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(a) /AEB > /DEC; 3 9 4 12 (b) /A > /A; 6 12 9 18 (c) /BAC > /ACD; 20 16 25 20
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7.21 Applying principle 5 In each part of Fig. 7-19, determine the proportion needed to prove the indicated triangles similar.
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Fig. 7-19
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CHAPTER 7 Similarity
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6 12 9 (b) 12 (a) 8 16 12 16 12 24 15 20 (c) 7 21 8 24 10 30
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7.22 Proportions obtained from similar triangles Find x in each part of Fig. 7-20.
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Fig. 7-20
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(a) Since BD i AC, /A > /B and /C > /D; hence, nAEC , nBED. Then (b) Since JK i GH, nFJK , nFGH by Principle 7. Hence, 6 x 6 4 and x 7 x 18 41. 2 10 and x 12 15.
7.23 Finding heights using ground shadows A tree casts a 15-ft shadow at a time when a nearby upright pole 6 ft high casts a shadow of 2 ft. Find the height of the tree if both tree and pole make right angles with the ground.
Solution
At the same time in localities near each other, the rays of the sun strike the ground at equal angles; hence, /B > /B in Fig. 7-21. Since the tree and the pole make right angles with the ground, /C > /C . Hence, h 15 nABC , nA B C , so 2 and h 45 ft. 6
Fig. 7-21
CHAPTER 7 Similarity
7.24 Proving a similar-triangle problem stated in words Prove that two isosceles triangles are similar if a base angle of one is congruent to a base angle of the other.
Solution
Given: Isosceles nABC (AB AC) Isosceles nA B C (A B A C ) /B > /B To Prove: nABC , nA B C Plan: Prove /C > /C and use Principle 3. PROOF:
Statements 1. 2. 3. 4. /B > /B /B > /C, /B > /C /C > /C nABC , nA B C 1. 2. 3. 4.
Reasons Given Base angles of an isosceles triangle are congruent. Things > to > things are > to each other. Two triangles are similar if two angles of one triangle are congruent to two angles of the other.
7.25 Proving a proportion problem involving similar triangles
Given: Parallelogram ABCD AE AF To Prove: BC BF Plan: Prove (AEF ,nBCF) PROOF:
Statements 1. 2. 3. 4. 5. 6. ABCD is a parallelogram. ED i BC /DEC > /ECB /EFA > /BFC nAEF , nBCF AE BC AF BF 1. 2. 3. 4. 5.
Reasons Given Opposite sides of a parallelogram are parallel. Alternate interior angles of parallel lines are congruent. Vertical angles are congruent. Two triangles are similar if two angles of one triangle are congruent to two angles of the other. 6. Corresponding sides of similar triangles are in proportion.
7.5 Extending A Basic Proportion Principle
PRINCIPLE 1: PRINCIPLE 2: PRINCIPLE 3:
Corresponding sides of similar triangles are in proportion. Corresponding segments of similar triangles are in proportion. Corresponding segments of similar polygons are in proportion.
When segments replaces sides, Principle 1 becomes the more general Principle 2. When polygons replaces triangles, Principle 2 becomes the even more general Principle 3. By segments we mean straight or curved segments such as altitudes, medians, angle bisectors, radii of inscribed or circumscribed circles, and circumferences of inscribed or circumscribed circles. The ratio of similitude of two similar polygons is the ratio of any pair of corresponding lines. Corollaries of Principles 2 and 3, such as the following, can be devised for any combination of corresponding lines: 1. Corresponding altitudes of similar triangles have the same ratio as any two correponding medians.
Thus if nABC , nA B C in Fig. 7-22, then h h m. m
CHAPTER 7 Similarity
Fig. 7-22
Fig. 7-23
2. Perimeters of similar polygons have the same ratio as any two corresponding sides.
Thus in Fig. 7-23, if quadrilateral I , quadrilateral I , then 34 17 4 2 6 3 10 5 14 . 7
SOLVED PROBLEMS
7.26 Line ratios from similar triangles (a) In two similar triangles, corresponding sides are in the ratio 3:2. Find the ratio of corresponding medians [see Fig. 7-24(a)]. (b) The sides of a triangle are 4, 6, and 7 [Fig. 7-24(b)]. If the perimeter of a similar triangle is 51, find its longest side. (c) In nABC of Fig. 7-24(c), BC 25 and the measure of the altitude to BC is 10. A line segment terminating in the sides of the triangle is parallel to BC and 3 units from A. Find its length.
Fig. 7-24
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