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Apply properties of arcs. Apply properties of chords.

Objectives. Apply properties of arcs. Apply properties of chords. A central angle is an angle formed from the middle of a circle and is inside a circle. (vertex is the center of a circle) An arc is the part of the actual circle that is inside the angle (the red line in the picture).

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Apply properties of arcs. Apply properties of chords.

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  1. Objectives Apply properties of arcs. Apply properties of chords.

  2. A central angleis an angle formed from the middle of a circle and is inside a circle. (vertex is the center of a circle) An arcis the part of the actual circle that is inside the angle (the red line in the picture)

  3. Writing Math Minor arcs may be named by TWO points. Major arcs and semicircles must be named by THREE points.

  4. mKLF = 0.65(360) mFJG = 0.20(360) Example 1: Data Application The circle graph shows the types of grass planted in the yards of one neighborhood. Find mKLF and mFJG = 234 = 72

  5. Adjacent arcs are arcs of the same circle that intersect at exactly one point. RS and ST are adjacent arcs.

  6. Find mBD. mBC = 97.4 mCD = 30.6 mBD = mBC + mCD Example 2: Using the Arc Addition Postulate Vert. s Thm. mCFD = 180 – (97.4 + 52) = 30.6 ∆Sum Thm. mCFD = 30.6 Arc Add. Post. = 97.4 + 30.6 Substitute. Simplify. = 128

  7. mJKL mKL = 115° mJKL = mJK + mKL Check It Out! Example 2a Find each measure. mKPL = 180° – (40 + 25)° Arc Add. Post. = 25° + 115° Substitute. = 140° Simplify.

  8. Within a circle or congruent circles, congruent arcs are two arcs that have the same measure. In the figure STUV.

  9. TV WS. Find mWS. TV  WS mTV = mWS mWS = 7(11) + 11 Example 3A: Applying Congruent Angles, Arcs, and Chords  chords have arcs. Def. of arcs 9n – 11 = 7n + 11 Substitute the given measures. 2n = 22 Subtract 7n and add 11 to both sides. n = 11 Divide both sides by 2. Substitute 11 for n. = 88° Simplify.

  10. PT bisects RPS. Find RT. mRT  mTS Check It Out! Example 3a RPT  SPT RT = TS 6x = 20 – 4x 10x = 20 Add 4x to both sides. x = 2 Divide both sides by 10. RT = 6(2) Substitute 2 for x. RT = 12 Simplify.

  11. Step 1 Draw radius PQ. PS QR , so PS bisects QR. Check It Out! Example 4 Find QR to the nearest tenth. PQ = 20 Radii of a are . Step 2 Use the Pythagorean Theorem. TQ2 + PT2 = PQ2 TQ2 + 102 = 202 Substitute 10 for PT and 20 for PQ. TQ2 = 300 Subtract 102 from both sides. TQ 17.3 Take the square root of both sides. Step 3 Find QR. QR= 2(17.3) = 34.6

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