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(a) If nABC , nA B C and s sr m 3 , then 2 mr 6 7 3 . 2 17. Since nABC , nA B C , 71. 2 s 7 51 and s 17 21.
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(b) The perimeter of nA B C is 4 (c) Since nADE , ABC, DE 25
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Line ratios from similar polygons Complete each of the following statements: (a) If corresponding sides of two similar polygons are in the ratio of 4:3, then the ratio of their perime _ ters is __ . (b) The perimeters of two similar quadrilaterals are 30 and 24. If a side of the smaller quadrilateral _ is 8, the corresponding side of the larger is __ . _ (c) If each side of a pentagon is tripled and the angles remain the same, then each diagonal is __ .
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CHAPTER 7 Similarity
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4 . 3 p s s 30 . Then (b) Since the quadrilaterals are similar, and s 10. sr pr 8 24 (c) Tripled, since polygons are similar if their corresponding angles are congruent and their corresponding sides are in proportion. (a) Since the polygons are similar,
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7.6 Proving Equal Products of Lengths of Segments
In a problem, to prove that the product of the lengths of two segments equals the product of the lengths of another pair of segments, it is necessary to set up the proportion which will lead to the two equal products.
SOLVED PROBLEM
7.28 Proving an equal-products problem Prove that if two secants intersect outside a circle, the product of the lengths of one of the secants and its external segment equals the product of the lengths of the other secant and its external segment.
Solution
Given: Secants AB and AC. To Prove: AB AD AC AE Plan: Prove nABE , nACD to obtain AB AE . AC AD PROOF:
Statements 1. 2. 3. 4. 5. Draw BE and CD. /A > /A /B > /C nAEB , nADC AE AD AD AC 1. 2. 3. 4.
Reasons A segment may be drawn between any two points. Reflexive property. Angles inscribed in the same arc are congruent. Two triangles are similar if two angles of one triangle are congruent respectively to two angles of the other. 5. Corresponding sides of similar triangles are in proportion. AE 6. In a proportion, the product of the means equals the product of the extremes.
AB AC 6. AB
7.7 Segments Intersecting Inside and Outside a Circle
PRINCIPLE 1:
If two chords intersect within a circle, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other.
EB CE ED.
Thus in Fig. 7-25, AE
Fig. 7-25
CHAPTER 7 Similarity
If a tangent and a secant intersect outside a circle, the tangent is the mean proportional between the secant and its external segment.
AB AP AP . AC
PRINCIPLE 2:
Thus in Fig. 7-26, if PA is a tangent, then
PRINCIPLE 3:
If two secants intersect outside a circle, the product of the lengths of one of the secants and its external segment equals the product of the lengths of the other secant and its external segment.
AD AC AE.
Thus in Fig. 7-27, AB
Fig. 7-26
Fig. 7-27
SOLVED PROBLEMS
7.29 Applying principle 1 Find x in each part of Fig. 7-28, if chords AB and CD intersect in E.
Fig. 7-28
Solutions
(a) ED (b) AE (c) CE 4. Then 16x EB 4(12), so that 16x 8(2), so x2 x. Then x2 48 or x 16 and x 27(3) or x2 3. 4. 81, and x 9.
x. Then x2 EB
3 and AE
7.30 Applying principle 2 Find x in each part of Fig. 7-29 if tangent AP and AB intersect at A.
Fig. 7-29
CHAPTER 7 Similarity
Solutions
(a) AB (b) AB (c) AB 9 2x x 9 6 24. Then x2 5) 6) 24(6) or x2 100 and x 16 or x2 6x 144, and x
1 72.
5. Then 5(2x 6. Then x(x
0. Factoring gives (x
8)(x
0 and x
7.31 Applying principle 3 Find x in each part of Fig. 7-30 if secants AB and AC intersect in A.
Fig. 7-30
Solutions
(a) AC (b) AC 12. Then 8x 2x 2 and AB 12(3) and x 42 . 2) 12(5) and x 14.
12. Then 2(2x
7.8 Mean Proportionals in a Right Triangle
PRINCIPLE 1:
The length of the altitude to the hypotenuse of a right triangle is the mean proportional between the lengths of the segments of the hypotenuse.
BD CD CD . DA
Thus in right nABC (Fig. 7-31),
Fig. 7-31
PRINCIPLE 2:
In a right triangle, the length of either leg is the mean proportional between the length of the hypotenuse and the length of the projection of that leg on the hypotenuse.
AB BC AB BC and BD AC AC . AD
Thus in right nABC (Fig. 7-31),
A proof of this principle is given in 16.
SOLVED PROBLEMS
7.32 Finding mean proportionals in a right triangle In each triangle in Fig. 7-32, find x and y.
CHAPTER 7 Similarity
Fig. 7-32
Solutions
3 (a) By Principle 1, x (b) By Principle 1, x 8 y 12 x or x2 27, and x 3 23. By Principle 2, y , so y2 36 and y 9 3 y 20 8 , so y2 80 and y 4 25. and x 16. By Principle 2, y 4 4 6.
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