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C hapter 11 Area of Polygons and Circles

C hapter 11 Area of Polygons and Circles. Chapter 11 Section 11.5 Areas of Circles and Sectors. Objectives. Bell Problem #. A REAS OF C IRCLES AND S ECTORS.

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C hapter 11 Area of Polygons and Circles

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  1. Chapter 11Area of Polygons and Circles

  2. Chapter 11 Section 11.5Areas of Circles and Sectors

  3. Objectives

  4. Bell Problem #

  5. AREAS OF CIRCLES AND SECTORS These regular polygons, inscribed in circles with radius r, demonstrate that as the number of sides increases, the area of the polygon approaches the value r2. 3-gon 4-gon 5-gon 6-gon

  6. AREAS OF CIRCLES AND SECTORS r THEOREM THEOREM 11.7 Area of a Circle The area of a circle is times the square of the radius, or A=r2

  7. Using the Area of a Circle . Find the area of P. 8 in. P SOLUTION Use r = 8 in the area formula. A = r2 = • 82 = 64  201.06 So, the area is 64, or about 201.06, square inches.

  8. Using the Area of a Circle Find the diameter of Z . • Z Area of Z = 96 cm2 • 96  = r2 SOLUTION The diameter is twice the radius. A = r2 96 = r2 30.56 r2 5.53 r Find the square roots. The diameter of the circle is about 2(5.53), or about 11.06, centimeters.

  9. Using the Area of a Circle In the diagram, sector APB is bounded by AP, BP, and AB. P The sector of a circle is the region bounded by two radii of the circle and their intercepted arc. A r B

  10. Using the Area of a Circle THEOREM A A P B mAB 360° mAB 360° A r2 • r2 , or A = = The following theorem gives a method for finding the area of a sector. THEOREM 11.8 Area of a Sector The ratio of the area A of a sector of a circle to the area of the circle is equal to the ratio of the measure of the intercepted arc to 360°.

  11. Finding the Area of a Sector C 4 ft 80° P D • r2 A = • •42 = m CD 360° 80° 360° Find the area of the sector shown at the right. SOLUTION Sector CPD intercepts an arc whose measure is 80°. The radius is 4 feet. Write the formula for the area of a sector. Substitute known values.  11.17 Use a calculator. So, the area of the sector is about 11.17 square feet.

  12. Finding the Area of a Region 5 m Area of shaded region Area of circle Area of hexagon = – USING AREAS OF CIRCLES AND REGIONS Find the area of a the shaded region shown. The diagram shows a regular hexagon inscribed in a circle with radius 5 meters. The shaded region is the part of the circle that is outside of the hexagon. SOLUTION

  13. Finding the Area of a Region r2 = – 5 m Area of shaded region Area of circle Area of hexagon = – • 52 – = The apothem of a hexagon is • side length • 1 2 • (6• 5) • 1 2 aP 75 2 25 – = 1 2 3 75 2 or about 13.59 square meters. So, the area of the shaded region is 25 – 5 2 , 3 3 3 USING AREAS OF CIRCLES AND REGIONS

  14. Finding the Area of a Region P P Complicated shapes may involve a number of regions. Notice that the area of a portion of the ring is the difference of the areas of two sectors.

  15. Finding the Area of a Region Area of rectangle Area of sector Area WOODWORKINGYou are cutting the front face of a clock out of wood, as shown in the diagram. What is the area of the front of the case? SOLUTION The front of the case is formed by a rectangle and a sector, with a circle removed. Note that the intercepted arc of the sector is a semicircle. = + – Area of circle

  16. Finding the Area of a Region 1 2 112 180°360° •• 32 6• • 4 2 • = + – Area of rectangle Area of sector Area of circle Area = + – 1 2 9 2 33 + • •9 – • (2)2 = 33 + – 4 = WOODWORKINGYou are cutting the front face of a clock out of wood, as shown in the diagram. What is the area of the front of the case? 34.57 The area of the front of the case is about 34.57 square inches.

  17. Homework p695 (10-20 all)

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