properties of a chord circle geometry homework: lesson 6.2/1-12, 18 quiz friday lessons 6.1 – 6.2...
TRANSCRIPT
Properties of a
Chord
Circle Geometry
Homework: Lesson 6.2/1-12, 18Quiz Friday Lessons 6.1 – 6.2Ying Yang Project Due Friday
What is a chord? A chord is a segment with
endpoints on a circle.
Any chord divides the circle into two arcs.
A diameter divides a circle into two semicircles.
Any other chord divides a circle into a minor arc and a major arc.
What is a chord? The diameter of a circle is the longest chord of any
circle since it passes through the center.
A diameter satisfies the definition of a chord, however, a chord is not necessarily a diameter.
Therefore, every diameter is a chord, but
not every chord is a diameter.
Chord Property #1, #2 and #3
A perpendicular line from the center of a chord to the center of a circle:
#1: Makes a 90° angle with the chord
#2: Creates two equal line segments RS and QR
#3: Must pass through the center of the circle O
J
L
K
M
JK is a diameter of the circle.
If one chord is a perpendicular bisector of another chord, then the first chord is a
diameter.
E
D
G
FDG
GF
, DE EF
F
G
E
B
A
C
D
AB CD if and only if EF EG.
Chord Arcs Conjecture
In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent.
ABBCif and only if
AB BC
C
B
A
Perpendicular Bisector of a Chord Conjecture
If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.
E
D
G
F
DG
GF
, DE EF
Perpendicular Bisector to a Chord Conjecture
If one chord is a perpendicular bisector of another chord, then the first chord passes through the center of the circle and is a diameter.
JK is a diameter of the circle.
J
L
K
M
Ex: Using Chord Arcs Conj.
Because AD = DC, and = . So, m = m
AD
DCAD
DC
2x = x + 40 Substitute x = 40 Subtract x from each
side.
BA
C
2x°
(x + 40)°
Ex: Finding the Center of a Circle
Perpendicular Bisector to a Chord Conjecture 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.
Ex: Finding the Center of a Circle con’t
Step 2: Draw the perpendicular bisector of each chord. These are the diameters.
Step 3: The perpendicular bisectors intersect at the circle’s center.
center
Ex: Finding the Center of a Circle con’t
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.
Chord Distance to the Center Conjecture
F
G
E
B
A
C
D
Ex: Find CF
AB = 8; DE = 8, and CD = 5. Find CF.
58
8 F
G
C
E
D
A
B
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, DG = =4.
58
8 F
G
C
E
D
A
B
2
8
Ex: Find CF con’t
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
58
8 F
G
C
E
D
A
B
Ex: Find CF con’t
An angle whose vertex is on the center of the circle and whose sides are radii of the circle
An angle whose vertex is ON the circle and whose sides are chords of the circle
Naming ArcsArcs are defined by their endpoints
Minor Arcs require the 2 endpoints
Major Arcs require the 2 endpoints AND a point the arc passes through
Semicircles also require the 2 endpoints (endpoints of the diameter) AND a point the arc passes through
EH F
G
EK
2x = x + 40 x = 40
BA
C
2x°
(x + 40)°D
Ex.3: Solve for the missing sides.
A
B
D C
7m
3m
BC =
AB =
AD ≈
7m
14m
7.6m
.AC
A
B
C
D
What can you tell me about segment AC if you know it is the perpendicular bisectors of segments DB?
It’s the DIAMETER!!!
Finding the Center of a Circle
If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.
This 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 diameters of the circle.
Finding the Center of a Circle
Step 3: The perpendicular bisectors intersect at the circle’s center.
center
Ex.6: QR = ST = 16. Find CU.
x = 3
Ex 7: AB = 8; DE = 8, and CD = 5. Find CF.
58
8 F
G
C
E
D
A
B
CG = CF
CG = 3 = CF
Ex.8: Find the length of
Tell what theorem you used.
.BF
BF = 10
Diameter is the perpendicular bisector of
the chordTherefore, DF = BF
Ex.9: PV = PW, QR = 2x + 6, and ST = 3x – 1. Find QR.
Congruent chords are equidistant from the center.
Congruent chords intercept congruent arcs
Ex.11:
Congruent chords are equidistant from the center.
Let’s get crazy…
c
3 8
12
Find c.
Let’s get crazy…
c
3 8
12
Step 1: What do we need to find?
We need a radius to complete this big triangle.
Find c.
Let’s get crazy…
c
3 8
12
Find c.
How do we find a radius?
We can draw multiple radii (radiuses).
Let’s get crazy…
c
38
12
Find c.
How do we find a radius?Now what do we have and what will we do?
We create this triangle, use the chord property to find on side
& the Pythagorean theorem find the missing side.
45
Let’s get crazy…
c
34
12
Find c.
!!! Pythagorean Theorem !!!
c2=a2+b2
c2=52+122
c2=169c = 13.
5
13