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No Slide Title 9.2 Polygons Copyright © 2005 Pearson Education, Inc. Polygons Polygons are names according to their number of sides. Number of Name Number of Name Sides Sides 3 Triangle 8 Octagon 4 Quadrilateral 9 Nonagon 5 Pentagon 10 Decagon 6 Hexagon 12 Dodecagon 7 Heptagon 20 Icosagon Copyright © 2005 Pearson Education, Inc. Slide 9-2 Triangles The sum of the measures of the interior angles of an n-sided polygon is (n − 2)180°. Example: A certain brick paver is in the shape of a regular octagon. Determine the measure of an interior angle and the measure of one exterior angle. Copyright © 2005 Pearson Education, Inc. Slide 9-3 Triangles continued Determine the sum of the The measure of one interior angles. interior angle is o 1080 o o = 135 Sn=−(2)180 8 o =−(8 2)(180 ) The exterior angle is = 6(180o ) supplementary to the interior angle, so = 1080o 180°−135° = 45° Copyright © 2005 Pearson Education, Inc. Slide 9-4 Types of Triangles Copyright © 2005 Pearson Education, Inc. Slide 9-5 Similar Figures Two polygons are similar if their corresponding angles have the same measure and their corresponding sides are in proportion. 9 6 4 4 66 3 4.5 Copyright © 2005 Pearson Education, Inc. Slide 9-6 Example Catherine Johnson wants to measure the height of a lighthouse. Catherine is 5 feet tall and determines that when her shadow is 12 feet long, the shadow of the lighthouse is 75 feet long. How tall is the lighthouse? x 5 12 75 Copyright © 2005 Pearson Education, Inc. Slide 9-7 Example continued ht. lighthouse lighthouse's shadow = ht. Catherine Catherine's shadow x75 = 512 12x = 375 x = 31.25 x 5 12 75 Therefore, the lighthouse is 31.25 feet tall. Copyright © 2005 Pearson Education, Inc. Slide 9-8 Congruent Figures If corresponding sides of two similar figures are the same length, the figures are congruent. Corresponding angles of congruent figures have the same measure. Copyright © 2005 Pearson Education, Inc. Slide 9-9 Quadrilaterals Quadrilaterals are four-sided polygons, the sum of whose interior angles is 360°. Quadrilaterals may be classified according to their characteristics. Copyright © 2005 Pearson Education, Inc. Slide 9-10 Classifications Trapezoid Parallelogram Both pairs of opposite Two sides are parallel. sides are parallel. Both pairs of opposite sides are equal in length. Copyright © 2005 Pearson Education, Inc. Slide 9-11 Classifications continued Rhombus Rectangle Both pairs of opposite sides Both pairs of opposite sides are parallel. The four sides are are parallel. Both pairs of equal in length. opposite sides are equal in length. The angles are right angles. Copyright © 2005 Pearson Education, Inc. Slide 9-12 Classifications continued Square Both pairs of opposite sides are parallel. The four sides are equal in length. The angles are right angles. Copyright © 2005 Pearson Education, Inc. Slide 9-13 9.3 Perimeter and Area Copyright © 2005 Pearson Education, Inc. Formulas Figure Perimeter Area Rectangle P = 2l + 2w A = lw Square P = 4s A = s2 Parallelogram P = 2b + 2w A = bh Triangle P = s + s + s 1 1 2 3 A = 2 bh P = s + s + b + b 1 Trapezoid 1 2 1 2 Ah= 2 ()b12+ b Copyright © 2005 Pearson Education, Inc. Slide 9-15 Example Marcus Sanderson needs to put a new roof on his barn. One square of roofing covers 100 ft2 and costs $32.00 per square. If one side of the barn roof measures 50 feet by 30 feet, determine a) the area of the entire roof. b) how many squares of roofing he needs. c) the cost of putting on the roof. Copyright © 2005 Pearson Education, Inc. Slide 9-16 Example continued a) The area of the roof is A = lw A = 30(50) A = 1500 ft2 1500(2 both sides of the roof) = 3000 ft2 b) Determine the number of squares area of roof 3000 ==30 area of one square 100 Copyright © 2005 Pearson Education, Inc. Slide 9-17 Example continued c) Determine the cost 30 squares × $32 per square $960 It will cost a total of $960 to roof the barn. Copyright © 2005 Pearson Education, Inc. Slide 9-18 Pythagorean Theorem Copyright © 2005 Pearson Education, Inc. Slide 9-19 Example Tomas is bringing his boat into a dock that is 12 feet above the water level. If a 38 foot rope is attached to the dock on one side and to the boat on the other side, determine the horizontal distance from the dock to the boat. 38 ft rope 12 ft Copyright © 2005 Pearson Education, Inc. Slide 9-20 Example continued abc22+= 2 1222+=b 38 3 144+=b2 1444 b2 = 1300 b2 = 1300 b ≈ 36.06 The distance is approximately 36.06 feet. Copyright © 2005 Pearson Education, Inc. Slide 9-21 Circles A circle is a set of points equidistant from a fixed point called the center. A radius, r, of a circle is a line segment from the center of the circle to any point on the circle. A diameter, d, of a circle is a line segment through the center of the circle with both end points on the circle. Copyright © 2005 Pearson Education, Inc. Slide 9-22 Example Terri is installing a new circular swimming pool in her backyard. The pool has a diameter of 27 feet. How much area will the pool take up in her yard? Ar= π 2 The radius of the pool is 13.5 feet. A = π(13.5)2 The pool will take up about A = 572.54 573 square feet. Copyright © 2005 Pearson Education, Inc. Slide 9-23 Next Steps Study Examples 1,3,4 in 9.2 Study Examples 1-6 in 9.3 Do Online homework Do Online quiz once you’ve studied both sections Copyright © 2005 Pearson Education, Inc. Slide 9-24.
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