Transcript 10-3 Arcs and Chords

Arcs and Chords
Section 11.4
Goal

Use properties of chords of circles.
Theorems



11.4 Diameter Perpendicular to a
Chord
11.5 Converse 11.4
11.6 Congruent Arcs
Arcs and Chords



A chord is a line
segment with
endpoints on the
circle.
The endpoints of
chords also create
arcs.
If a chord is not a
diameter, then its
endpoints divide a
circle into a major
and a minor arc.
minor arc AB
Arc
A
B
O
C
major arc ACB
Congruent Arcs and Chords

In a circle or in congruent circles,
two minor arcs are congruent if
and only if their corresponding
chords are congruent.

EF  AB
Example 1
TV  WS. Find mWS.
TV  WS
mTV = mWS
9n – 11 = 7n + 11
2n = 22
 chords have  arcs.
Def. of  arcs
Substitute the given measures.
Subtract 7n and add 11 to both sides.
n = 11
Divide both sides by 2.
mWS = 7(11) + 11 Substitute 11 for n.
= 88°
Simplify.
Your Turn:
PT bisects RPS. Find RT.
RPT  SPT
mRT  mTS
RT = TS
6x = 20 – 4x
10x = 20
x=2
Add 4x to both sides.
Divide both sides by 10.
RT = 6(2)
Substitute 2 for x.
RT = 12
Simplify.
Example 2
Jewelry A circular piece of jade is hung from a
chain by two wires around the stone.
JM  KL and
= 90. Find
.
Example 2
Your Turn:
A. 42.5
B. 85
C. 127.5
D. 170
Example 3
Example 3
WX = YZ
segments
Definition of congruent
7x – 2 = 5x + 6
Substitution
2x = 8
x=4
Add 2 to each side.
Divide each side by 2.
So, WX = 7x – 2 = 7(4) – 2 or 26.
Answer: WX = 26
Your Turn:
A. 6
B. 8
C. 9
D. 13
Theorem 11.4 Diameter
Perpendicular to a Chord

If a diameter (or
radius)of a circle
is perpendicular
to a chord, then it
bisects the chord
and its arc.
F
o
E
G
If diameter OG is  to chord DF ,
then DE  EF and DG  GF .
D
Theorem 11.5 Diameter
Perpendicular to a Chord Converse


The perpendicular
bisector of a chord
is a diameter (or
radius) of the
circle.
Converse of
Theorem 10.3
J
M
O
K
If JK is a  bisector of chord LM ,
then JK is a diameter of O.
L
Example 4
In C the diameter AF is perpendicular to BD.
Use the diagram to find the length of BD.
SOLUTION
Because AF is a diameter that is perpendicular to BD,
you can use Theorem 11.4 to conclude that AF bisects
BD. So, BE = ED = 5.
BD = BE + ED
Segment Addition Postulate
=5+5
= 10
Substitute 5 for BE and ED.
ANSWER
Simplify.
The length of BD is 10.
Your Turn
Find the Length of a Segment
1. Find the length of JM.
ANSWER
12
ANSWER
30
2. Find the length of SR.
Finding the Center of a Circle


Theorem 11.5
can be used to
locate a circle’s
center as shown
in the next few
slides.
Step 1: Draw
any two chords
that are not
parallel to each
other.
Finding the Center of a Circle

Step 2: Draw
the perpendicular
bisector of each
chord. These are
the diameters.
Finding the Center of a Circle

Step 3: The
perpendicular
bisectors
intersect at the
circle’s center.
center
Example 4
Find the value of x.
a.
b.
SOLUTION
a.
Because QP  RS , it follows that QP  RS .
So, mQP = mRS = 60°, and x = 60.
b. Because AB  DE, it follows that AB  DE . So,
x = DE = 3.
Your Turn
Find Measures of Angles and Chords
Find the value of x.
1.
ANSWER
4
ANSWER
3
ANSWER
30
2.
3.
Example 5
Answer:
Your Turn:
A. 14
B. 80
C. 160
D. 180
Solving Problems
Draw a circle with a chord that is 15 inches long and 8 inches from
the center of the circle.
 Draw a radius so that it forms a right triangle.
 How could you find the length of the radius?
Solution: ∆ODB is a right triangle and OD bi sec ts AB

AB 15
DB=
= =7.5 cm
2
2
OD=8 cm
OB2 =OD 2 +DB2
OB2 =82 +(7.5) 2 =64+56.25=120.25
OB= 120.25  11cm
A
15cm
D
8cm
O
B
x
Example 6
CERAMIC TILE In the ceramic stepping
stone below, diameter AB is 18 inches long
and chord EF is 8 inches long. Find CD.
Example 6
Step 1 Draw radius CE.
This forms right ΔCDE.
Example 6
Step 2
Find CE and DE.
Since AB = 18 inches, CB = 9 inches.
All radii of a circle are congruent, so
CE = 9 inches.
9
4
Since diameter AB is perpendicular to
EF, AB bisects chord EF by Theorem
10.3. So, DE =1/2(8) or 4 inches.
Example 6
Step 3
Use the Pythagorean Theorem to find CD.
CD2 + DE2 = CE2 Pythagorean Theorem
9
4
Answer:
CD2 + 42 = 92
Substitution
CD2 + 16 = 81
Simplify.
CD2 = 65
Subtract 16 from each
side.
Take the positive
square root.
Your Turn:
In the circle below, diameter QS is 14 inches long
and chord RT is 10 inches long. Find VU.
A. 3.87
B. 4.25
C. 4.90
D. 5.32
Theorem 11.6 Congruent Arcs


In the same circle,
or in congruent
circles, two chords
are congruent if
and only if they are
equidistant from
the center.
AB  CD if and only
if EF  EG.
C
G
D
E
B
F
A
Example 7
Since chords EF and GH are congruent, they are
equidistant from P. So, PQ = PR.
Example 7
PQ = PR
4x – 3 = 2x + 3
x=3
So, PQ = 4(3) – 3 or 9
Answer: PQ = 9
Substitution
Simplify.
Your Turn:
A. 7
B. 10
C. 13
D. 15
Example 8
AB = 8; DE = 8,
and CD = 5. Find
CF.
A
8 F
B
C
E
5
8
G
D
Example 8
Because AB and DE
are congruent
chords, they are
equidistant from
the center. So CF
 CG. To find CG,
first find DG.
CG  DE, so CG
bisects DE.
Because
DE = 8,
8
DG = 2 =4.
A
8 F
B
C
E
5
8
G
D
Example 8
Then use DG to find
CG. DG = 4 and
CD = 5, so ∆CGD
is a 3-4-5 right
triangle.
So CG = 3.
Finally, use CG to find
CF. Because CF 
CG, CF = CG = 3
A
8 F
B
C
E
5
8
G
D
Assignment

Pg. 610 – 612;#1 - 19 odd, 23 – 31
odd.