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Warm Up Section 3.2 Draw and label each of the following in a circle with center P. (1). Radius: (2). Diameter: (3). Chord that is NOT a diameter: (4). Secant: (5). Tangent: . Warm Up Section 3.2 answers

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slide1

Warm Up Section 3.2

Draw and label each of the following in a circle with center P.

(1). Radius: (2). Diameter:

(3). Chord that is NOT a diameter:

(4). Secant:

(5). Tangent:

slide2

Warm Up Section 3.2 answers

Draw and label each of the following in a circle with center P.

(1). Radius:

(2). Diameter:

(3). Chord that is NOT

a diameter:

(4). Secant:

(5). Tangent:

T

A

E

P

R

D

C

slide3

Properties of Chords

Section 3.2

Standard: MCC9-12.G.C.2

Essential Question: Can I understand and use properties of chords to solve problems?

slide4

Parts of circles are called Arcs.

If the part of the circle is less than half the circle it is called a minor arc. If the part of the circle is more than half the circle it is called a major arc. And if it is exactlyhalf the circle it is called a semicircle.

slide5

The circle below has center P. Draw a diameter on your circle. Label the endpoints of the diameter H and K. Put another point on your circle and label it T.

There are two semicircles pictured in your drawing. One can be symbolized HK and the other HTK .

K

T

P

H

slide6

K

T

P

Name two minor arcs(you only need two letters to name a minor arc because you always travel the shortest distance unless told otherwise.) _______ , _______

Name two major arcs(you must use three letters to name a major arc because you always travel the shortest distance unless told otherwise.) _______ , _______

H

HT

TK

KHT

HKT

slide7

The circle below has center M. Mark two points on your circle and label them R and S. Now draw the

following rays: and

Color the inside of RMS.

Name the arc that is inside the colored part of the angle. This is called the intercepted arc. Arc: ________

The angle is called acentral anglebecause its vertex is the center of the circle.

S

R

M

RS

slide8

S

R

M

Arcs are measured in degrees. The measure of the intercepted arc is equal to the measure of the central angle that forms that arc.

So, if mRMS = 60o, then m RS = 60o.

slide9

In this section you will learn to use relationships

of arcs and chords in a circle.

In the same circle, or congruent circles, two minor

arcs are congruent if and only if their corresponding

chords are congruent.

C

B

AB  DC if and only if

_____  ______.

CD

AB

D

A

slide10

1. In the diagram, A  D, , and

m EF = 125o. Find m BC.

E

B

A

D

F

C

chords

Because BC and EF are congruent ________ in

congruent _______, the corresponding minor arcs

BC and EF are __________ .

So, m ______ = m ______ = ______o.

circles

congruent

125

BC

EF

slide11

If one chord is a perpendicular bisector of another

chord, then the first chord is a diameter.

If QS is a perpendicular bisector of TR, then ____

is a diameter of the circle.

T

S

P

Q

R

QS

slide12

If a diameter of a circle is perpendicular to a chord,

then the diameter bisects the chord and its arc.

If EG is a diameter and EG  DF, then HD  HF

and ____  ____ .

F

E

H

G

D

GF

GD

slide13

2. If m TV = 121o, find m RS.

T

6

S

V

6

R

If the chords are congruent, then the arcs are congruent. So, m RS =

121o

slide14

3. Find the measure of CB, BE, and CE.

C

Since BD is a diameter,

it bisects, the chord and the arcs.

4xo

A

B

D

4x = 80 – x

5x = 80

x = 16

(80 – x)o

E

4(16) = 64 so mCB = m BE = 64o

mCE = 2(64o) = 128o

slide15

In the same circle, or in congruent circles, two chords

are congruent if and only if they are equidistant

from the center.

if and only if

_____  ____

C

GE

FE

G

E

D

A

F

B

slide16

4. In the diagram of F, AB = CD = 12. Find EF.

Chords and are

congruent, so they are

equidistant from F.

Therefore EF = 6

G

A

B

7x – 8

F

3x

D

E

C

7x – 8 = 3x

4x = 8

x = 2

So, EF = 3x = 3(2) = 6

slide17

In the diagram of F, suppose AB = 27 and

  • EF = GF = 7. Find CD.

Since and are

both 7 units from the

center, they are

congurent.

G

A

B

F

D

E

C

So, AB = CD = 27.

slide18

In S, SP = 5, MP = 8, ST = SU,  and

  • NRQ is a right angle. Show that PTS  NRQ.

N

Step 1: Look at PTS !!

Since MP = 8, and NQ bisects

MP, we know MT= PT = 4.

Since  , PTN is

a right angle.

4

T

4

M

P

U

3

5

S

R

Q

Use the Pythagorean

Theorem to find TS.

42 + TS2 = 52

TS2 = 25 – 16

TS2 = 9 So, TS = 3.

slide19

In F, SP = 5, MP = 8, ST = SU,  and

  • NRQ is a right angle. Show that PTS  NRQ.

N

Step 2: Now, look at NRQ!

Since the radius of the circle

is 5, QN = 10.

Since ST = SU, MP and RN

are equidistantfrom the center.

Hence, MP = RN = 8.

8

T

M

P

U

10

S

R

6

Q

Use the Pythagorean

Theorem to find RQ.

RQ2 + 82 = 102

RQ2 = 100 – 64

RQ2 = 36 So, RQ = 6.

slide20

In F, SP = 5, MP = 8, ST = SU,  and

  • NRQ is a right angle. Show that PTS  NRQ.

N

Step 3: Identify ratios of

corresponding sides. In PTS,

PT = 4, TS = 3, and PS = 5.

In NRQ, NR = 8, RQ = 6,

and QN = 10.

Find the corresponding ratios:

T

M

P

U

S

R

Q

slide21

Because the corresponding sides lengths are proportional,

(all have a ratio of ½), PTS  NRQ by SSS  Thrm.

slide22

7. In S, QN = 26, NR = 24, ST = SU,  and

NRQ is a right angle. Show that PTS  NRQ.

N

N

24

T

M

P

U

26

S

R

R

10

Q

Q

242 + RQ2 = 262

RQ2 = 676 – 576

RQ2 = 100

So, RQ = 10.

slide23

7. In S, QN = 26, NR = 24, ST = SU,  and

NRQ is a right angle. Show that PTS  NRQ.

N

12

T

P

5

T

M

P

13

U

S

S

Cords equidistant from center 

MP = RN = 24. So, PT = ½(24).

SP is half of the diameter QN,

so SP = ½(26) = 13.

R

Q

122 + ST2 = 132

ST2 = 169 –144

ST2 = 25 So, ST = 5.

slide24

Step 3: Identify ratios of

corresponding sides. In PTS,

PT = 12, TS = 5, and PS = 13,

In NRQ, NR = 24, RQ = 10,

and QN = 26.

Find the corresponding ratios:

N

T

M

P

U

S

R

Q

Because the corresponding sides lengths are proportional

(all have a ratio of ½), PTS  NRQ by SSS  Thrm.

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