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11.3 Perimeters and Areas of Similar Polygons Geometry Mrs. Spitz Spring 2006

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Objectives/Assignment Compare perimeters and areas of similar figures. Use perimeters and areas of similar figures to solve real-life problems. Assignment: pp. 679-680 #1-27 all

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Comparing Perimeter and Area For any polygon, the perimeter of the polygon is the sum of the lengths of its sides and the area of the polygon is the number of square units contained in its interior.

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Comparing Perimeter and Area In lesson 8.3, you learned that if two polygons are similar, then the ratio of their perimeters is the same as the ratio of the lengths of their corresponding sides. In activity 11.3 on pg. 676, you may have discovered that the ratio of the areas of two similar polygons is NOT this same ratio.

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Thm 11.5 – Areas of Similar Polygons If two polygons are similar with the lengths of corresponding sides in the ratio of a:b, then the ratio of their areas is a 2 :b 2

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Thm 11.5 continued Side length of Quad I Side length of Quad II = a b Area of Quad I Area of Quad II = a2a2 b2b2 III ka kb Quad I ~ Quad II

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Ex. 1: Finding Ratios of Similar Polygons Pentagons ABCDE and LMNPQ are similar. a.Find the ratio (red to blue) of the perimeters of the pentagons. b.Find the ratio (red to blue) of the areas of the pentagons 5 10

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Ex. 1: Solution a.Find the ratio (red to blue) of the perimeters of the pentagons. The ratios of the lengths of corresponding sides in the pentagons is 5:10 or ½ or 1:2. The ratio is 1:2. So, the perimeter of pentagon ABCDE is half the perimeter of pentagon LMNPQ. 5 10

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Ex. 1: Solution b.Find the ratio (red to blue) of the areas of the pentagons. Using Theorem 11.5, the ratio of the areas is 1 2 : 2 2. Or, 1:4. So, the area of pentagon ABCDE is one fourth the area of pentagon LMNPQ. 5 10

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Using perimeter and area in real life Ex. 2: Using Areas of Similar Figures Because the ratio of the lengths of the sides of to rectangular pieces is 1:2, the ratio of the areas of the pieces of paper is 1 2 : 2 2 or, 1:4

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Using perimeter and area in real life Ex. 2: Using Areas of Similar Figures Because the cost of the paper should be a function of its area, the larger piece of paper should cost about 4 times as much, or $1.68.

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Using perimeter and area in real life Ex. 3: Finding Perimeters and Areas of Similar Polygons Octagonal Floors. A trading pit at the Chicago Board of Trade is in the shape of a series of octagons. One octagon has a side length of about 14.25 feet and an area of about 980.4 square feet. Find the area of a smaller octagon that has a perimeter of about 76 feet.

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Using perimeter and area in real life All regular octagons are similar because all corresponding angles are congruent and corresponding side lengths are proportional. First – Draw and label a sketch. Ex. 3: Solution

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Using perimeter and area in real life Ex. 3: Solution FIND the ratio of the side lengths of the two octagons, which is the same as the ratio of their perimeters. perimeter of ABCDEFGH perimeter of JKLMNPQR = a b 76 8(14.25) = 76 114 = 2 3

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Using perimeter and area in real life Ex. 3: Solution CALCULATE the area of the smaller octagon. Let A represent the area of the smaller octagon. The ratio of the areas of the smaller octagon to the larger is a 2 :b 2 = 2 2 :3 2, or 4:9. 9 A 980.4 = 4 9 9A = 980.4 4 A = 3921.6 A 435.7 Write the proportion. Cross product property. Divide each side by 9. Use a calculator. The area of the smaller octagon is about 435.7 square feet.

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Upcoming Quiz after 11.3. There are no other quizzes for this chapter. 11.4 Friday 11.5 Monday 11.6 Wednesday Test Friday, May 12

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