Tuesday, August 31, 2010

Problem 517: Right Triangle, Six Squares, Areas

Geometry Problem
Click the figure below to see the complete problem 517 about Right Triangle, Six Squares, Areas.

Problem 517: Right Triangle, Six Squares, Areas


See also:
Complete Problem 517

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 5:34 PM
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2 comments:

  1. c .t . e. oSeptember 1, 2010 at 1:42 AM

    name AB = a, BC = b, AC = c
    S1 + S2 + S3 = a² + b² + (a² + b²) = 2a² + 2b²
    extend BA,it meet side of A5 in midpoint
    Str on the right of A5 is (a∙b)/2 =>
    Str/2 = (a∙b)/4 =>
    side of A5 = 4a² + b²
    Str on the left of A6 is a² + 4b²
    S5 + S4 + S6 = (4a + b²) + (a² + b²) + (a² + 4b²)
    = 6a² + 6b²

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  2. AnonymousSeptember 4, 2010 at 4:08 PM

    In general, 3x(A4+A1+A2)=A3+A5+A6.
    When triangle ABC is a right triangle, A3=A4.
    So, 3x(A1+A2+A3)=A4+A5+A6.

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