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Math Help - [SOLVED] Square Problem

  1. #1
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    [SOLVED] Square Problem

    Please help me with this problem.
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  2. #2
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    Quote Originally Posted by learnmath View Post
    Please help me with this problem.
    You know that:

    BC² + CD² = BD²
    1² + 1² = BD²
    2 = BD²
    BD = √2

    Since BE = BC, and BC =1, then BE = 1, per transitivity.

    Have you arrived this far?
    Last edited by Sodapop; May 7th 2009 at 06:47 PM.
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  3. #3
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    Yes, I can understand this part. Please help me with the other steps to arrive at the final answer.
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  4. #4
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    Hi

    Consider triangle BPE. PR is one height of this triangle. The area of BPE is A(BPE) = \frac12\:PR \times BE

    The same for triangle BPC : A(BPC) = \frac12\:PQ\times BC

    The area of triangle BCE is the sum of the areas above : A(BCE) = A(BPE) + A(BPC) = \frac12\:PR\times BE + \frac12\:PQ\times BC

    Since BE = BC, PQ + PR)\times BC" alt="A(BCE) = \frac12\PQ + PR)\times BC" />

    On the other hand the area of triangle BCE can be calculated using the projection of E on [BC], say H

    A(BCE) = \frac12\:EH\times BC

    Since ABCDE is a square, angle CBE is 45 degrees. Therefore EH = BE\times \sin(45) = BC \:\frac{\sqrt{2}}{2}

    A(BCE) = \frac12\:\frac{\sqrt{2}}{2}\times BC^2

    Hence PQ + PR)\times BC = \frac12\:\frac{\sqrt{2}}{2}\times BC^2" alt="A(BCE) = \frac12\PQ + PR)\times BC = \frac12\:\frac{\sqrt{2}}{2}\times BC^2" />

    And finally PQ + PR = \frac{\sqrt{2}}{2}\times BC
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  5. #5
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    Thank you!
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