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Find the Area. 10-5: Area of Regular Polygons p. 543-550 Primary: M(G&M)–10–6 Solves problems involving perimeter, circumference, or area of two-dimensional.

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Presentation on theme: "Find the Area. 10-5: Area of Regular Polygons p. 543-550 Primary: M(G&M)–10–6 Solves problems involving perimeter, circumference, or area of two-dimensional."— Presentation transcript:

1 Find the Area

2 10-5: Area of Regular Polygons p. 543-550 Primary: M(G&M)–10–6 Solves problems involving perimeter, circumference, or area of two-dimensional figures (including composite figures) or surface area or volume of three-dimensional figures (including composite figures) within mathematics or across disciplines or contexts. Secondary: GSE’s M(G&M)–10–9 Solves problems on and off the coordinate plane involving distance, midpoint, perpendicular and parallel lines, or slope.

3 Area of any Regular Polygon = perimeter of the polygon a = apothem (the segment from the center of the polygon to the side (where it is perpendicular) a **All regular polygons can be inscribed within a circle a = apothem

4 Find the area of the regular polygon described. A triangle with a side length of 14 inches

5 A regular hexagon with a perimeter of 36 meters

6 Ex: Find the area of the shaded region. 4 c m  r 2 4 c m 30o 2 2ð3 4ð3  (4) 2 16  16 - 12ð3 = 29.5 cm2 http://guilford.rps205.com/departments/Math/Links/Honors%20Geometry/Honors%20Geometry%20Power%20Points/11.2%20%20Area%20of%20reg%20polygons.ppt#7

7 Ex: A regular octagon has a radius of 4 in. Find its area. First, we have to find the apothem length. 4sin67.5 = a 3.7 = a Now, the side length. Side length=2(1.53)=3.06 4 a 135 o 67.5 o 3.7 x 4cos67.5 = x 1.53 = x A = ½ Pa = ½ (24.48)(3.7) = 45.288 in 2 http:// guilford.rps205.com/departments/Math/Links/Honors%20Geometry/Honors%20Geometry%20Power%20Points/11.2%20%20Area%20of%20reg%20polygons.ppt#7

8 Homework


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