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Objectives: Use congruent chords, arcs and central angles

Section 11-2 Chords and Arcs SPI 32B: Identify chords of circles given a diagram SPI 33A: Solve problems involving the properties of arcs, tangents, chords. Objectives: Use congruent chords, arcs and central angles. Chord of a circle

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Objectives: Use congruent chords, arcs and central angles

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  1. Section 11-2 Chords and Arcs SPI 32B: Identify chords of circles given a diagramSPI 33A: Solve problems involving the properties of arcs, tangents, chords • Objectives: • Use congruent chords, arcs and central angles Chord of a circle A segment whose endpoints are on a circle A Arc B Central Angle

  2. AOXBOX by the definition of an angle bisector. AX BX because congruent central angles have congruent chords. AX BX because congruent chords have congruent arcs. Relate Central Angles, Chords and Arcs Theorem 11-4: Within a circle or in congruent circles: (1) Congruent central angles have congruent chords. (2) Congruent chords have congruent arcs. (3) Congruent arcs have congruent central angles. In the diagram, radius OX bisects AOB. What can you conclude?

  3. Relate Central Angles, Chords and Arcs Theorem 11-5: Within a circle or in congruent circles: (1) Chords equidistant from the center are congruent. (2) Congruent chords are equidistant from the center Do Now Find AB. QS = QR + RSSegment Addition Postulate QS = 7 + 7 Substitute. QS = 14 Simplify. AB = QSChords that are equidistant from the center of a circle are congruent. AB = 14 Substitute 14 for QS.

  4. Lines Through the Center of a Circle Theorem 11-6: In a circle, a diameter that is perpendicular to a chord bisects the chord and its arcs. Theorem 11-7: In a circle, a diameter that bisects a chord (that is NOT a diameter) is perpendicular to the chord. Theorem 11-8: In a circle, the perpendicular bisector of a chord contains the center of the circle.

  5. Do Now . . . . Draw a diagram to represent the situation. The distance from the center of O to PQ is measured along a perpendicular line. 1 2 PM = PQ A diameter that is perpendicular to a chord bisects the chord. 1 2 PM = (16) = 8 Substitute. The radius of O is 17 in. Relate Central Angles, Chords and Arcs P and Q are points on O. The distance from O to PQ is 15 in., and PQ = 16 in. Find the radius of O. (Hint: Draw a diagram) OP 2 = PM 2 + OM 2Use the Pythagorean Theorem. r 2 = 82 + 152Substitute. r 2 = 289 Simplify. r = 17 Find the square root of each side.

  6. Do Now Find each missing length to the nearest tenth.

  7. Do Now Find each missing length to the nearest tenth.

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