properties of arcs and chords
DESCRIPTION
Properties of arcs and chords. Warm Up 1. What percent of 60 is 18? 2. What number is 44% of 6? 3. Find m WVX. 30. 2.64. 104.4. Objectives. Apply properties of arcs. Apply properties of chords. Vocabulary. central anglesemicircle arcadjacent arcs minor arccongruent arcs - PowerPoint PPT PresentationTRANSCRIPT
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Properties of arcs and chords
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Warm Up1. What percent of 60 is 18?2. What number is 44% of 6?
3. Find mWVX.
302.64
104.4
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Apply properties of arcs.Apply properties of chords.
Objectives
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central angle semicirclearc adjacent arcsminor arc congruent arcsmajor arc
Vocabulary
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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.
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Minor arcs may be named by two points. Major arcs and semicircles must be named by three points.
Writing Math
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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°
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Check It Out! Example 1 Use the graph to find each of the following.
a. mFMCmFMC = 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 .
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Adjacent arcs are arcs of the same circle that intersect at exactly one point. RS and ST are adjacent arcs.
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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.mCFD = 30.6Arc Add. Post.Substitute.Simplify.
Find mBD.
mCD = 30.6
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Check It Out! Example 2a
Find each measure. mJKL
mKPL = 180° – (40 + 25)°
= 25° + 115°
mKL = 115°mJKL = mJK + mKL
= 140°
Arc Add. Post.Substitute.Simplify.
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Check It Out! Example 2b
Find each measure. mLJN
= 295°mLJN = 360° – (40 + 25)°
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Within a circle or congruent circles, congruent arcs are two arcs that have the same measure. In the figure ST UV.
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Example 3A: Applying Congruent Angles, Arcs, and Chords
TV WS. Find mWS.
9n – 11 = 7n + 112n = 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
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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
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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 = 3Substitute 3 for t.
C J, and mGCD mNJM. Find NM.
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Check It Out! Example 3a
PT bisects RPS. Find RT.
6x = 20 – 4x10x = 20
x = 2RT = 6(2)RT = 12
Add 4x to both sides.
Divide both sides by 10.
Substitute 2 for x.Simplify.
RPT SPT
RT = TS mRT mTS
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Check It Out! Example 3b
A B, and CD EF. Find mCD. Find each measure.
25y = (30y – 20)20 = 5y4 = 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.
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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.
Step 1 Draw radius RN.
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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.
Step 1 Draw radius PQ.
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Lesson Quiz: Part I
1. The circle graph shows the types of cuisine available in a city. Find mTRQ.
158.4
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Lesson Quiz: Part II
2. NGH 139
Find each measure.
3. HL 21
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Lesson Quiz: Part III
12.9
4. T U, and AC = 47.2. Find PL to the nearest tenth.