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ssrs barcode font Solutions in ObjectiveC
Solutions QRCode Scanner In ObjectiveC Using Barcode Control SDK for iPhone Control to generate, create, read, scan barcode image in iPhone applications. QR Creation In ObjectiveC Using Barcode creator for iPhone Control to generate, create Denso QR Bar Code image in iPhone applications. (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. QR Code Decoder In ObjectiveC Using Barcode scanner for iPhone Control to read, scan read, scan image in iPhone applications. Create Bar Code In ObjectiveC Using Barcode encoder for iPhone Control to generate, create barcode image in iPhone applications. (b) The perimeter of nA B C is 4 (c) Since nADE , ABC, DE 25
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GS1 128 Creator In None Using Barcode creation for Online Control to generate, create EAN 128 image in Online applications. Code39 Generation In None Using Barcode maker for Online Control to generate, create Code 3 of 9 image in Online applications. 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 equalproducts 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. 725, AE
Fig. 725 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. 726, 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. 727, AB
Fig. 726 Fig. 727 SOLVED PROBLEMS
7.29 Applying principle 1 Find x in each part of Fig. 728, if chords AB and CD intersect in E.
Fig. 728 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. 729 if tangent AP and AB intersect at A.
Fig. 729 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. 730 if secants AB and AC intersect in A.
Fig. 730 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. 731), Fig. 731 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. 731), 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. 732, find x and y.
CHAPTER 7 Similarity
Fig. 732 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.

