Saturday, November 28, 2009

Problem 395: Square, 15 Degree, Equilateral triangle

Proposed Problem
Click the figure below to see the complete problem 395 about Square, 15 Degree, Equilateral triangle, Congruence.

Problem 395: Square, 15 Degree, Equilateral triangle.
See also:
Complete Problem 395

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 7:04 AM
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7 comments:

  1. http://www.wolframalpha.com/input/?i=%281-1%2F2*tg%2815%29%29%5E2%2B1%2F4

    ReplyDelete

  2. http://geometri-problemleri.blogspot.com/2009/11/problem-52-ve-cozumu.html

    ReplyDelete

  3. solution plz some1

    ReplyDelete

  4. in triangle ABE ang(ABE)=75;BE=AB/(2cos15)
    with the law of cosine
    AE²=AB²+BE²-2AB.BE cos75=
    AB²(1+(1/(2cos15)²-cos75/cos15)= AB² then
    cos(75)=cos(45+30)=(sqr6-sqr2)/4
    cos(15)=cos(45-30)=(sqr6+sqr2)/4
    same result for CE
    CE=AE=AD
    ADE is equilateral

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  5. build on each side of the square (from the inside) isosceles triangle with 15 degrees..from there its easy..

    ReplyDelete

  6. Consider the equilateral triangle BCF at the exterior of the square ABCD say of side a. The quandrilateral FEDC and FEAB are rhombuses, therefore

    ED = FC = a and EA = FB = a

    (This holds because ang(FEC) = 75 = ang(ECF), so FE = FC = a and FE parallel to CD = a)

    ReplyDelete

  7. See
    http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=472125

    ReplyDelete

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