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Presentation transcript:

Splash Screen

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

Concept

A. 42.5 B. 85 C. 127.5 D. 170 Example 1

A. 42.5 B. 85 C. 127.5 D. 170 Example 1

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

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

Concept

Use a Radius Perpendicular to a Chord Answer: Example 3

Use a Radius Perpendicular to a Chord Answer: Example 3

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

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

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

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

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 10.3. So, DE = (8) or 4 inches. __ 1 2 Example 4

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 CD2 + 42 = 92 Substitution CD2 + 16 = 81 Simplify. CD2 = 65 Subtract 16 from each side. Take the positive square root. Answer: Example 4

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 CD2 + 42 = 92 Substitution CD2 + 16 = 81 Simplify. CD2 = 65 Subtract 16 from each side. Take the positive square root. Answer: Example 4

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

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

Concept

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

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

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

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

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

End of the Lesson