- 66 Views
- Uploaded on
- Presentation posted in: General

Warm Up Section 3.2 Draw and label each of the following in a circle with center P.

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

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

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

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?

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.

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

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

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

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 mRMS = 60o, then m RS = 60o.

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

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

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

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

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

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

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

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

- 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.

- 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.

- 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.

- 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

Because the corresponding sides lengths are proportional,

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

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.

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.

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.