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Chapter 6 Proportions and Similarity. 6.1 Proportions.

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Presentation on theme: "Chapter 6 Proportions and Similarity. 6.1 Proportions."— Presentation transcript:

1 Chapter 6 Proportions and Similarity

2 6.1 Proportions

3 Ratios A ratio is a comparison of two quantities. The these quantities must be integers, must be in simplified form and must be of the same unit of measure. If comparison is not of the same unit of measure, then it is called a rate. Ratios can be written three different ways: a:b a to b a/b

4 Proportions Proportions are equations that state two (or more) ratios are equal. Cross product rule – the product of the means equals the product of the extremes. Product of extremes = ad Product of means = bc Cross Product Rule: ad = bc

5 Example Cross Product Rule Distribution Add 10 to both sides Divide both sides by 6

6 Rates Remember, Rates are like ratios in that they have integers as numbers and are written in lowest form. The thing that is different is that the numbers have different units of measure. MPH? MPG? $/hr?

7 Example: Say you have a model car that has a wheel diameter of 1” and the wheel diameter of the real car is 18”. If you measure the length of the model and you find it is 9” long, how long is the car? Set up the ratio of model wheel/real wheel = model length/real length Plug in the numbers and go….

8 6.2 Similar Polygons

9 Similar Polygons When two polygons have the same shape but are not congruent, they may be similar. By definition, a similar polygon is where all the corresponding angles are congruent, but all the ratios of corresponding sides are equal. If quad ABCD ~ EFGH, then….

10 Scale Factor The ratio of the corresponding sides is called the scale factor. The scale factor is a ratio – that means that the numbers must be integers and the ratio must be in lowest, simplified form. If the scale factor is 1 to 2 that means that the 2 nd figure is two times larger than the first. So, for every unit of length in the first figure, if you multiply it by 2, you’ll get the length of the corresponding side in the 2 nd figure. Or the length of the first figure = (1/2) length of the 2 nd figure.

11 Example: If you have a picture and you “blow it up” on a copy machine by 20%, what is the scale factor of original to new? Blowing up by 20% is the same as multiplying the new figure by 120% or 1.2. So, the ratio is 1 to 1.2 which simplifies to 5 to 6. Is the new figure larger? So, that is why it is 5 to 6.

12 Another Example You have a photo that is 3” x 5” – you want to “blow it up” as large as it can get to fit on a 8.5” x 11” of paper, BUT you want to keep the same aspect of length and width. How big can it be without cutting any of the photo off? What is “Aspect”? That is keeping the ratio of the H vs L the same….

13 Example Continued 5” 3” 11” 8.5” (3/5) ≠ (8.5/11) Set up two ratios: (3/5) = (x/11) and (3/5) = (8.5/y) Solving for x and y we get… x = 6.6, y = 14 1/6 So, y is too big – we’ll cut off something

14 Continued 11” 8.5” 6.6” So, 3 x 5 picture gets blown up to 6.6 x 11 to get the bird as big as it can get w/o cutting things off.

15 6.3 Similar Triangles

16 Similar Triangles Remember when we said that the only way to prove triangles congruent was to prove all 3 sets of sides and the 3 sets of angles were congruent? Then we told you that there were four short cuts? SSS, SAS, ASA and AAS? Well, there are 3 short cuts for Similar Triangles.

17 Similar Triangle Short Cuts SSS~ This says that when you have the three ratios of corresponding sides equal, then you have similar triangles. SAS~ This says that when you have congruent, included angles and the ratios of the two sets of sides equal, then you have similar triangles. AA This says that when you have two sets of corresponding angles congruent, then the triangles are similar.

