = 5

3. The triangle is a 30°-60°-90° triangle, so the hypotenuse is twice the

length of the side opposite the 30° angle: 2(14) = 28

12-2