confidential 1 geometry arcs and chords. confidential 2 warm up 1)90° 2)9

CONFIDENTIAL 1

GeometryGeometryArcs and chordsArcs and chords

CONFIDENTIAL 2

Warm UpWarm Up

In the figure ,QP are QM tangent to N . Find each measure.

1) mNMQ 2) MQ

2x

4x - 9 108 N

M

Q

P

1) 90°2) 9

CONFIDENTIAL 3

Arcs and ChordsArcs and Chords

A center angle is an angle whose vertex isthe center of a circle. An arc is an unbroken part of a circle consisting of two points calledthe end points and all the points on the circle

between them.

CONFIDENTIAL 4

Arcs and Their MeasureArcs and Their Measure

ARC MEASURE DIAGRAM

A minor arc is an arcwhose points are on or in the interior of a

center angle.

The measure of a minor arcis equal to the measure of

its center angle.

A major arc is an arcwhose points are on or in the exterior of a

center angle.

The measure of a major arcis equal to minus the

measure of its center angle.

If the endpoints of an arc lie on a diameter,

the arc is a semicircle

The measure of a semicircleis equal to

mAC = mABC = x

360

mADC = 360 - mABC = 360 - x

180

mEFG = 180

A

BC

x

A

BC

x

G

F

E

CONFIDENTIAL 5

The circle graph shows the types of music sold during one week at a music store.

Find

mBC

A

B

CDE

FG

H

Other 26%

M

Rock 25%

Rap 13%

Country 10%Pop 13%

Jazz 13%

Classical 3%

R & B 11%

mBC = mBMC m of arc = m of central mBMC = 0.13(360 ) central is 13% =46.8 of the

CONFIDENTIAL 6

Now you try

Use the graph to find each of the following.

Classical 3%

B

CDE

FG

H

Other 26%

M

Rock 25%

Rap 13%

Country 10%Pop 13%

Jazz 13%

R & B 11%

1) mFMC 2) mAHB 3) mEMD

1) 108°2) 270°3) 36°

CONFIDENTIAL 7

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

R

S

T

CONFIDENTIAL 8

R

S

T

Arc Addition Postulate

The measures of an arc formed by two adjacent arcs is

the sum of the measures of the two arcs.

mABC = mAB + mBC

CONFIDENTIAL 9

Using the Arc Addition Postulate

Find mCDE

mCD = 90 mCFD = 90 mDFE = 18 vert.s Thm.

mDE = 18 mDFE = 18

mCE = mCD + mDE Arc Add.Post = 90 + 18 = 108 Substitute and simlify

18

E

D

C

B

A

CONFIDENTIAL 10

Now you try!

Find each measure.

1) mJKL 2) mLJN

2540

PN

M

L

K

J

1) 140°2) 250°

CONFIDENTIAL 11

With in a circle or congruent circles, congruent arcs are

two arcs that have the same measure. In the figure, ST UV

x

x

V

U

T

S

CONFIDENTIAL 12

THEOREM A HYPOTHESIS CONCLUSION

In a circle or congruent circles

1) Congruent central angles have

congruent chords

2) Congruent chords have congruent arcs

3) Congruent arcs havecongruent central

angles.

E

D C

B

A

E

D C

B

A

E

D C

B

A

EAD BAC

ED BC

ED BC

DE BC

DE BC

DAE DAC

CONFIDENTIAL 13

The converses of the parts of previous theorems are also true. For example, with PART A, congruent chords have congruent

central angles

CONFIDENTIAL 14

E

D C

B

A

PART A

Proof:

Given: BAC DAEProve: BC DE

Statements Reasons1) BAC DAE

2) AB AD ,AC AE

3) BAC DAE

4) BE DE

1) Given

2) All radii of a are

3) SAS Steps 2,1

4) CPCTC

CONFIDENTIAL 15

Applying Congruent Angles, Arcs, and Chords

Find each measure.

RS TU. Find mRS.

RS TU chords have arcs.

mRS = mTU Def. of arcs

3x = 2x+27 Substitute the given measures.

x = 27 Substract 2x from both sides.

mRS = 3(27) Substitute 27 for x.

= 81 Simplify.

(2x + 27)

3x

U

T

S

R

1)

CONFIDENTIAL 16

B E,and AC DF. Find mDEF.

ABC DEF arcs have central s

mABC = mDEF Def. of s

5y + 5 = 7y - 43 Substitute the given measures. 5 = 2y - 43 substract 5y from both sides. 48 = 2y Add 43 to both sides. 24 = y Divide both sides by 2.

mDEF = 7(24) - 43 Substitute 24 for y =125 Simplify.

2)

(7y - 43)

(5y + 5)

FE

D

CB

A

CONFIDENTIAL 17

Now you try!

Find each measure

PT bisects RPS . Find RT1)

20 - 4x

6x

R

T

S

P

2) A B, and CD EF.

Find mCD.

(30y - 20)

25yFE

D

C

BA

1) 122) 100°

CONFIDENTIAL 18

THEOREM B HYPOTHESIS CONCLUSION1) In a circle, if a radius

(or diameter) isperpendicular toa chord, then it

bisects the chordand its arc

2) In a circle, theperpendicular

bisector of a chordis a radius (or diameter).

CD EF

E F

D

C

JK is bisector of GH

G

K

J

H

A

CD bisects EF and EF

JK is a diameter of A

CONFIDENTIAL 19

Using Radii and Chords

Find BD.

Step 1 Draw radius AD. AD = 5 Radii of a are .

Step 2 Use the Pythagorean Theorem. CD2 + AC2 = AD2

CD2 + 32 = 52 Substitute 3 for AC and 5 for AD. CD2 = 16 Substract 32 from both sides.

