Geometry Problem
Click the figure below to see the complete problem 483 about Square, Angle, 90 degrees, Measurement.
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Complete Problem 483
Level: High School, SAT Prep, College geometry
Tuesday, July 27, 2010
Problem 483: Square, Angle, 90 degrees, Triangle, Measurement
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Note that ABEC is cyclic quadrilateral ( Angle ABC=Angle AEC =90)
ReplyDeleteAC= ½ *SQRT(a^2+b^2) and AB=BC= SQRT(2)/2 *SQRT(a^2+b^2)
Apply Ptolemy’s theorem in the cyclic quadrilateral ABEC we have:
AE.BC=EC.AB+AC.BE
Replace AE=a , BE=x , EC=b , and AC, AB , BC from values above we get x= (a-b)* SQRT(2)/2
Peter Tran
D'aprés la théorème d'alkachi on a dans le triangle [ABE] : [BE^2=AB^2+AE^2-2AB\times AE\times cos\widehat{BAE}] et on pose [BE=x] et [AE=a] et [AB=Y] et [\widehat{BAE}=\alpha] alors: [x^2=a^2+y^2-2ay\times cos\alpha] de la mème manière on a dans le triangle [BEC] on a: [x^2=b^2+y^2-2by\times cos\alpha] ( [\widehat{BAE}=\widehat{BCE}] parce que [ABCE] est un quadrilatère circulaire) alors: [\frac{a^2+y^2-x^2}{2ay}=\frac{b^2+y^2-x^2}{2by}] on simplifie et on trouve: [x^2=y^2-ab] et on sais [y^2=\frac{a^2+b^2}{2}] alors: [x^2=\frac{a^2+b^2}{2}-ab=\frac{(a-b)^2}{2}] [\Rightarrow x=\frac{a-b}{2}\sqrt2]
ReplyDeleteil faut installer LATEX.
ReplyDeleteMark F on AE such that Tr. FEC is isoceles. Now ABFC is cyclic hence easily we can see that Tr. s BEC & AFC are similar
ReplyDeleteSo a - b = x sqrt2
Sumith Peiris
Moratuwa
Sri Lanka
https://photos.app.goo.gl/jgaUR1LSxEYsudCZ7
ReplyDeletedraw BF perpendicular to BE ( F on AE)
note that ABEC is cyclic and ^(BEA)= 45
triangle BAF congruent to BCE ( case ASA)
so AF=EC= band EF= a-b
in isosceles right triangle BEF we have EF= BE. sqrt(2)=> (a-b)=x.sqrt(2)
Complete the rectangle AECF, F being outside the side AD.
ReplyDeleteABECDF is cyclic. Connect BF and denote its point of intersection with AE as G.
Observe that ABG is congruent to CBE (ASA) with CBE being ABG rotated by 90º on B.
Hence BG=x ,AG=b=> EG=a-b and GBE is right isosceles
Hence x=(a-b)/Sqrt(2)