GeometryGeometry 9.3 Arcs and Chords. Geometry Geometry Objectives/Assignment Use properties of arcs...

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9.3 Arcs and Chords

Objectives/Assignment

• Use properties of arcs of circles.• Use properties of chords of circles.

Using Arcs of Circles

• In a plane, an angle whose vertex is the center of a circle is a central angle of the circle.

central angle

minorarcmajor

Using Arcs of Circles

• minor arc of the circle is less than 180

• major arc of the circle is greater than 180

• semicircle if the endpoints of an arc are the endpoints of a diameter.

central angle

minorarcmajor

Naming Arcs

The measure of a minor arc is defined to be the measure of its central angle.

m = m GHF = 60°. GF

Naming Arcs

• The measure of a major arc is defined as the difference between 360° and the measure of its associated minor arc. For example, m = 360° - 60° = 300°. The measure of the whole circle is 360°.

Finding Measures of Arcs

• Find the measure of each arc of R.

PMN PR

Ex. 1: Finding Measures of Arcs

Solution:

is a minor arc, so m = mMRN = 80°

PMN PR

Solution:

is a major arc, so m = 360° – 80° = 280°

PMN PR

Solution:

is a semicircle, so m = 180°

PMN PR

Theorem 9.1

• 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

Theorem 9.2

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

, DE EF

Theorem

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

the circle.

Using thisTheorem

• You can use Theorem 9.3 to find m .

• Because AD DC, and . So, m = m

2x = x + 40 Substitute x = 40 Subtract x from each

(x + 40)°D

• 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.

Theorem 9.3

Using Theorem 9.4

AB = 8; DE = 8, and CD = 5. Find CF.

Using Theorem 9.4

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.

Using Theorem 9.4

Then use DG to find CG. DG = 4 and CD = 5, so ∆CGD is right triangle.

So CG = 3CF = CG = 3

Practice

• 360 – 120 =240• 240/3 = 80

Practice

• 12 * 2 = 24

Practice

• BY² = 20² - 12²• BY = 16 • XY = 16*2 = 32

• BC = 360 / 5 = 72

• MX = 12• XN = 12• MN = PQ• YQ = 12

• Radius² = 8² + 6²• R² = 100• R = 10 inches

• Chord ² =13² - 5²• Chord ² = 144• C= 12 + 12• C= 24 inches

• Using Pythagorean Formula

• Distance ² = 17 ² - 15 ²• D ² = 64• Distance = 8 inches

• Using Pythagorean Formula

• Radius = 6 inches• APB = 120°• PAB isosceles triangle• A= (180 ° - 120 °)/2• A = 30 °

• Cos 30 = x/ 6• X = 6*Cos 30 = 5.2

• AB = 5.2 +5.2 = 10.4

• PD = PC = PB = Radius• CE = 2• EP = 5 -2 = 3• Using Pythagorean Formula for EPB• EB ² = PB ² - PE ²• EB ² = 5 ² - 3 ²• EB ²= 16• EB = 4• AB = AE + EB = 4+ 4 = 8

GeometryGeometry 9.3 Arcs and Chords. Geometry Geometry Objectives/Assignment Use properties of arcs...

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