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Published byEdwin Bond Modified over 8 years ago
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Circle Segments and Volume
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Chords of Circles Theorem 1
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In the same circle, or in congruent circles two minor arcs are congruent if and only if their corresponding chords are congruent.
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Chord Arcs Conjecture In the same circle, two minor arcs are congruent if and only if their corresponding chords are congruent. IFF and G
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8x – 7 3x + 3 8x – 7 = 3x + 3 Solve for x. x = 2
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Find WX. Example
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Find 130º Example
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Chords of Circles Theorem 2
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If a diameter is perpendicular to a chord, then it also bisects the chord and its arc. This results in congruent arcs too. Sometimes, this creates a right triangle & you’ll use Pythagorean Theorem.
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Perpendicular Bisector of a Chord Conjecture If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc. H
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IN Q, KL LZ. If CK = 2x + 3 and CZ = 4x, find x. K Q C L Z x = 1.5
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In P, if PM AT, PT = 10, and PM = 8, find AT. T A M P MT = 6 AT = 12
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Chords of Circles Theorem 3
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Perpendicular Bisector to a Chord Conjecture If one chord is a perpendicular bisector of another chord, then the bisecting chord is a diameter. is a diameter of the circle.
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If one chord is a perpendicular bisector of another chord, then the first chord is a diameter. ,
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Chords of Circles Theorem 4
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In the same circle or in congruent circles two chords are congruent when they are equidistant from the center.
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Chord Distance to the Center Conjecture
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In K, K is the midpoint of RE. If TY = -3x + 56 and US = 4x, find the length of TY. Y T S K x = 8 U R E TY = 32
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Example CE = 30
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Example LN = 96
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Segment Lengths in Circles
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part Go down the chord and multiply
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9 2 6 x x = 3 Solve for x.
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Find the length of DB. 8 12 2x 3x x = 4 A B C D DB = 20
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Find the length of AC and DB. x = 8 x 5 x – 4 10 A B C D AC = 13 DB = 14
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E A B C D 7 13 4 x 7(7 + 13) 4(4 + x) = Ex: 3 Solve for x. 140 = 16 + 4x 124 = 4x x = 31
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E A B C D 8 5 6 x 6(6 + 8) 5(5 + x) = Ex: 4 Solve for x. 84 = 25 + 5x 59 = 5x x = 11.8
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E A B C D 4 x 8 10 x(x + 10) 8 (8 + 4) = Ex: 5 Solve for x. x 2 +10x = 96 x 2 +10x – 96 = 0 x = 6
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24 12 x 24 2 =12 (12 + x) 576 = 144 + 12x x = 36 Ex: 5 Solve for x.
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15 5 x x2x2 =5 (5 + 15) x 2 = 100 x = 10 Ex: 6
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