Tracy Kompelien's 2-D Shapes Are Behind the Drapes! PDF

By Tracy Kompelien

ISBN-10: 159928507X

ISBN-13: 9781599285078

Publication annotation now not on hand for this title.
Title: 2-D Shapes Are in the back of the Drapes!
Author: Kompelien, Tracy
Publisher: Abdo Group
Publication Date: 2006/09/01
Number of Pages: 24
Binding kind: LIBRARY
Library of Congress: 2006012570

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M 2 ) . 7 29 Pythagoras Dltrerence5 Let X Thus = AD. b2 -h2 = x2. ThenPDAc = x 2+b2_h 2 = 2x2. • x = (b 2 + ~ - P DAC / P BAC a 2 )/(2c). 59 In triangle ABC, let BC the bisector C F in terms of a, b. and c. - 2 a ), h 2 • we obtain the final result: = a, C A = b, and AB = c. Express the length of Solution. We assume that LA '" 90°. 56 to triangles BAC and F AC P FAC / P BAC = (AF · b)/b · c = AF/c. F A B Figure 1-39 = b/a; hence AF = (bc)/(a + b) . Let x = CF . We have AF/c = P FAC / PBAC = (AF2 + b2 - x 2)/W + ~ - a2).

L. 65; Then P ADN SMDA SBDA SAJC P MDA SADN SADC SABJ ---- = ---- = BD DC DC BD == = 1 . 71 1. 69, show that BG = C E. 2. 69, let M be the midpoint of BC. EG and EG = 2AM. 3. The sum of the squares of the diagonals of a parallelogram is equal to sum of the squares of the four sides of the given parallelogram. 8 TrtCoaomebic Functio... 4. :JuIJ. = CP AP . 8 = LC DA = 90° and P is the intersection of AC Trigonometric Functions We have discussed two major geometric quantities: the area and the Pythagoras difference.

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2-D Shapes Are Behind the Drapes! by Tracy Kompelien

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