Use Inscribed Angles and Polygons

MM2G3 Students will understand properties of circles.

MM2G3 b Understand and use properties of central, inscribed, and related angles. MM2G3 d Justify measurements and relationships in circles and geometric and algebraic properties.

Use Inscribed Angles and Polygons

How do we use inscribed angles of circles?

M2 Unit 3: Day 4

Lesson 6.4

Friday, April 21, 2023


MM2G3 Students will understand properties of circles.Daily Homework Quiz

1.

Find the value of x in C. Explain. .

ANSWER

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

Daily Homework Quiz



2.

Find the value of x in C. Explain. .

ANSWER

4; In the same circle, if two chords are equidistant from the center, then they are .=~

Daily Homework Quiz



3. Determine whether RS is a diameter.

ANSWER

Yes. Sample answer: RS is the bisector of TU by Theorem 5.3. Then RS is a diameter of the circle by Theorem 10.4.

Daily Homework Quiz



ANSWER 25

1. The measure of the interior angles of a quadrilateralare 80º, 100º, 55º, and 5xº. Find the value of x.

2. Two supplementary angles have measures 6xº and12xº. Find each angle measure.

ANSWER 60º; 120º

Warm Ups



3. Solve 3x = ( 4x + 12).1 2

4. Solve 80 = ( 360 – 2x).1 2

Warm Ups



7

Inscribed AngleInscribed Angle: An angle whose vertex lies on a circle and whose sides are chords of the circle (or one side tangent to the circle).

1 42 3

No! No!Yes! Yes!

Examples:

Inscribed Angle

Intercepted Arc

B

A

CCB is an inscribed angleA

» is the intercepted arc of ÐAB C



Theorem 6.9 Inscribed Angle Theorem

The measure of an inscribed angle is half the measure of its intercepted arc.

»1BC

2Ð =m A mAC B

C

A

O



EXAMPLE Use inscribed angles

a. m T mQRb.

Find the indicated measure in P.

SOLUTION

12

12

M T = mRS = (48o) = 24oa.

mQR = 180o mTQ = 180o 100o = 80o. So, mQR = 80o. – –

mTQ = 2m R = 2 50o = 100o. Because TQR is a semicircle,b.


MM2G3 Students will understand properties of circles.GUIDED PRACTICE

Find the measure of the red arc or angle.

1.

SOLUTION

12

12

M G = mHF = (90o) = 45oa.

Guided Practice




2.

SOLUTION

mTV = 2m U = 2 38o = 76o. b.

Guided Practice



Theorem 6.10

If two inscribed angles of a circle intercept the same arc, then the angles are congruent.

A

B

D

C

ABD ACD


MM2G3 Students will understand properties of circles.EXAMPLE 2

Find the measure of an intercepted arc

Find mRS and m STR. What do you notice about STR and RUS?

SOLUTION

From Theorem 6.9, you know that mRS = 2m RUS = 2 (31o) = 62o.

Also, m STR = mRS = (62o) = 31o. So, STR RUS.12

12

EXAMPLE



Standardized Test Practice

SOLUTION

Notice that JKM and JLM intercept the same arc, and so JKM JLM by Theorem 6.10 Also, KJL and KML intercept the same arc, so they must also be congruent. Only choice C contains both pairs of angles.

So, by Theorem 10.8, the correct answer is C.




3.

SOLUTION

ZYN ZXN

ZXN 72°

Notice that ZYN and ZXN intercept the same arc, and so ZYN by Theorem 6.10. Also, KJL and KML intercept the same arc, so they must also be congruent.

ZXN

Guided Practice



4. Find mLJM.

mLJM = 5(3.5) – 7 = 10.5 Substitute 3.5 for b.

5b – 7 = 3b Substitute the given values.

2b – 7 = 0 Subtract 3b from both sides.

2b = 7 Add 7 to both sides.

b = 3.5 Divide both sides by 2.

mLJM = mLKM mLJM and mLKM

both intercept LM.

Guided Practice



5. Find mEDF.

2x + 3 = 75 – 2x Substitute the given values.

4x = 72 Add 2x and subtract 3 from both sides.

x = 18 Divide both sides by 4.

mEDF = 2(18) + 3 = 39°

mEDF = mEGF mEGF and mEDF

both intercept EF.

Guided Practice



18

An angle inscribed in a semicircle is a right angle.

R

P 180

S90

Theorem 6.11 Inscribed Right Angle Theorem



Find a.

WZY is a right angle WZY is inscribed in a semicircle.

mWZY = 90 Def of rt.

5a + 20 = 90 Substitute 5a + 20 for mWZY.

5a = 70 Subtract 20 from both sides.

a = 14 Divide both sides by 5.

EXAMPLE



6. Find z.

8z – 6 = 90 Substitute.

8z = 96 Add 6 to both sides.

z = 12 Divide both sides by 8.

ABC is a right angle ABC is inscribed in a semicircle.

mABC = 90 Def of rt.

Guided Practice



21

mDAB + mDCB = 180

mADC + mABC = 180

If a quadrilateral is inscribed in a circle, then the opposite angles are supplementary.

AB

C

D

Theorem 6.12 Inscribed Quadrilateral Theorem



Find the value of each variable.

a.

SOLUTION

PQRS is inscribed in a circle, so opposite angles are supplementary.

a.

m P + m R = 180o

75o + yo = 180o

y = 105

m Q + m S = 180o

80o + xo = 180o

x = 100

EXAMPLE



b. JKLM is inscribed in a circle, so opposite angles are supplementary.

m J + m L = 180o

2ao + 2ao = 180o

a = 45

m K + m M = 180o

4bo + 2bo = 180o

b = 30

4a = 180 6b = 180


b.

SOLUTION

EXAMPLE



7.


SOLUTION

ABCD is inscribed in a circle, so opposite angles are supplementary.

m A + m C = 180o

y°+ 68o = 180o

y = 112

m D+ m B = 180o

82o + xo = 180o

x = 98

Guided Practice



8.


SOLUTION

STUV is inscribed in a circle, so opposite angles are supplementary.

m S + m U = 180o

c°+ (2c – 6 )o = 180o

c = 62

m V+ m T = 180o

8xo + 10xo = 180o

x = 10

(3c)o = 186o 18xo = 180o

Guided Practice



9. Find the angle measures of GHJK.

mG + mJ = 180 GHJK is inscribed in a .

3b + 25 + 6b + 20 = 180 Substitute the given values.

9b + 45 = 180 Simplify.

9b = 135 Subtract 45 from both sides.

b = 15 Divide both sides by 9.

mG = 3(15) + 25 = 70 Substitute 15 for b

mJ = 6(15) + 20 = 110 in each expression.

mK = 10(15) – 69 = 81mH + mK = 180 H and K are supp.

mH + 81 = 180 Substitute 81 for mK.

mH = 99 Subtract 81 from both sides

Guided Practice



Step 2 Find the measure of each angle.

Example 4 Continued

mG = 3(15) + 25 = 70 Substitute 15 for b

mJ = 6(15) + 20 = 110 in each expression.

mK = 10(15) – 69 = 81mH + mK = 180 H and K are supp.

mH + 81 = 180 Substitute 81 for mK.

mH = 99 Subtract 81 from both sides



35o

xo

yoQ

L

K

J Find x and y.

90 QJL m 90o

55x

125180 y

1 2

m y JQ

35y

Guided Practice

70o

110o

180o



Homework: pg 207-208 (#1, 2, 5, 6, 8-18, 22)

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Use Inscribed Angles and Polygons