Geometry 9 3 Arcs and Chords Geometry ObjectivesAssignment

Geometry 9. 3 Arcs and Chords

Geometry Objectives/Assignment • Use properties of arcs of circles. • Use properties of chords of circles.

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

Geometry 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.

Geometry Naming Arcs The measure of a minor arc is defined to be the measure of its central angle. m 60° = m GHF = 60°. 180°

Geometry 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°. 60° 180°

Geometry Finding Measures of Arcs • a. b. c. Find the measure of each arc of R. 80°

Geometry Ex. 1: Finding Measures of Arcs • Find the measure of each arc of R. a. b. c. Solution: is a minor arc, so m = m MRN = 80°

Geometry Ex. 1: Finding Measures of Arcs • Find the measure of each arc of R. a. b. c. Solution: is a major arc, so m = 360° – 80° = 280°

Geometry Ex. 1: Finding Measures of Arcs • Find the measure of each arc of R. a. b. c. Solution: is a semicircle, so m = 180°

Geometry 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. if and only if

Geometry Theorem 9. 2 • If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc. ,

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

Using this. Theorem Geometry D • You can use Theorem 9. 3 to find m. 2 x° • Because AD DC, and . So, m =m 2 x = x + 40 x = 40 Substitute Subtract x from each side. (x + 40)°

Geometry Theorem 9. 3 • 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.

Geometry Using Theorem 9. 4 AB = 8; DE = 8, and CD = 5. Find CF.

Geometry 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.

Geometry Using Theorem 9. 4 Then use DG to find CG. DG = 4 and CD = 5, so ∆CGD is right triangle. So CG = 3 CF = CG = 3

Geometry Practice • 360 – 120 =240 • 240/3 = 80

Geometry Practice • 12 * 2 = 24

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

Geometry • BC = 360 / 5 = 72

Geometry • MX = 12 • XN = 12 • MN = PQ • YQ = 12

Geometry • Radius² = 8² + 6² • R² = 100 • R = 10 inches 16 6 •

Geometry • Chord ² =13² - 5² • Chord ² = 144 • C= 12 + 12 • C= 24 inches 5 13 •

Geometry • Using Pythagorean Formula • Distance ² = 17 ² - 15 ² • D ² = 64 • Distance = 8 inches 30 17 •

Geometry • Using Pythagorean Formula • Radius = 6 inches • APB = 120° • PAB isosceles triangle • A= (180 ° - 120 °)/2 • A = 30 ° • AB = 5. 2 +5. 2 = 10. 4 • Cos 30 = x/ 6 • X = 6*Cos 30 = 5. 2

Geometry • 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