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11-3 Areas of Regular Polygons and Circles

11-3 Areas of Regular Polygons and Circles. Objectives. Find areas of regular polygons Find areas of circles. Find Areas of Regular Polygons. A Definition…. Apothem – a segment that is drawn from the center of a regular polygon perpendicular to the side of the polygon.

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11-3 Areas of Regular Polygons and Circles

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  1. 11-3Areas of Regular Polygons and Circles

  2. Objectives • Find areas of regular polygons • Find areas of circles

  3. Find Areas of Regular Polygons

  4. A Definition… • Apothem – a segment that is drawn • from the center of a regular polygon • perpendicular to the side of the polygon. • If all radii were drawn, there would be 6 congruent isosceles triangles. • An area of a hexagon is determined by adding areas of the triangles inside. A B H F G C E D

  5. A Formula… Area of a triangle = ½bh = ½sa If the area of a triangle = ½sa, then the area of a hexagon = 6(½sa)

  6. Example #1… Find the area of a regular pentagon with a perimeter of 40 cm. K P J L P N Q M

  7. We first need to find the apothem. Since it is regular, all of the central angles are congruent, or equal to 360/5 or 72°. That makes angle NPQ equal to 36°. Side NM = 8 since the perimeter is 40, so NQ = 4. Now, we can solve the problem: tan NPQ = QN / PQ tan36 = 4 / PQ PQ = 4 / TAN 36 PQ = 5.5 Area = ½ Pa or ½ (40)(5.5) = 110 So the area of the pentagon is 110 cm. K P J L P N Q M

  8. Find Areas of Circles

  9. Area of a Circle… Area of a circle = r²

  10. Example #2… A caterer has a 48-inch diameter table that is 34 inches tall. She wants a tablecloth that will touch the floor. Find the area of the tablecloth. 34 + 48 + 34 = ll6 divided by 2 = 58 so the radius = 58. A = pi(58)² = 10,568 inches2 48 in 34in

  11. Example #3… Find the area of the shaded region. Assume that the triangle is equilateral and the radius of the circle is 4. 4m

  12. A = pi * r² A = pi(4)² = 50.3  Area of Circle Construct ΔABC, a right Δ . Use 30-60-90 rules to find the lengths of the sides and then to find the height of the equilateral triangle. AB = 2 BC = 2√3 DB = 2√3(√3) = 6  Height of ΔCDE One side of ΔCDE is equal to 2(2√3) = 4√3 So, A = ½ (4√3)(6) = 20.8  Area of ΔCDE Now, area of shaded region is 50.3 – 20.8 which is equal to 29.5 m². D A E B C 4m

  13. Assignment… • Pre-AP Pg. 613 #8 – 22, 24, 30, 32, 39 - 44 • Geometry Pg. 613 #8 – 22, 24, 30, 32, 39 - 41

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