Bell work 1Bell work 1Find the measure of Arc ABC, if Arc AB = 3x, Find the measure of Arc ABC, if Arc AB = 3x, Arc BC = (x + 80Arc BC = (x + 80º), and º), and __ ____ __AB AB BCBC

AB = 3xº

A

B

C

BC = ( x + 80º )

Bell work 1 AnswerBell work 1 AnswerSince,Since,

__ ____ __AB AB BC , then AB BC , then AB BC, thus BC, thus3x = x + 80º3x = x + 80º2x = 80º2x = 80ºx = 40º = AB & BCx = 40º = AB & BCTherefore AB + BC = Therefore AB + BC = ABC = 240ºABC = 240º

AB = 3xº

BC = (x + 80º)

AC

B

Bell work 2Bell work 2You are standing at point X. Point X is 10 You are standing at point X. Point X is 10

feet from the center of the circular water feet from the center of the circular water

tank and 8 feet from point Y. Segment XY is tank and 8 feet from point Y. Segment XY is

tangent to the circle P at point Y. What is the tangent to the circle P at point Y. What is the

radius, r,radius, r, of the circular water tank? of the circular water tank?

••

•• X

r 10 ft

8 ftY

P

Bell work 2Bell work 2

Answer Answer Use the Pythagorean Theorem since Use the Pythagorean Theorem since segment XY is tangent to circle P at segment XY is tangent to circle P at Point Y, then it is perpendicular to the Point Y, then it is perpendicular to the radius, rradius, r at point Y. at point Y.

r =r = 6 ft6 ft

Unit 3 : Circles: Unit 3 : Circles: 10.3 Arcs and Chords10.3 Arcs and Chords

Objectives: Students will:Objectives: Students will:

1. Use inscribed angles and properties 1. Use inscribed angles and properties of inscribed angles to solve of inscribed angles to solve problems related to circlesproblems related to circles

Words for CirclesWords for Circles

1.1. Inscribed AngleInscribed Angle

2.2. Intercepted ArcIntercepted Arc

3.3. Inscribed PolygonsInscribed Polygons

4.4. Circumscribed Circumscribed CirclesCircles

Are there any words/terms that you are unsure of?

Inscribed AnglesInscribed AnglesInscribed angleInscribed angle – is an angle whose – is an angle whose vertex is on the circle and whose sides vertex is on the circle and whose sides contain chords of the circle.contain chords of the circle.

INSCRIBED ANGLE

A

B

INTERCEPTED ARC,

AB

Vertex on the circle

Intercepted ArcIntercepted ArcIntercepted ArcIntercepted Arc – is the arc that lies in – is the arc that lies in the interior of the the interior of the inscribed angleinscribed angle and and has endpoints on the angle. has endpoints on the angle.

INSCRIBED ANGLE

A

B

INTERCEPTED ARC,

AB

Vertex on the circle

(p. 613) Theorem 10. 8 (p. 613) Theorem 10. 8 Measure of the Inscribed AngleMeasure of the Inscribed Angle

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

Central Angle

•

CENTER P

P

A

B

•

•Inscribed angle

C

m ∕_ ABC

= ½ m AC

m ∕_ ABC

= ½ mAC = 30º

Example 1Example 1

Central Angle

60º•

A

B

•

•

Measure of theINTERCEPTED ARC = the measure of theCentral Angle

AC = 60ºC

•

The measure of the inscribed angle ABC = ½ the The measure of the inscribed angle ABC = ½ the

measure of the intercepted AC.measure of the intercepted AC.

30º

Example 2Example 2

T

R

•

•U

•

Find the measure of the intercepted TU, if the Find the measure of the intercepted TU, if the

inscribed inscribed angle Rangle R is a right angle. is a right angle.

Example 2 AnswerExample 2 Answer

T

R

•

•U

•

The measure of the intercepted TU = 180º, if the The measure of the intercepted TU = 180º, if the

inscribed angle R is a right angle.inscribed angle R is a right angle.