18 SSS ~ 12 24 8 16 12 18 Is 8/12 = 12/18 = 16/24? Yes, SF = 2/3 A C B D F E So ΔABC ~ ΔDEF by SSS ~ Because of Def of Similar Polygons, <A <D

19 SAS ~ 12 8 18 A C B D F E 33° Is 8/12 = 12/18? Yes, SF = 2/3 Do we have congruent included angles? Yes So ΔABC ~ ΔDEF by SAS ~

20 AA A C B D F E 33° 36° Here we have two sets of congruent angles so, the triangles are similar by AA. ΔABC ~ ΔDEF

21 Similarity of Triangles Similar Triangles are symmetric, reflexive and transitive: Reflexive – ΔABC ~ ΔABC Symmetric – If ΔABC ~ ΔDEF, then ΔDEF ~ ΔABC. Transitive – If ΔABC ~ ΔDEF and ΔDEF ~ ΔGHI, then ΔABC ~ ΔGHI

22 Common Example A E D CB Given: Prove: ΔABE ~ΔACD Corresponding Angle Theorem. Reflexive Prop ΔABE ~ ΔACD by AA

23 Another Example A E D CB 2 x 3 4 1.5 y+2 Solve for x and y. ΔABE ~ ΔACD by AA Because the two triangles are similar we can set up the proportions. Now solve for x & y x = 6 & y = 2.5

24 6.4 Parallel Lines and Proportional Parts

25 Short Cuts Side Splitter – If you have a triangle with a segment parallel to one of the three sides, you can use side splitter. Converse of Side Splitter – If you have a triangle where two sides are split proportionally, then the segment is parallel to the sides. Midsegment Theorem – If you go from the MP of one side to the MP of the other side, then the 3 rd side is 2x the segment.

26 Side Splitter (Triangle Proportionality Theorem) A E D CB Here we have a triangle with a segment parallel to a side – classic side splitter case. You don’t need to set up similar triangles and do the proportions, just set up the proportion AB/AE = BC/ED. Notice – you don’t use it for the parallel sides, only the sides that are split! The proof of this theorem is found on page 307 in the book if you care to see it.

27 Example A E D CB 2 x 3 4 Here we have a triangle with a segment parallel to a side – classic side splitter case. Just set up the proportion AB/AE = BC/ED. 2/3 = 4/x …….. solving for x we get x = 6 This is so much easier, but you have to be careful that you’re only working with the sides that have been split by parallel lines.

28 Converse of Side Splitter A E D CB 2 6 3 4 Just set up the proportion AB/AE = BC/ED. 2/3 = 4/6 …….. Since this is true we can conclude that BE and CD are parallel.

29 Midsegment Theorem A E D CB Since B is MP of AC and E is MP of AD we can use Midsegment Theorem Since segment BE connects the two MP’s of the sides, we can say that 2BE = CD or BE = (½)CD Remember this only works when we are connecting the two MP’s of the two sides.

30 Corollaries H G F E A B C D Here you have two coplanar lines that are cut by multiple parallel lines. AB/HG = BC/GF = CD/FE AND AB/HG =AC/HF = BD/GE = AD/HE

31 Continued H G F E A B C Here you have two coplanar lines that are cut by multiple parallel lines. D Here if AB = BC = CD then HG = GF = FE.

32 6.5 Parts of Similar Triangles

33 Proportional Perimeters Theorem A B C D E F 12 20 12 18 30 18 Are these two Δ’s similar? Yes SSS~ Find the perimeters of ΔABC then ΔDEF. P of ΔABC = 44, P of ΔDEF = 66 What is the SF?SF is 2/3 Find the ratio of the corresponding perimeters It is the same as the SF -- 2/3 Proportional Perimeters Theorem – If two triangles are similar, then the Perimeters are proportional to the measures of the corresponding sides.

34 Special Segments of Δ’s All the special segments of similar triangles are proportional to the corresponding sides (Same as the SF) It works for all special segments, Altitudes, Angle Bisectors, Perpendicular Bisectors and Medians.

35 Angle Bisectors of Δ’s Full Definition – An Angle Bisector of a Triangle is a segment that is drawn from a vertex to the opposite side that divides the vertex angle into two congruent angles and it divides the opposite side proportionally. Example is on the next slide.

36 Angle Bisectors of Δ’s A C B D is an Angle Bisector of ΔABC. 1 2 The side that is split by the <bis is split proportionally.


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