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How can I find a segment length for a piece of a chord? GPS Geometry Day 25 (9-11-13) UNIT QUESTION: What special properties are found with the parts of a circle? Standard: MMC9-12.G.C.1-5,G.GMD.1-3 Today’s Question: How can I find a segment length for a piece of a chord? Standard: MMC9-12.G.C.2

How do you know when two chords are congruent? corresponding arcs are congruent B A M P L b. equidistant from the center C Discuss that these are “if and only if statements”. Give some basic examples. LP  PM ALP = BMP = 90 D

2x x + 40

In K, K is the midpoint of RE In K, K is the midpoint of RE. If TY = -3x + 56 and US = 4x, find the length of TY. U T x = TY = K E R S Y

IF AC is the perpendicular bisector of segment DB, then… It’s the DIAMETER!!! Arcs DC and BC are congruent!!! A C Discuss “if and only if” and two things must be true to prove the third is true: diameter, perpendicular, bisects. Give some basic examples. B

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

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

Your turn!  ÐUTV  ÐXTW. Find WX.___________ Find ___________

Your turn!  Find the length of each chord. CE = _______ LN = _______

Segment Lengths in Circles Find the lengths of segments of chords Find the lengths of segments of tangents and secants

ab = cd Two chords intersect Type 1: INSIDE the circle a d c b Emphasize that the chords are NOT congruent or bisected!

Solve for x. 9 6 x x = 2

x = DB = Find the length of DB. 12 2x 8 3x A D C B What can we conclude about AC? C B

Find the length of each chord. x = AC = DB = A x - 4 x 5 C 10 B

EA • EB = EC • ED Two secants intersect Type 2: OUTSIDE the circle E A

Ex: 3 Solve for x. B 13 A 7 E 4 C x D 7 (7 + 13) = 4 (4 + x) x = 31 140 = 16 + 4x 124 = 4x

Ex: 4 Solve for x. B x A 5 D 8 6 C E 6 (6 + 8) = 5 (5 + x) 84 = 25 + 5x x = 11.8 59 = 5x

Type 2 (with a twist): Secant and Tangent C B E A EA2 = EB • EC

x = 36 242 = 12 (12 + x) 576 = 144 + 12x x 12 24 Ex: 5 Solve for x. C B x 12 E 24 A 242 = 12 (12 + x) x = 36 576 = 144 + 12x

Ex: 6 5 B E 15 C x A x2 = 5 (5 + 15) x2 = 100 x = 10

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