MM2G3 Students will understand properties of circles. MM2G3 d Justify measurements and relationships in circles using geometric and algebraic properties

MM2G3 Students will understand properties of circles.

MM2G3 d Justify measurements and relationships in circles using geometric and algebraic properties.

Apply Properties of Chords

Essential Question:

How do we use relationships of arcs and chords in a circle?

M2 Unit 3: Day 3

Lesson 6.3

Wednesday, April 19, 2023


MM2G3 a Understand and use properties of chords, tangents, and secants as an application of triangle similarity.

Daily Homework Quiz

Describe each figure as a minor arc, major arc or a semicircle. Find the arc measure.

1. BC

ANSWER minor arc, 32o

Daily Homework Quiz



Daily Homework Quiz

2.


CBE

ANSWER major arc, 212o

Daily Homework Quiz



Daily Homework Quiz

3.


BCE

ANSWER semicircle, 180o

Daily Homework Quiz



4.


BCAEExplain why =~ .

ANSWER

BCAE =~

m AFE = m BFC because the angles are vertical angles, so AFE BFC.Then arcs and are arcs that have the same measure in the same circle. By definition .

=~AE BC

Daily Homework Quiz



5.ACD AC

Two diameters of P are AB and CD.If m = 50 , find m and m .

. ADo

ANSWERo

310 ; 130o

Daily Homework Quiz



radius1. DC

Tell whether the segment is best described as a radius,chord, or diameter of C.

Warm Ups

diameter2. BD

3. DEchord

4. AE

5. Solve 4x = 8x – 12. 6. Solve 3x + 2 = 6x – 4.

x = 3 x = 2

chord



Theorem 6.5 In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent.

C

B

A» ¼@ @ if and only if AB BC AB BC



Use congruent chords to find an arc measure

In the diagram, P Q, FG JK , and mJK = 80o. Find mFG

SOLUTION

Because FG and JK are congruent chords in congruent circles, the corresponding minor arcs FG and JK are congruent.

So, mFG = mJK = 80o.



SOLUTION

Because AB and BC are congruent chords in the same circle, the corresponding minor arcs AB and BC are congruent.

Use the diagram of D.

1. If mAB = 110°, find mBC

So, mBC = mAB = 110o.



GUIDED PRACTICEUse the diagram of D.

2. If mAC = 150°, find mABBecause AB and BC are congruent chords in the same circle, the corresponding minor arcs AB and BC are congruent.

Subtract

Substitute

mAB = 105° Simplify

So, mBC = mAB

And, mBC + mAB + mAC = 360°

So, 2 mAB + mAC = 360° 2 mAB + 150° = 360°

2 mAB = 360 – 150 2 mAB = 210

mAB = 105° ANSWER



Theorem 6.6

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

is a diameter of the circleJK

J

L

K

M



Theorem 6.7

If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.

E

D

G

FDE EF¼ »@DG GF



Use a diameter

SOLUTION

Use the diagram of E to find the length of AC . Tell what theorem you use.

Diameter BD is perpendicular to AC . So, by Theorem 6.7, BD bisects AC , and CF = AF. Therefore, AC = 2 AF = 2(7) = 14.



3. CD

So 9x° = (80 – x)° So 10x° = 80° x = 8°

So mCD = 9x° = 72°

From the diagramDiameter BD is perpendicular to CE . So, by Theorem 6.7, BD bisects CE , Therefore mCD = mDE.

Find the measure of the indicated arc in the diagram.

SOLUTION



4. DE

mCD = mDE.

So mDE = 72°

5. CE

mCE = mDE + mCD

So mCE = 72° + 72° = 144°

Find the measure of the indicated arc in the diagram.

SOLUTION

SOLUTION



Theorem 6.8In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.

F

G

E

B

A

C

D

if and only if GE = FEAB CD



SOLUTION

Chords QR and ST are congruent, so by Theorem 6.8 they are equidistant from C. Therefore, CU = CV.

CU = CV

2x = 5x – 9

x = 3

So, CU = 2x = 2(3) = 6.

Use Theorem 6.8

Substitute.

Solve for x.

In the diagram of C, QR = ST = 16. Find CU.



Since CU = CV. Therefore Chords QR and ST are equidistant from center and from theorem 6.8 QR is congruent to ST

SOLUTION

QR = STQR = 32

Use Theorem 6.8.

Substitute.

6. QR

Suppose ST = 32, and CU = CV = 12. Find the given length.



Since CU is the line drawn from the center of the circle to the chord QR it will bisect the chord.

SOLUTION

QU = 16

Substitute.

7. QU

2So QU = QR1

2So QU = (32)1




Join the points Q and C. Now QUC is right angled triangle. Use the Pythagorean Theorem to find the QC which will represent the radius of the C

SOLUTION

8. The radius of C




SOLUTION

Suppose ST = 32, and CU = CV = 12. Find the given length.8. The radius of C

So QC2 = 162 + 122

So QC2 = 256 + 144So QC2 = 400So QC = 20

So QC2 = QU2 + CU2 By Pythagoras Thm

Substitute

Square

Add

Simplify

ANSWER The radius of C = 20



Homework

Page 203-204 # 4 – 21 all.

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MM2G3 Students will understand properties of circles. MM2G3 d Justify measurements and relationships in circles using geometric and algebraic properties