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Pages 382-385 Exercises. 1. m 1 = 120; m 2 = 60; m 3 = 30 2. m 4 = 90; m 5 = 45; m 6 = 45 3. m 7 = 60; m 8 = 30; m 9 = 60 4. 2144.475 cm 2 5. 2851.8 ft 2. 14. 384 3 in. 2 15. 300 3 ft 2 16. 162 3 m 2 17. 75 3 m 2 18. 12 3 in. 2
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Pages 382-385 Exercises 1.m 1 = 120; m 2 = 60; m 3 = 30 2.m 4 = 90; m 5 = 45; m 6 = 45 3.m 7 = 60; m 8 = 30; m 9 = 60 4. 2144.475 cm2 5. 2851.8 ft2 14. 384 3 in.2 15. 300 3 ft2 16. 162 3 m2 17. 75 3 m2 18. 12 3 in.2 19. a. 72 b. 54 20. a. 45 b. 67.5 GEOMETRY LESSON 7-5 6. 12,080 in.2 7. 2475 in.2 8. 1168.5 m2 9. 2192.4 cm2 10. 841.8 ft2 11. 27.7 in.2 12. 93.5 m2 13. 72 cm2
24. (continued) d. Answers may vary. Sample: About 4 in.; the length of a side of a pentagon should be between 3.7 in. and 6 in. 25.m 1 = 36; m 2 = 18; m 3 = 72 26. The apothem is one leg of a rt. and the radius is the hypotenuse. 27. 73 cm2 28. 130 in.2 29. 27 m2 30. 103 ft2 31. 220 cm2 32. a–c. regular octagon GEOMETRY LESSON 7-5 21. a. 40 b. 70 22. a. 30 b. 75 23. 310.4 ft2 24. a. 9.1 in. b. 6 in. c. 3.7 in.
32. (continued) d. Construct a 60° angle with vertex at circle’s center. 33. 600 3 m2 34. Check students’ work. 35. 128 cm2 36. 24 3 cm2, 41.6 cm2 37. 900 3 m2, 1558.8 m2 38. 100 ft2 39. 16 3 in.2, 27.7 in.2 40. m2, 70.1 m2 41. a.b = s; h = s A = bh A = s • s A = s2 3 41. (continued) b. apothem = ; A = ap = (3s) A = s2 3 1 2 1 2 1 2 81 3 2 s 3 6 s 3 6 1 2 1 2 3 2 3 2 GEOMETRY LESSON 7-5
7.6 Circles and Arcs Objective: To find the measures of central angles and arcs To find the circumference and arc length
Definitions • Circle – the set of all points equidistant from a given point called the center • CenterofaCircle – the point from which all points are equidistant • Radius – a segment that has one endpoint at the center and the other endpoint on the circle
Definitions • CongruentCircles – circles that have congruent radii • Diameter – a segment that contains the center of a circle and has both endpoints on the circle • CentralAngle – an angle whose vertex is the center of the circle
Definitions • Circumference – the distance around the circle • Pi (∏) – the ration of the circumference of a circle to its diameter
Examples • What if we are given a pie chart that represents data that have been collected? • How can we find the measure of the arc or the measure of the angle?
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. Circles and Arcs GEOMETRY LESSON 7-6 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
132 1320 Circles and Arcs • Some info to really help • The measure of the arc is the same as the measure of the central angle which creates that arc
R R T S P T S P R T S P Arcs A minor arc is smaller than a semicircle A major arc is greater than a semicircle A semicircle is half of a circle
. Identify the minor arcs, major arcs, and semicircles in P with point A as an endpoint. 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 GEOMETRY LESSON 7-6
Arcs • AdjacentArcs – arcs of the same circle that have exactly one point in common • CongruentArcs – arcs that have the same measure and are in the same circle or in congruent circles • ConcentricCircles – circles that lie in the same plane and have the same center • Arclength – a fraction of a circle’s circumference
Postulate 7-1: Arc Addition Postulate • The measure of the arc formed by two adjacent arcs is the sum of the measures of the two arcs. • mABC = mAB + mBC
. 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 GEOMETRY LESSON 7-6
Circumference • The circumference of a circle is the product of pi (∏) and the diameter. • C = ∏d or C = 2∏r
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 next whole number. 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 GEOMETRY LESSON 7-6 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. About 76 ft of fencing material is needed. 7-6
Arc Length • The length of an arc of a circle is the product of the ratio measure of the arc 360 and the circumference of the circle • Length of AB = mAB/360 *2∏r
. Find the length of ADB in M in terms of . Because mAB = 150, mADB = 360 – 150 = 210. Arc Addition Postulate 210 360 length of ADB = • 2 (18) Substitute. mADB 360 length of ADB = • 2 rArc Length Formula The length of ADB is 21 cm. length of ADB = 21 Circles and Arcs GEOMETRY LESSON 7-6 7-6
1. A circle graph has a section marked “Potatoes: 28%.” What is the measure of the central angle of this section? 2. Explain how a major arc differs from a minor arc. Use O for Exercises 3–6. 3. Find mYW. 4. Find mWXS. 5. Suppose that P has a diameter 2 in. greater than the diameter of O. How much greater is its circumference? Leave your answer in terms of . 6. Find the length of XY. Leave your answer in terms of . . . . 2 9 Circles and Arcs GEOMETRY LESSON 7-6 100.8 A major arc is greater than a semicircle. A minor arc is smaller than a semicircle. 30 270
Assignment • P. 389-390 • #1-32 odd, 34-39