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(a) 2b:3s s: x or 2b :3 s2 : x or 2b : s2 3: x (b) x : y a:b
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7.11 Changing proportions into new proportions Use each of the following to form a new proportion whose first term is x: 15 (a) x
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x 15 x (b) By Principle 5, 6 (a) By Principle 3, 4 3 8 3 x 8 x (d) By Principle 4, 2 (c) By Principle 6, 1 3 x 15 ; thus, 5 2
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7.12 Combining numerators and denominators of proportions Use Principle 8 to find x in each of the following proportions: (a) x 9 2 2 3 (b) x 8 y x 4 y 2 3 (c) 3x y 15 y 10 3 3 5
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CHAPTER 7 Similarity
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(a) Adding numerators and denominators yields (b) Here we have (x y) 8 (x 4 y) (3x x 9 2 3 2 x 2 or 3 12 2 , so x 3 3x 3 or 5 30 4. 3 , so x 5 6. 2 , so x 3 8.
2 2x , which gives 3 12 y) 15 (y 10 3) 5 3
(c) We use all three ratios to get
7.3 Proportional Segments
If two segments are divided proportionately, (1) the corresponding new segments are in proportion, and (2) the two original segments and either pair of corresponding new segments are in proportion.
a c Thus if AB and AC in Fig. 7-1 are divided proportionately by DE, we may write a proportion such as using b d c a the four segments; or we may write a proportion such as using the two original segments and two of their AB AC new segments.
Fig. 7-1
7.3A Obtaining the Eight Arrangements of Any Proportion
a c A proportion such as can be arranged in eight ways. To obtain the eight variations, we let each term b d of the proportion represent one of the new segments of Fig. 7-1. Two of the possible proportions are then obtained from each direction, as follows:
7.3B Principles of Proportional Segments
PRINCIPLE 1:
If a line is parallel to one side of a triangle, then it divides the other two sides proportionately.
a b c . d
Thus in nABC of Fig. 7-2, if DE i BC, then
CHAPTER 7 Similarity
Fig. 7-2
PRINCIPLE 2:
If a line divides two sides of a triangle proportionately, it is parallel to the third side. (Principles 1 and 2 are converses.)
a b c , then DE i BC. d a b c . d
Thus in nABC (Fig. 7-2), if
PRINCIPLE 3:
4 4 4
Three or more parallel lines divide any two transversals proportionately.
Thus if AB i EF i CD in Fig. 7-3, then
PRINCIPLE 4:
A bisector of an angle of a triangle divides the opposite side into segments which are proportional to the adjacent sides.
a b c . d
Thus in nABC of Fig. 7-4, if CD bisects /C, then
Fig. 7-3
Fig. 7-4
SOLVED PROBLEMS
7.13 Applying principle 1 Find x in each part of Fig. 7-5.
Fig. 7-5
Solutions
(a) DE i BC; hence (b) We have EC x 12 28 , so that x 14 24. x x 4 5 . Then 7x 7 5x 20 and x 10.
7 and DE i BC; hence,
7.14 Applying principle 3 Find x in each part of Fig. 7-6.
CHAPTER 7 Similarity
Fig. 7-6
Solutions
(a) We have EC
4 4 4
4 and AB iEF i CD; hence, 5x 2x 5 1
4 and x 6 20
6. 14x 7. Then 6x 27 and x 42.
(b) AB iCD iE F; hence,
7 , from which 20x 4
7.15 Applying principle 4 Find x in each part of Fig. 7-7.
Fig. 7-7
Solutions
(a) BD bisects /B; hence, x 18 and x 12. 10 15 3x 1 10 30 . Thus, 21x (b) BD bisects /B; hence, 2x 21 7
20x and x
7.16 Proving a proportional-segments problem
Given: EG i BD, EF i BC To Prove: FG i CD Plan: Prove that FG divides AC and AD proportionately. PROOF:
Statements 1. EG i BD, EF i BC AG AE 2. AE , EB GD EB AF AG 3. FC GD 4. FG i CD AF FC 1. Given
Reasons
2. A line (segment) parallel to one side of a triangle divides the other two sides proportionately. 3. Substitution postulate 4. If a line divides two sides of a triangle proportionately, it is parallel to the third side.
CHAPTER 7 Similarity
7.4 Similar Triangles
Similar polygons are polygons whose corresponding angles are congruent and whose corresponding sides are in proportion. Similar polygons have the same shape although not necessarily the same size. The symbol for similar is ,. The notation nABC , nA B C is read triangle ABC is similar to triangle A-prime B-prime C-prime. As in the case of congruent triangles, corresponding sides of similar triangles are opposite congruent angles. (Note that corresponding sides and angles are usually designated by the same letter and primes.) In Fig. 7-8 nABC , nA B C because m/A and m/A 37 a ar m/B b br m/B c cr or 53 6 3 m/C 8 4 m/C 10 5 90
Fig. 7-8
Fig. 7-9
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