TU = 180 º

Example 3Example 3

T

R

•

•

TU = 86º

U

•

Find the measure of the inscribed angles Q , R ,and Find the measure of the inscribed angles Q , R ,and

S, given that their common intercepted TU = 86ºS, given that their common intercepted TU = 86º

Q

S

Example 3 AnswersExample 3 Answers

T

R

•

•

TU = 86º

U

•

Angles Q, R, and S = ½ their common intercepted arc TUAngles Q, R, and S = ½ their common intercepted arc TUSince their intercepted Arc TU = 86º, then Since their intercepted Arc TU = 86º, then Angle QAngle Q = = Angle RAngle R = = Angle SAngle S = = 43º43º

Q

S

(p .614) Theorem 10.9(p .614) Theorem 10.9

T•

•

IF ∕_ Q and ∕_ S both

intercepted TU, then

∕_ Q ∕_ S

U

•

If two inscribed angles of a circle intercepted If two inscribed angles of a circle intercepted

the same arc, then the angles are congruentthe same arc, then the angles are congruent

Q

S

Inscribed vs. CircumscribedInscribed vs. Circumscribed

Inscribed polygonInscribed polygon – is when all of its – is when all of its vertices lie on the circle and the vertices lie on the circle and the polygon is inside the circle. The Circle polygon is inside the circle. The Circle then is then is circumscribedcircumscribed about the about the polygonpolygon

Circumscribed circleCircumscribed circle – lies on the – lies on the outside of the outside of the inscribed polygoninscribed polygon intersecting intersecting all the vertices of the polygon.all the vertices of the polygon.

Inscribed vs. CircumscribedInscribed vs. Circumscribed

The Circle is circumscribed about the The Circle is circumscribed about the polygon.polygon.

Circumscribed Circle

Inscribed Polygon

(p. 615) Theorem 10.10(p. 615) Theorem 10.10

If a right triangle is inscribed in a If a right triangle is inscribed in a

circle, then the hypotenuse is the circle, then the hypotenuse is the

diameter of the circle. diameter of the circle.

•

Hypotenuse= Diameter

(p. 615) Converse of (p. 615) Converse of Theorem 10.10Theorem 10.10

If one side of an inscribed triangle is a If one side of an inscribed triangle is a diameter of the circle, then the triangle is a diameter of the circle, then the triangle is a right triangle and the angle opposite the right triangle and the angle opposite the diameter is a right angle. diameter is a right angle.

•

Diameter= Hypotenuse

B

The triangle is inscribed in the circle and one of its sides is the diameter Angle B is a right angleand measures 90º

Example Example

Triangle ABC is inscribed in the circle Triangle ABC is inscribed in the circle

Segment AC = the diameter of the Segment AC = the diameter of the

circle. Angle B = 3x. Find the value of x.circle. Angle B = 3x. Find the value of x.

•

B

A C

3xº

AnswerAnswer

Since the triangle is inscribed in the circle and Since the triangle is inscribed in the circle and one of its sides is the diameter = hypotenuse one of its sides is the diameter = hypotenuse side, then its opposite angle, Angle B, measures side, then its opposite angle, Angle B, measures 9090ºº Thus, Thus, 3x = 903x = 90ºº

x = x = 3030ºº

(p. 615) Theorem 10.11(p. 615) Theorem 10.11

A quadrilateral can be inscribed in a circle iff its A quadrilateral can be inscribed in a circle iff its

opposite angles are supplementary.opposite angles are supplementary.

••

••••X

Y

P

••Z

The Quadrilateral WXYZ is inscribed in the circle iff

/ X + / Z = 180º, and

/ W + / Y = 180ºW

••

ExampleExample

A quadrilateral WXYZ is inscribed in circle P, if A quadrilateral WXYZ is inscribed in circle P, if ∕ ∕__ X = 103º and X = 103º and ∕_ ∕_ Y = 115º , Find the measures of Y = 115º , Find the measures of ∕ ∕__ W = ? and W = ? and ∕_ ∕_ Z = ? Z = ?

••

••••X

Y

P

••Z

The Quadrilateral WXYZ is inscribed in the circle iff

/ X + / Z = 180º, and

/ W + / Y = 180ºW

••

103º 115º

ExampleExample

From Theorem 10.11From Theorem 10.11 ∕ ∕__ W = 180º – 115º = 65º and W = 180º – 115º = 65º and ∕ ∕__ Z = 180º – 103º = 77º Z = 180º – 103º = 77º

••

••••X

Y

P

••Z

The Quadrilteral WXYZ is inscribed in the circle iff

/ X + / Z = 180º, and

/ W + / Y = 180ºW

••

103º 115º

Home workHome work

PWS 10.3 APWS 10.3 A

P. 617 (9 -22) allP. 617 (9 -22) all

JournalJournal

Write two things about “Inscribed Write two things about “Inscribed

Angles” or “Inscribed Polygons” related Angles” or “Inscribed Polygons” related

to circles from this lesson.to circles from this lesson.