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Geometry 6.2 Arcs and Chords Objectives/Assignment • Use properties of arcs of circles.. Geometry • Use properties of chords of circles. • HOMEWORK Assignment: WS 6.2 Using Arcs of Circles • In a plane, an angle central angle whose vertex is the Geometry center of a circle is A a central angle of the circle. If the minor arc measure of a major P arc central angle, APB is less than 180°, B C then A and B and the points of P Using Arcs of Circles • in the interior of APB central angle Geometry form a minor arc of the A circle. The points A and B and the points of P in the exterior of APB minor arc major P form a major arc of the arc circle. If the endpoints B of an arc are the C endpoints of a diameter, then the arc is a semicircle. Naming Arcs G • Arcs are named by 60° Geometry their endpoints. For example, the minor E 60° H F arc associated with APB above is AB. Major arcs and semicircles are E named by their 180° endpoints and by a point on the arc. Naming Arcs G • For example, the 60° Geometry major arc 60° associated with E H F APB is A CB . EGF here on the right is a semicircle. The measure of a E 180° minor arc is defined to be the measure of its central angle. Naming Arcs • For instance, m GF= G 60° Geometry mGHF = 60°. • m GF is read “the E 60° H F measure of arc GF.” You can write the measure of an arc next to the arc. The E 180° measure of a semicircle is always 180°. Naming Arcs • The measure of a GF G 60° Geometry major arc is defined as the difference between E 60° H F 360° and the measure of its associated minor arc. For example, m GEF = 360° - 60° = 300°. E The measure of the 180° whole circle is 360°. Ex. 1: Finding Measures of Arcs • Find the measure Geometry of each arc of R. a. M N b. MPN c. PMN N 80° P R M Ex. 1: Finding Measures of Arcs • Find the measure Geometry of each arc of R. a. M N b. MPN c. PMN N 80° P R Solution: M N is a minor arc, so m M N = mMRN = 80° M Ex. 1: Finding Measures of Arcs • Find the measure Geometry of each arc of R. a. M N b. MPN c. PMN N 80° P R Solution: MPN is a major arc, so m MPN = 360° – 80° = 280° M Ex. 1: Finding Measures of Arcs • Find the measure Geometry of each arc of R. a. M N b. MPN c. PMN N 80° P R Solution: P MN is a semicircle, so m PMN= 180° M Note: C A • Two arcs of the same Geometry circle are adjacent if they intersect at exactly one point. You can add the measures of adjacent areas. • Arc Addition Conjecture B The measure of an arc formed by two adjacent arcs is the sum of the measures of the two m A BC = m AB+ m BC arcs. Ex. 2: Finding Measures of Arcs G • Find the measure of Geometry each arc. H a. GE 40° b. GEF 80° R c. GF 110° m GE = m GH+ m HE = F E 40° + 80° = 120° Ex. 2: Finding Measures of Arcs G • Find the measure of Geometry each arc. H a. GE 40° b. GEF 80° R c. GF 110° m GEF = m GE + m EF = F E 120° + 110° = 230° Ex. 2: Finding Measures of Arcs G • Find the measure of Geometry each arc. H a. GE 40° b. GEF 80° R c. GF 110° m GF = 360° - m GEF = F E 360° - 230° = 130° Ex. 3: Identifying Congruent Arcs A • Find the measures Geometry of the blue arcs. D Are the arcs 45° B congruent? 45° • AB and DC are in C the same circle and m AB = m DC= 45°. So, AB DC Ex. 3: Identifying Congruent Arcs • Find the measures Geometry of the blue arcs. Are the arcs 80° congruent? Q P • P Q and RS are in congruent circles and m P Q = m RS = 80°. So, P Q RS R 80° S Ex. 3: Identifying Congruent Arcs • Find the measures of the blue arcs. Are the Geometry arcs congruent? Z X • m XY = m ZW= 65°, but 65° XYand ZW are not arcs of the Y same circle or of congruent circles, so XY and ZW are NOT W congruent. Using Chords of Circles • A point Y is called Geometry the midpoint of if XY YZ . Any line, segment, or ray that contains Y bisects XYZ . Chord Arcs Conjecture • In the same circle, or in A Geometry congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent. AB BC if and only if B C AB BC Perpendicular Bisector of a Chord Conjecture • If a diameter of a Geometry circle is perpendicular to a F chord, then the diameter bisects the chord and its arc. E G DE EF , DG GF D Perpendicular Bisector to a Chord Conjecture • If one chord is a Geometry perpendicular J bisector of another M chord, then the first chord passes through the center of the circle and is a K diameter. JK is a diameter of L the circle. Ex. 4: Using Chord Arcs Conj. (x + 40)° • You can use Geometry 2x° Theorem 10.4 to C find m AD . A B • Because AD DC, and AD DC . So, m AD = m DC 2x = x + 40 Substitute x = 40 Subtract x from each side. Ex. 5: Finding the Center of a Circle • Theorem 10.6 can Geometry be used to locate a circle’s center as shown in the next few slides. • Step 1: Draw any two chords that are not parallel to each other. Ex. 5: Finding the Center of a Circle • Step 2: Draw the Geometry perpendicular bisector of each chord. These are the diameters. Ex. 5: Finding the Center of a Circle • Step 3: The Geometry perpendicular bisectors intersect at the circle’s center center. Ex. 6: Using Properties of Chords • Masonry Hammer. A Geometry masonry hammer has a hammer on one end and a curved pick on the other. The pick works best if you swing it along a circular curve that matches the shape of the pick. Find the center of the circular swing. Ex. 6: Using Properties of Chords • Draw a segment AB, Geometry from the top of the masonry hammer to the end of the pick. Find the midpoint C, and draw perpendicular bisector CD. Find the intersection of CD with the line formed by the handle. So, the center of the swing lies at E. Chord Distance to the Center Conjecture C • In the same circle, G Geometry or in congruent D circles, two chords are congruent if and E only if they are equidistant from the B center. F A • AB CD if and only if EF EG. Ex. 7: AB = 8; DE = 8, and A Geometry CD = 5. Find CF. 8 F B C E 5 8 G D Ex. 7: Because AB and DE A Geometry are congruent 8 F chords, they are B equidistant from the center. So CF C CG. To find CG, E first find DG. 5 8 CG DE, so CG G bisects DE. D Because DE = 8, 8 DG = 2 =4. Ex. 7: Then use DG to find A Geometry CG. DG = 4 and 8 F CD = 5, so ∆CGD is B a 3-4-5 right C triangle. So CG = 3. Finally, use CG to 5 E find CF. Because G 8 CF CG, CF = CG D =3 Reminders: • Quiz on Friday on 6.1- 6.2 Geometry • Yin Yang Due Friday

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