Geometry Arcs and Chords. Geometry Mr. Austin

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1 10.2 Arcs and Chords Mr. Austin

2 Objectives/Assignment Use properties of arcs of circles, as applied. Use properties of chords of circles. Assignment: pp #3-47 Reminder Quiz after 10.3 and 10.5

3 Using Arcs of Circles In a plane, an angle whose vertex is the center of a circle is a central angle of the circle. If the measure of a central angle, APB is less than 180, then A and B and the points of P major arc C central angle P A B minor arc

4 Using Arcs of Circles in the interior of APB form a minor arc of the circle. The points A and B and the points of P in the exterior of APB form a major arc of the circle. If the endpoints of an arc are the endpoints of a diameter, then the arc is a semicircle. major arc C central angle P A B minor arc

5 Naming Arcs G Arcs are named by their endpoints. For example, the minor arc associated with APB above is AB. Major arcs and semicircles are named by their endpoints and by a point on the arc. E E H F

6 Naming Arcs G For example, the major arc associated with APB is ACB. EGFhere on the right is a semicircle. The measure of a minor arc is defined to be the measure of its central angle. E E H F

7 Naming Arcs G For instance, m GF = m GHF = 60. m GF is read the measure of arc GF. You can write the measure of an arc next to the arc. The measure of a semicircle is always 180. E E H F

8 Naming Arcs G The measure of a GF major arc is defined as the difference between 360 and the measure of its associated minor arc. For example, m = = 300. The measure of the whole circle is 360. GEF E E H F

9 Ex. 1: Finding Measures of Arcs Find the measure of each arc of R. a. b. MN MPN c. PMN N 80 R P M

10 Ex. 1: Finding Measures of Arcs Find the measure of each arc of R. a. b. MN MPN c. PMN Solution: MN is a minor arc, so m MN = m MRN = 80 N 80 M R P

11 Ex. 1: Finding Measures of Arcs Find the measure of each arc of R. a. b. MN MPN c. PMN Solution: MPN is a major arc, so m MPN = = 280 N 80 M R P

12 Ex. 1: Finding Measures of Arcs Find the measure of each arc of R. a. b. MN MPN c. PMN Solution: is a semicircle, so m = 180 PMN PMN N 80 M R P

13 Note: C A Two arcs of the same circle are adjacent if they intersect at exactly one point. You can add the measures of adjacent areas. Postulate 26 Arc Addition Postulate. The measure of an arc formed by two adjacent arcs is the sum of the measures of the two arcs. m = m AB + m ABC B BC

14 Ex. 2: Finding Measures of Arcs Find the measure of each arc. a. GE b. GEF c. GF m GE = m GH + m HE = = 120 F G R 110 H E

15 Ex. 2: Finding Measures of Arcs Find the measure of each arc. a. GE b. GEF c. m GEF= m GE + m = = 230 GF EF 40 F G 80 R 110 H E

16 Ex. 2: Finding Measures of Arcs Find the measure of each arc. a. GE 40 b. GEF R c. GF 110 m GF = m GEF = F = 130 G 80 H E

17 Ex. 3: Identifying Congruent Arcs Find the measures of the blue arcs. Are the arcs congruent? ABand DC are in the same circle and m AB = m = 45. So, AB DC DC B A D C

18 Ex. 3: Identifying Congruent Arcs Find the measures of the blue arcs. Are the arcs congruent? PQ and RS are in congruent circles and m PQ = m = 80. So, PQ RS RS P 80 Q R 80 S

19 Ex. 3: Identifying Congruent Arcs Find the measures of the blue arcs. Are the arcs congruent? m XY = m ZW = 65, but XY and ZW are not arcs of the same circle or of congruent circles, so XY and ZW are NOT congruent. 65 Y X W Z

20 Using Chords of Circles A point Y is called the midpoint of if XY YZ. Any line, segment, or ray that contains Y bisects XYZ.

21 Theorem 10.4 In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent. AB BC if and only if B A C AB BC

22 Theorem 10.5 If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc. GF DE EF, E F G DG D

23 Theorem 10.5 If one chord is a perpendicular bisector of another chord, then the first chord is a diameter. JK is a diameter of the circle. J M K L

24 Ex. 4: Using Theorem 10.4 You can use Theorem 10.4 to find m AD. Because AD DC, and AD DC. So, m AD = m DC 2x A B (x + 40) C 2x = x + 40 x = 40 Substitute Subtract x from each side.

25 Ex. 5: Finding the Center of a Circle Theorem 10.6 can 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.

26 Ex. 5: Finding the Center of a Circle Step 2: Draw the perpendicular bisector of each chord. These are the diameters.

27 Ex. 5: Finding the Center of a Circle Step 3: The perpendicular bisectors intersect at the circle s center. center

28 Ex. 6: Using Properties of Chords Masonry Hammer. A 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.

29 Ex. 6: Using Properties of Chords Draw a segment AB, 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.

30 Theorem 10.7 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. C A E F G D B

31 Ex. 7: Using Theorem 10.7 AB = 8; DE = 8, and CD = 5. Find CF. A 5 C 8 F G 8 B E D

32 Ex. 7: Using Theorem 10.7 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, 8 DG = =4. 2 A D 5 C 8 F G 8 B E

33 Ex. 7: Using Theorem 10.7 Then use DG to find CG. DG = 4 and CD = 5, so CGD is a right triangle. So CG = 3. Finally, use CG to find CF. Because CF CG, CF = CG = 3 A D 5 C 8 F G 8 B E

34 Homework 10.2 Page 607 #12-34

Geometry Arcs and Chords. Geometry Mr. Peebles Spring 2013

Geometry Arcs and Chords. Geometry Mr. Peebles Spring 2013 10.2 Arcs and Chords Geometry Mr. Peebles Spring 2013 Bell Ringer: Solve For r. B 16 ft. A r r 8 ft. C Bell Ringer B 16 ft. Answer A r r 8 ft. C c 2 = a 2 + b 2 Pythagorean Thm. (r + 8) 2 = r 2 + 16 2