Presentation is loading. Please wait.

Presentation is loading. Please wait.

Daily Check. Homework Review 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.

Similar presentations


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

1 Daily Check

2 Homework Review

3 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 GPS Geometry Day 25 (9-11-13)

4 How do you know when two chords are congruent? LP  PM ALP = BMP = 90 a.corresponding arcs are congruent A B C D M L P b. equidistant from the center

5 Flow Chart Proof

6 2x x + 40 2x = x + 40 x = 40

7 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 TY = 32 U R E

8 A B C D IF AC is the perpendicular bisector of segment DB, then… It’s the DIAMETER!!! Arcs DC and BC are congruent!!!

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

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

11 Your turn!  UTV   XTW. Find WX.___________ Find ___________ 11 130º

12 Your turn! Find the length of each chord. CE = _______ LN = _______ 30 96

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

14

15 a b c d ab = cd

16 9 2 6 x x = 3 Solve for x.

17 Find the length of DB. 8 12 2x 3x x = 4 DB = 20 A B C D

18 Find the length of each chord. x = 8 AC = 13 DB = 14 x 5 x - 4 10 A B C D

19 EAB C D EA EB = EC ED

20 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

21 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

22 E A B C EA 2 = EB EC

23 E A B C 24 12 x 24 2 =12 (12 + x) 576 = 144 + 12x x = 36 Ex: 5 Solve for x.

24 E A B C 15 5 x x2x2 =5 (5 + 15) x 2 = 100 x = 10 Ex: 6

25


Download ppt "Daily Check. Homework Review 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."

Similar presentations


Ads by Google