Tuesday, July 27, 2010

Problem 483: Square, Angle, 90 degrees, Triangle, Measurement

Geometry Problem
Click the figure below to see the complete problem 483 about Square, Angle, 90 degrees, Measurement.

Problem 483: Triangle, Square, Angle, 90 degrees, Measurement
See also:
Complete Problem 483

Level: High School, SAT Prep, College geometry

4 comments:

  1. Note that ABEC is cyclic quadrilateral ( Angle ABC=Angle AEC =90)
    AC= 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

    ReplyDelete
  2. 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]

    ReplyDelete
  3. Mark 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
    So a - b = x sqrt2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete