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Math Help - Triangle problem and one more

  1. #1
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    Triangle problem and one more

    Triangle problem and one more-screen-shot-2010-09-09-6.31.18-pm.pngTriangle problem and one more-screen-shot-2010-09-09-6.22.27-pm.png

    I've been working on these for 2 hours, and still cant get the answer, thanks!!!

    Nevermind the picture below, the problem is with the two problems in the link
    Attached Thumbnails Attached Thumbnails Triangle problem and one more-screen-shot-2010-09-09-6.22.31-pm.png  
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  2. #2
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    h(h+7) / 2 = 60
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  3. #3
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    Quote Originally Posted by Wilmer View Post
    h(h+7) / 2 = 60
    which problem is that the answer to?

    please people i need some answers quickly!
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  4. #4
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    Answer to the third problem:

    We know that the area of a triangle is 1/2 * bh

    Area = bh/2

    Area = 20
    h = h
    b = h+6

    20 = h*(h+6)/2
    20 = (h^2 + 6h)/ 2

    20 = \frac{1}{2}h^2 + 3h

    0 = \frac{1}{2}h^2 + 3h - 20

    h = ...

    Use the same method to solve the other one that is similar.
    If you know how to solve the third one, why not the first one? They are exactly the same. Unless you did trial and error to get the answer.
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  5. #5
    Senior Member Educated's Avatar
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    For the second one, use the pythagorus theorum:

    a^2 = b^2 + c^2

    One leg is 2000ft long, b = 2000
    The hypotenuse is 400ft longer than the other leg, a-400 = c

    We substitute these known values into the equation and get:

    a^2 = 2000^2 + (a-400)^2

    a^2 = 2000^2 + a^2 - 800a + 160000

    Now simplify and use the quadratic formula to solve for a, which is the length of the hypotenuse. (You might get 2 answers. In that case, use the one most appropriate)
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