Geometry Section 10-3 1112
Upcoming SlideShare
Loading in...5
×
 

Geometry Section 10-3 1112

on

  • 746 views

Arcs and Chords

Arcs and Chords

Statistics

Views

Total Views
746
Views on SlideShare
573
Embed Views
173

Actions

Likes
0
Downloads
5
Comments
0

1 Embed 173

http://mrlambmath.wikispaces.com 173

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Geometry Section 10-3 1112 Geometry Section 10-3 1112 Presentation Transcript

  • Section 10-3 Arcs and ChordsMonday, May 14, 2012
  • Essential Questions • How do you recognize and use relationships between arcs and chords? • How do you recognize and use relationships between arcs, chords, and diameters?Monday, May 14, 2012
  • Theorems 10.2 - Congruent Minor Arcs: 10.3 - Perpendicularity: 10.4 - Perpendicularity: 10.5 - Congruent Chords:Monday, May 14, 2012
  • Theorems 10.2 - Congruent Minor Arcs: In the same or congruent circles, two minor arcs are congruent IFF their corresponding chords are congruent 10.3 - Perpendicularity: 10.4 - Perpendicularity: 10.5 - Congruent Chords:Monday, May 14, 2012
  • Theorems 10.2 - Congruent Minor Arcs: In the same or congruent circles, two minor arcs are congruent IFF their corresponding chords are congruent 10.3 - Perpendicularity: If a diameter or radius of a circle is perpendicular to a chord, then it bisects the chord and its arc 10.4 - Perpendicularity: 10.5 - Congruent Chords:Monday, May 14, 2012
  • Theorems 10.2 - Congruent Minor Arcs: In the same or congruent circles, two minor arcs are congruent IFF their corresponding chords are congruent 10.3 - Perpendicularity: If a diameter or radius of a circle is perpendicular to a chord, then it bisects the chord and its arc 10.4 - Perpendicularity: The perpendicular bisector of a chord is a diameter or radius of the circle 10.5 - Congruent Chords:Monday, May 14, 2012
  • Theorems 10.2 - Congruent Minor Arcs: In the same or congruent circles, two minor arcs are congruent IFF their corresponding chords are congruent 10.3 - Perpendicularity: If a diameter or radius of a circle is perpendicular to a chord, then it bisects the chord and its arc 10.4 - Perpendicularity: The perpendicular bisector of a chord is a diameter or radius of the circle 10.5 - Congruent Chords: In the same or congruent circles, two chords are congruent IFF they are equidistant from the centerMonday, May 14, 2012
  • Example 1  = 90°. Find mAB. In  X, AB ≅ CD and mCD Monday, May 14, 2012
  • Example 1  = 90°. Find mAB. In  X, AB ≅ CD and mCD   = 90° mABMonday, May 14, 2012
  • Example 2  ≅ YZ . Find WX. In the figure,  A ≅ B and WX Monday, May 14, 2012
  • Example 2  ≅ YZ . Find WX. In the figure,  A ≅ B and WX  7x − 2 = 5x + 6Monday, May 14, 2012
  • Example 2  ≅ YZ . Find WX. In the figure,  A ≅ B and WX  7x − 2 = 5x + 6 2x = 8Monday, May 14, 2012
  • Example 2  ≅ YZ . Find WX. In the figure,  A ≅ B and WX  7x − 2 = 5x + 6 2x = 8 x=4Monday, May 14, 2012
