1. Splash Screen

2. You used the relationships between arcs and angles to find measures.
Recognize and use relationships between arcs and chords.
Recognize and use relationships between arcs, chords, and diameters.
Then/Now

3. Concept

4. A. 42.5
B. 85 C
D. 170
Example 1

5. A. 42.5
B. 85 C
D. 170
Example 1

6. A. 6
B. 8
C. 9
D. 13
Example 2

7. A. 6
B. 8
C. 9
D. 13
Example 2

8. Concept

9. Use a Radius Perpendicular to a Chord
Answer:
Example 3

10. Use a Radius Perpendicular to a Chord
Answer:
Example 3

11. A. 14
B. 80
C. 160
D. 180
Example 3

12. A. 14
B. 80
C. 160
D. 180
Example 3

13. Use a Diameter Perpendicular to a Chord
CERAMIC TILE In the ceramic stepping stone below, diameter AB is 18 inches long and chord EF is 8 inches long. Find CD.
Example 4

14. Step 1 Draw radius CE. This forms right ΔCDE.
Use a Diameter Perpendicular to a Chord
Step 1 Draw radius CE.
This forms right ΔCDE.
Example 4

15. Use a Diameter Perpendicular to a Chord
Step 2 Find CE and DE.
Since AB = 18 inches, CB = 9 inches. All radii of a circle are congruent, so CE = 9 inches.
Since diameter AB is perpendicular to EF, AB bisects chord EF by Theorem So, DE = (8) or 4 inches. __ 1 2
Example 4

16. Step 3 Use the Pythagorean Theorem to find CD.
Use a Diameter Perpendicular to a Chord
Step 3 Use the Pythagorean Theorem to find CD.
CD2 + DE2 = CE2 Pythagorean Theorem
CD = 92 Substitution
CD = 81 Simplify.
CD2 = 65 Subtract 16 from each side.
Take the positive square root.
Answer:
Example 4

17. Step 3 Use the Pythagorean Theorem to find CD.
Use a Diameter Perpendicular to a Chord
Step 3 Use the Pythagorean Theorem to find CD.
CD2 + DE2 = CE2 Pythagorean Theorem
CD = 92 Substitution
CD = 81 Simplify.
CD2 = 65 Subtract 16 from each side.
Take the positive square root.
Answer:
Example 4

18. In the circle below, diameter QS is 14 inches long and chord RT is 10 inches long. Find VU.
Example 4

19. In the circle below, diameter QS is 14 inches long and chord RT is 10 inches long. Find VU.
Example 4

20. Concept

21. Chords Equidistant from Center
Since chords EF and GH are congruent, they are equidistant from P. So, PQ = PR.
Example 5

22. PQ = PR 4x – 3 = 2x + 3 Substitution x = 3 Simplify.
Chords Equidistant from Center
PQ = PR
4x – 3 = 2x + 3 Substitution
x = 3 Simplify.
So, PQ = 4(3) – 3 or 9
Answer:
Example 5

23. PQ = PR 4x – 3 = 2x + 3 Substitution x = 3 Simplify.
Chords Equidistant from Center
PQ = PR
4x – 3 = 2x + 3 Substitution
x = 3 Simplify.
So, PQ = 4(3) – 3 or 9
Answer: PQ = 9
Example 5

24. A. 7
B. 10
C. 13
D. 15
Example 5

25. A. 7
B. 10
C. 13
D. 15
Example 5

26. End of the Lesson