CD = 4 Take the square root of both sides.

Step 3 Find BD. BD = 2(4) = 8 AE BD , so AE bisects BD.

32

E

DCB

A

CONFIDENTIAL 20

Now you try!

4) Find QR to the nearest tenth.

1010

TS

R

Q

P

4) 46.6

CONFIDENTIAL 21

Now some problems for you to practice.

CONFIDENTIAL 22

ASSESSMENT

1) The circle graph shows how a typical household spends money on energy. Find each of the following.

6) m UPT5) m RQ4) mUT

3) m SAQ2) m VAU1) m PAQ

1) 198°2) 19.8°3) 61.2°4) 365) 39.66) 324

CONFIDENTIAL 23

Find each measure.

51

F

E

D

C

B

A

2) mDF 3) mDEB

2) 139°3) 270°

CONFIDENTIAL 24

4) mJL 5) mHLK

30

72

LK

J

H

G

4) 108°5) 210°

CONFIDENTIAL 25

6y

8y - 8

S

RQ

P

6) QPR RPS.Find QR

6) 24

CONFIDENTIAL 26

- 9X

(45 - 6X)

F

E

DC

BA

7) A B, and CD EF . Find EBF

7) 135°

CONFIDENTIAL 27

Arcs and Their MeasureArcs and Their Measure

ARC MEASURE DIAGRAM

A minor arc is an arcwhose points are on or in the interior of a

center angle.

The measure of a minor arcis equal to the measure of

its center angle.

A major arc is an arcwhose points are on or in the exterior of a

center angle.

The measure of a major arcis equal to minus the

measure of its center angle.

If the endpoints of an arc lie on a diameter,

the arc is a semicircle

The measure of a semicircleis equal to

mAC = mABC = x

360

mADC = 360 - mABC = 360 - x

180

mEFG = 180

A

BC

x

A

BC

x

G

F

E

REVIEWArcs and ChordsArcs and Chords

CONFIDENTIAL 28

The circle graph shows the types of music sold during one week at a music store.

Find

mBC

A

B

CDE

FG

H

Other 26%

M

Rock 25%

Rap 13%

Country 10%Pop 13%

Jazz 13%

Classical 3%

R & B 11%

mBC = mBMC m of arc = m of central mBMC = 0.13(360 ) central is 13% =46.8 of the

CONFIDENTIAL 29

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

R

S

T

CONFIDENTIAL 30

R

S

T

Arc Addition Postulate

The measures of an arc formed by two adjacent arcs is

the sum of the measures of the two arcs.

mABC = mAB + mBC

CONFIDENTIAL 31

Using the Arc Addition Postulate

Find mCDE

mCD = 90 mCFD = 90 mDFE = 18 vert.s Thm.

mDE = 18 mDFE = 18

mCE = mCD + mDE Arc Add.Post = 90 + 18 = 108 Substitute and simlify

18

E

D

C

B

A

CONFIDENTIAL 32

With in a circle or congruent circles, congruent arcs are

two arcs that have the same measure. In the figure, ST UV

x

x

V

U

T

S

CONFIDENTIAL 33

THEOREM A HYPOTHESIS CONCLUSION

In a circle or congruent circles

1) Congruent central angles have

congruent chords

2) Congruent chords have congruent arcs

3) Congruent arcs havecongruent central

angles.

E

D C

B

A

E

D C

B

A

E

D C

B

A

EAD BAC

ED BC

ED BC

DE BC

DE BC

DAE DAC

CONFIDENTIAL 34

E

D C

B

A

PART A

Proof:

Given: BAC DAEProve: BC DE

Statements Reasons1) BAC DAE

2) AB AD ,AC AE

3) BAC DAE

4) BE DE

1) Given

2) All radii of a are

3) SAS Steps 2,1

4) CPCTC

CONFIDENTIAL 35

Applying Congruent Angles, Arcs, and Chords

Find each measure.

RS TU. Find mRS.

RS TU chords have arcs.

mRS = mTU Def. of arcs

3x = 2x+27 Substitute the given measures.

x = 27 Substract 2x from both sides.

mRS = 3(27) Substitute 27 for x.

= 81 Simplify.

(2x + 27)

3x

U

T

S

R

1)

CONFIDENTIAL 36

B E,and AC DF. Find mDEF.

ABC DEF arcs have central s

mABC = mDEF Def. of s

5y + 5 = 7y - 43 Substitute the given measures. 5 = 2y - 43 substract 5y from both sides. 48 = 2y Add 43 to both sides. 24 = y Divide both sides by 2.

mDEF = 7(24) - 43 Substitute 24 for y =125 Simplify.

2)

(7y - 43)

(5y + 5)

FE

D

CB

A

CONFIDENTIAL 37

THEOREM B HYPOTHESIS CONCLUSION1) In a circle, if a radius

(or diameter) isperpendicular toa chord, then it

bisects the chordand its arc

2) In a circle, theperpendicular

bisector of a chordis a radius (or diameter).

CD EF

E F

D

C

JK is bisector of GH

G

K

J

H

A

CD bisects EF and EF

JK is a diameter of A

CONFIDENTIAL 38

Using Radii and Chords

Find BD.

Step 1 Draw radius AD. AD = 5 Radii of a are .

Step 2 Use the Pythagorean Theorem. CD2 + AC2 = AD2

CD2 + 32 = 52 Substitute 3 for AC and 5 for AD. CD2 = 16 Substract 32 from both sides.

CD = 4 Take the square root of both sides.

Step 3 Find BD. BD = 2(4) = 8 AE BD , so AE bisects BD.

32

E

DCB

A

CONFIDENTIAL 39

You did a great job today!You did a great job today!