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Math Help - [SOLVED] SAT page 718 #7

  1. #1
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    [SOLVED] SAT page 718 #7

    Note: Figure not drawn to scale

    In the figure below, line segment AE and line segment CD are each perpendicular to line segment CE. If x=y, the length of line segment AB is 4, and the length of line segment BD is 8, what is the length of line segment CE?

    Thanks in advance
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fabxx View Post
    Note: Figure not drawn to scale

    In the figure below, line segment AE and line segment CD are each perpendicular to line segment CE. If x=y, the length of line segment AB is 4, and the length of line segment BD is 8, what is the length of line segment CE?

    Thanks in advance
    note that x and y are 45 degrees. note further that we have similar triangles here. the sides of triangle BCD are all twice the corresponding sides of ABE. angle CDB = angle BAE and angle CBD = angle ABE

    now, two questions: how did i know all these facts? and can you finish the problem now that you know them?
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  3. #3
    eri
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    Possible answer

    Quote Originally Posted by Jhevon View Post
    note that x and y are 45 degrees. note further that we have similar triangles here. the sides of triangle BCD are all twice the corresponding sides of ABE. angle CDB = angle BAE and angle CBD = angle ABE

    now, two questions: how did i know all these facts? and can you finish the problem now that you know them?
    HI,
    so knowing that an isoceles rt triangle with angles 45-45-90
    have sides 1-1-sqrt2,
    would you then take the hypotenuse of 4 and work it back to get sqrt2
    and then take this multiplying factor and multiply the other sides by it?
    you could have 2 sqrt2 = 4 or 4/sqrt2 = 4 (same thing)
    the other 2 sides are 2*sqrt2


    Then 8 is 2 times 4, triangles are the same type, so multiply 2 times
    the smaller triangels sides to get 4sqrt2
    add and get 6*sqrt2
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by eri View Post
    HI,
    so knowing that an isoceles rt triangle with angles 45-45-90
    have sides 1-1-sqrt2,
    would you then take the hypotenuse of 4 and work it back to get sqrt2
    and then take this multiplying factor and multiply the other sides by it?
    you could have 2 sqrt2 = 4 or 4/sqrt2 = 4 (same thing)
    the other 2 sides are 2*sqrt2


    Then 8 is 2 times 4, triangles are the same type, so multiply 2 times
    the smaller triangels sides to get 4sqrt2
    add and get 6*sqrt2
    it is even easier. knowing these are similar triangles (since all the angles are proportional) we know that CE = 3BE. and we can find BE using trigonometry. \sin x = \frac {BE}{AB} \implies BE = AB \sin x = 4 \sin 45 = 4 \cdot \frac {\sqrt{2}}2 = 2 \sqrt{2}

    so finally, CE = 3 BE = 6 \sqrt{2}
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    Thanks to Jhevon and eri!
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