Warm Up Section 4.3

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:

PA AC

DE

DR

ET

Answers to Warm Up Section 4.3

P

A

C

E

D

R

T

Properties of ChordsSection 4.3

Standard: MM2G3 ad

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

In this section you will learn to use relationshipsof 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.

A

C

D

B AB CD if and only if _____ ______.AB CD

1. In the diagram, A D, , and m EF = 125o. Find m BC.

A

C

D

B

EFBC

F

E

Because BC and EF are congruent ________ in congruent _______, the corresponding minor arcsBC and EF are __________ .

So, m ______ = m ______ = ______o.

chordscircles

congruent

125BC EF

If one chord is a perpendicular bisector of anotherchord, then the first chord is a diameter.

If QS is a perpendicular bisector of TR, then ____ is a diameter of the circle.

R

P

S

Q

T

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 TR DF, then HD HFand ____ ____ .

D

H

E

G

F

GD GF

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

R

S

V

T

6

6

If the chords are congruent, then the arcs are congruent. So, m RS = 121o

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

E

DB

C

(80 – x)o

Since BD is a diameter,it bisects, the chord and the arcs.

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

mCE = 2(64o) = 128o

A

4xo

4x = 80 – x 5x = 80 x = 16

In the same circle, or in congruent circles, two chordsare congruent if and only if they are equidistantfrom the center.

E D

A

C

BF

G

if and only if _____ ____

CDAB FE GE

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

E

B

D

A

C

7x – 8

G AB

3xF

Chords and arecongruent, so they areequidistant from F.

Therefore EF = 6

CD

7x – 8 = 3x 4x = 8 x = 2

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

5. In the diagram of F, suppose AB = 27 and EF = GF = 7. Find CD.

E

B

D

A

C

G AB

F

Since and are both 7 units from the center, they are congurent.

DC

So, AB = CD = 27.

6. In F, SP = 5, MP = 8, ST = SU, and NRQ is a right angle. Show that PTS NRQ.

S

N

TM P

Q

QN

U

Step 1: Look at PTS !!Since MP = 8, and NQ bisectsMP, we know MT= PT = 4.

Since , PTN isa right angle.

MP

R

5

44

QN MP

Use the Pythagorean Theorem to find TS.

42 + TS2 = 52

TS2 = 25 – 16 TS2 = 9 So, TS = 3.

3

6. In F, SP = 5, MP = 8, ST = SU, and NRQ is a right angle. Show that PTS NRQ.

S

N

TM P

Q

QN

U

Step 2: Now, look at NRQ!Since the radius of the circleis 5, QN = 10.

Since ST = SU, MP and RNare equidistant from the center.Hence, MP = RN = 8.

MP

R

10

Use the Pythagorean Theorem to find RQ.

RQ2 + 82 = 102

RQ2 = 100 – 64 RQ2 = 36 So, RQ = 6.

8

6

6. In F, SP = 5, MP = 8, ST = SU, and NRQ is a right angle. Show that PTS NRQ.

S

N

TM P

Q

QN

U

MP

R

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:

8

4

NR

PT

6

3

RQ

TS

10

5

NQ

PS

8

4

NR

PT

6

3

RQ

TS

10

5

NQ

PS

Because the corresponding sides lengths are proportional(all have a ratio of ½), PTS NRQ by SSS.

7. In S, QN = 26, NR = 24, ST = SU, and NRQ is a right angle. Show that PTS NRQ.

S

N

TM P

Q

QN

U

MP

R

242 + RQ2 = 262

RQ2 = 676 – 576 RQ2 = 100 So, RQ = 10.

N

Q

R

26

24

7

7. In S, QN = 26, NR = 24, ST = SU, and NRQ is a right angle. Show that PTS NRQ.

S

N

TM P

Q

QN

U

MP

R

122 + ST2 = 132

ST2 = 169 –144 ST2 = 25 So, RQ = 5.

S

TP

12

MP = RN = 24. So, PT = ½(24).

SP is half of the diameter QN,so SP = ½(26) = 13.

135

S

N

TM P

Q

U

R

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:

24

12

NR

PT10

5

RQ

TS

26

13

NQ

PS

Because the corresponding sides lengths are proportional(all have a ratio of ½), PTS NRQ by SSS.