Circle Segments and Volume

Chords of Circles Theorem 1

In the same circle, or in congruent circles two minor arcs are congruent if and only if their corresponding chords are congruent.

Chord Arcs Conjecture In the same circle, two minor arcs are congruent if and only if their corresponding chords are congruent. IFF and G

8x – 7 3x + 3 8x – 7 = 3x + 3 Solve for x. x = 2

Find WX. Example

Find 130º Example

Chords of Circles Theorem 2

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.

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

IN  Q, KL  LZ. If CK = 2x + 3 and CZ = 4x, find x. K Q C L Z x = 1.5

In  P, if PM  AT, PT = 10, and PM = 8, find AT. T A M P MT = 6 AT = 12

Chords of Circles Theorem 3

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.

If one chord is a perpendicular bisector of another chord, then the first chord is a diameter. , 

Chords of Circles Theorem 4

In the same circle or in congruent circles two chords are congruent when they are equidistant from the center.

Chord Distance to the Center Conjecture

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

Example CE = 30

Example LN = 96

Segment Lengths in Circles

part Go down the chord and multiply

9 2 6 x x = 3 Solve for x.

Find the length of DB x 3x x = 4 A B C D DB = 20

Find the length of AC and DB. x = 8 x 5 x – 4 10 A B C D AC = 13 DB = 14

E A B C D x 7(7 + 13) 4(4 + x) = Ex: 3 Solve for x. 140 = x 124 = 4x x = 31

E A B C D x 6(6 + 8) 5(5 + x) = Ex: 4 Solve for x. 84 = x 59 = 5x x = 11.8

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

24 12 x 24 2 =12 (12 + x) 576 = x x = 36 Ex: 5 Solve for x.

15 5 x x2x2 =5 (5 + 15) x 2 = 100 x = 10 Ex: 6