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Daily Check.

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Presentation on theme: "Daily Check."— Presentation transcript:

1 Daily Check

2 Homework Review

3 How can I find a segment length for a piece of a chord?
GPS Geometry Day 25 ( ) 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

4 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

5 2x x + 40

6 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

7 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

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

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

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

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

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

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

14 Solve for x. 9 6 x x = 2

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

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

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

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

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

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

21 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 = x

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

23 Homework Practice Worksheet


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