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Because there are 360° in a circle, multiply each percent by 360 to

Circles and Arcs. LESSON 10-6. Additional Examples. A researcher surveyed 2000 members of a club to find their ages. The graph shows the survey results. Find the measure of each central angle in the circle graph. Because there are 360° in a circle, multiply each percent by 360 to

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Because there are 360° in a circle, multiply each percent by 360 to

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  1. Circles and Arcs LESSON 10-6 Additional Examples A researcher surveyed 2000 members of a club to find their ages. The graph shows the survey results. Find the measure of each central angle in the circle graph. Because there are 360° in a circle, multiply each percent by 360 to find the measure of each central angle. 65+ : 25% of 360 = 0.25 • 360 = 90 45–64: 40% of 260 = 0.4 • 360 = 144 25–44: 27% of 360 = 0.27 • 360 = 97.2 Under 25: 8% of 360 = 0.08 • 360 = 28.8 Quick Check

  2. . Minor arcs are smaller than semicircles. Two minor arcs in the diagram have point A as an endpoint, AD and AE. Major arcs are larger than semicircles. Two major arcs in the diagram have point A as an endpoint, ADE and AED. Two semicircles in the diagram have point A as an endpoint, ADB and AEB. Circles and Arcs LESSON 10-6 Additional Examples Identify the minor arcs, major arcs, and semicircles in P with point A as an endpoint. Quick Check

  3. . Find mXY and mDXM in C. mXY = mXD + mDYArc Addition Postulate mXY = m XCD + mDYThe measure of a minor arc is the measure of its corresponding central angle. mXY = 56 + 40 Substitute. mXY = 96 Simplify. mDXM = mDX + mXWMArc Addition Postulate mDXM = 56 + 180 Substitute. mDXM = 236 Simplify. Circles and Arcs LESSON 10-6 Additional Examples Quick Check

  4. Draw a diagram of the situation. C = dFormula for the circumference of a circle C = (24) Substitute. C 3.14(24) Use 3.14 to approximate . C 75.36 Simplify. Circles and Arcs LESSON 10-6 Additional Examples A circular swimming pool with a 16-ft diameter will be enclosed in a circular fence 4 ft from the pool. What length of fencing material is needed? Round your answer to the nearest whole number. The pool and the fence are concentric circles. The diameter of the pool is 16 ft, so the diameter of the fence is 16 + 4 + 4 = 24 ft. Use the formula for the circumference of a circle to find the length of fencing material needed. Quick Check About 75 ft of fencing material is needed.

  5. . Find the length of ADB in M in terms of . Because mAB = 150, mADB = 360 – 150 = 210. Arc Addition Postulate 210 360 mADB 360 length of ADB = • 2 rArc Length Formula length of ADB = • 2 (18) Substitute. length of ADB = 21 The length of ADB is 21 cm. Circles and Arcs LESSON 10-6 Additional Examples Quick Check

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