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Holt McDougal Geometry Arcs and Chords Arcs and Chords Holt Geometry Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz Holt McDougal Geometry

# Holt McDougal Geometry Arcs and Chords Holt Geometry Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson Quiz Holt McDougal Geometry

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Arcs and Chords Arcs and Chords

Holt Geometry

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt McDougal Geometry Holt McDougal Geometry

Arcs and Chords

Warm Up

1. What percent of 60 is 18?

2. What number is 44% of 6?

3. Find mWVX.

30

2.64

104.4 Holt McDougal Geometry

Arcs and Chords

Apply properties of arcs.

Apply properties of chords.

Objectives Holt McDougal Geometry

Arcs and Chords

central angle semicirclearc adjacent arcsminor arc congruent arcsmajor arc

Vocabulary Holt McDougal Geometry

Arcs and Chords

A central angle is an angle whose vertex is the center of a circle. An arc is an unbroken part of a circle consisting of two points called the endpoints and all the points on the circle between them. Holt McDougal Geometry

Arcs and Chords Holt McDougal Geometry

Arcs and Chords

Minor arcs may be named by two points. Major arcs and semicircles must be named by three points.

Writing Math Holt McDougal Geometry

Arcs and Chords

Example 1: Data Application

The circle graph shows the types of grass planted in the yards of one neighborhood. Find mKLF.

= 234

mKLF = 360° – mKJF

mKJF = 0.35(360)= 126

mKLF = 360° – 126° Holt McDougal Geometry

Arcs and Chords

Check It Out! Example 1

Use the graph to find each of the following.

a. mFMC

mFMC = 0.30(360)

= 108

= 270

b. mAHB c. mEMD = 0.10(360)= 36= 360° – 0.25(360)mAHB

= 360° – mAMB

Central is 30% of the .

Central is 10% of the . Holt McDougal Geometry

Arcs and Chords

Adjacent arcs are arcs of the same circle that intersect at exactly one point. RS and ST are adjacent arcs. Holt McDougal Geometry

Arcs and Chords

Example 2: Using the Arc Addition Postulate

mCFD = 180 – (97.4 + 52)= 30.6

= 97.4 + 30.6= 128

mBD = mBC + mCD

mBC = 97.4 Vert. s Thm.

∆ Sum Thm.

Substitute.Simplify.

Find mBD.

mCD = 30.6 Holt McDougal Geometry

Arcs and Chords

Check It Out! Example 2a

Find each measure.

mJKL

mKPL = 180° – (40 + 25)°

= 25° + 115°

mKL = 115°

mJKL = mJK + mKL

= 140°

Substitute.

Simplify. Holt McDougal Geometry

Arcs and Chords

Check It Out! Example 2b

Find each measure.

mLJN

= 295°

mLJN = 360° – (40 + 25)° Holt McDougal Geometry

Arcs and Chords

Within a circle or congruent circles, congruent arcs are two arcs that have the same measure. In the figure ST UV. Holt McDougal Geometry

Arcs and Chords Holt McDougal Geometry

Arcs and Chords

Example 3A: Applying Congruent Angles, Arcs, and Chords

TV WS. Find mWS.

9n – 11 = 7n + 11

2n = 22

n = 11

= 88°

chords have arcs.

Def. of arcs

Substitute the given measures.

Subtract 7n and add 11 to both sides.

Divide both sides by 2.

Substitute 11 for n.

Simplify.

mTV = mWS

mWS = 7(11) + 11

TV WS Holt McDougal Geometry

Arcs and Chords

Example 3B: Applying Congruent Angles, Arcs, and Chords

C J, and mGCD mNJM. Find NM.

GD = NM

arcs have chords.GD NM

GD NM GCD NJM

Def. of chords Holt McDougal Geometry

Arcs and Chords

Example 3B Continued

14t – 26 = 5t + 1

9t = 27

NM = 5(3) + 1

= 16

Substitute the given measures.

Subtract 5t and add 26 to both sides.

Divide both sides by 9.

Simplify.

t = 3

Substitute 3 for t.

C J, and mGCD mNJM. Find NM. Holt McDougal Geometry

Arcs and Chords

Check It Out! Example 3a

PT bisects RPS. Find RT.

6x = 20 – 4x

10x = 20

x = 2

RT = 6(2)

RT = 12

Divide both sides by 10.

Substitute 2 for x.

Simplify.

RPT SPT

RT = TS

mRT mTS Holt McDougal Geometry

Arcs and Chords

Check It Out! Example 3b

A B, and CD EF. Find mCD.

Find each measure.

25y = (30y – 20)

20 = 5y

4 = y

CD = 25(4)

Subtract 25y from both sides. Add 20 to both sides.Divide both sides by 5.

Substitute 4 for y.

Simplify.mCD = 100

mCD = mEF chords have arcs.Substitute. Holt McDougal Geometry

Arcs and Chords Holt McDougal Geometry

Arcs and Chords

Find NP.

Example 4: Using Radii and Chords

Step 2 Use the Pythagorean Theorem.

Step 3 Find NP.

RN = 17 Radii of a are .

SN2 + RS2 = RN2

SN2 + 82 = 172

SN2 = 225SN = 15

NP = 2(15) = 30

Substitute 8 for RS and 17 for RN.Subtract 82 from both sides.Take the square root of both sides.

RM NP , so RM bisects NP. Holt McDougal Geometry

Arcs and Chords

Check It Out! Example 4

Find QR to the nearest tenth.

Step 2 Use the Pythagorean Theorem.

Step 3 Find QR.

PQ = 20 Radii of a are .

TQ2 + PT2 = PQ2

TQ2 + 102 = 202

TQ2 = 300TQ 17.3

QR = 2(17.3) = 34.6

Substitute 10 for PT and 20 for PQ.Subtract 102 from both sides.Take the square root of both sides.

PS QR , so PS bisects QR. Holt McDougal Geometry

Arcs and Chords

Lesson Quiz: Part I

1. The circle graph shows the types of cuisine available in a city. Find mTRQ.

158.4 Holt McDougal Geometry

Arcs and Chords

Lesson Quiz: Part II

2. NGH 139

Find each measure.

3. HL 21 Holt McDougal Geometry

Arcs and Chords

Lesson Quiz: Part III

12.9

4. T U, and AC = 47.2. Find PL to the nearest tenth. ##### Holt McDougal Geometry 2-5 Algebraic Proof 2-5 Algebraic Proof Holt Geometry Warm Up Warm Up Lesson Presentation Lesson Presentation Lesson Quiz Lesson
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