  • Example 2  ≅ YZ . Find WX. In the figure,  A ≅ B and WX  7x − 2 = 5x + 6 2x = 8 x=4 WX = 7(4) − 2Monday, May 14, 2012
  • Example 2  ≅ YZ . Find WX. In the figure,  A ≅ B and WX  7x − 2 = 5x + 6 2x = 8 x=4 WX = 7(4) − 2 WX = 28 − 2Monday, May 14, 2012
  • Example 2  ≅ YZ . Find WX. In the figure,  A ≅ B and WX  7x − 2 = 5x + 6 2x = 8 x=4 WX = 7(4) − 2 WX = 28 − 2 WX = 26Monday, May 14, 2012
  • Example 3  =150°. Find mDE. In  G, mDEF Monday, May 14, 2012
  • Example 3  =150°. Find mDE. In  G, mDEF   = 1 mDEF mDE  2Monday, May 14, 2012
  • Example 3  =150°. Find mDE. In  G, mDEF   = 1 mDEF mDE  2  = 1 (150) mDE 2Monday, May 14, 2012
  • Example 3  =150°. Find mDE. In  G, mDEF   = 1 mDEF mDE  2  = 1 (150) mDE 2  = 75° mDEMonday, May 14, 2012
  • Example 4 In  C, AB =18 inches and EF = 8 inches. Find CD.Monday, May 14, 2012
  • Example 4 In  C, AB =18 inches and EF = 8 inches. Find CD. CF is a radius.Monday, May 14, 2012
  • Example 4 In  C, AB =18 inches and EF = 8 inches. Find CD. CF is a radius. a +b =c 2 2 2Monday, May 14, 2012
  • Example 4 In  C, AB =18 inches and EF = 8 inches. Find CD. CF is a radius. a +b =c 2 2 2 4 +b =9 2 2 2Monday, May 14, 2012
  • Example 4 In  C, AB =18 inches and EF = 8 inches. Find CD. CF is a radius. a +b =c 2 2 2 4 +b =9 2 2 2 16 + b = 81 2Monday, May 14, 2012
  • Example 4 In  C, AB =18 inches and EF = 8 inches. Find CD. CF is a radius. a +b =c 2 2 2 4 +b =9 2 2 2 16 + b = 81 2 b = 65 2Monday, May 14, 2012
  • Example 4 In  C, AB =18 inches and EF = 8 inches. Find CD. CF is a radius. a +b =c 2 2 2 4 +b =9 2 2 2 16 + b = 81 2 b = 65 2 b = 65Monday, May 14, 2012
  • Example 4 In  C, AB =18 inches and EF = 8 inches. Find CD. CF is a radius. a +b =c 2 2 2 4 +b =9 2 2 2 16 + b = 81 2 b = 65 2 b = 65 inches or ≈ 8.06 inchesMonday, May 14, 2012
  • Example 5 In  P, EF = GH = 24, PQ = 4x − 3, and PR = 2x + 3. Find PQ.Monday, May 14, 2012
  • Example 5 In  P, EF = GH = 24, PQ = 4x − 3, and PR = 2x + 3. Find PQ. 4x − 3 = 2x + 3Monday, May 14, 2012
  • Example 5 In  P, EF = GH = 24, PQ = 4x − 3, and PR = 2x + 3. Find PQ. 4x − 3 = 2x + 3 2x = 6Monday, May 14, 2012
  • Example 5 In  P, EF = GH = 24, PQ = 4x − 3, and PR = 2x + 3. Find PQ. 4x − 3 = 2x + 3 2x = 6 x=3Monday, May 14, 2012
  • Example 5 In  P, EF = GH = 24, PQ = 4x − 3, and PR = 2x + 3. Find PQ. 4x − 3 = 2x + 3 2x = 6 x=3 PQ = 4(3) − 3Monday, May 14, 2012
  • Example 5 In  P, EF = GH = 24, PQ = 4x − 3, and PR = 2x + 3. Find PQ. 4x − 3 = 2x + 3 2x = 6 x=3 PQ = 4(3) − 3 PQ = 12 − 3Monday, May 14, 2012
  • Example 5 In  P, EF = GH = 24, PQ = 4x − 3, and PR = 2x + 3. Find PQ. 4x − 3 = 2x + 3 2x = 6 x=3 PQ = 4(3) − 3 PQ = 12 − 3 PQ = 9Monday, May 14, 2012
  • Check Your Understanding p. 704 #1 - 6Monday, May 14, 2012
  • Problem SetMonday, May 14, 2012
  • Problem Set p. 705 #7-33 odd, 45, 49, 51 "I may not have gone where I intended to go, but I think I have ended up where I needed to be." - Douglas AdamsMonday, May 14, 2012