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Math Help - Triangle areas?

  1. #1
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    Triangle areas?

    Two noncongruent triangles have equal area. One triangle has sides 5, 5, and 4. The other has sides 5, 5, and x. Find the value of x.

    I managed to get a quadratic equation, but I have no idea how to solve it. Any tips here?
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  2. #2
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    Quote Originally Posted by Rinnie View Post
    Two noncongruent triangles have equal area. One triangle has sides 5, 5, and 4. The other has sides 5, 5, and x. Find the value of x.

    I managed to get a quadratic equation, but I have no idea how to solve it. Any tips here?
    what is your equation? how did you arrive at it?
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    my equation is x^2 - 10x + 8*sqrt21.
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    This equation can be solved by completing the square or using the quadratic formula

    Completing the square

    x^2-10x+8\sqrt{21}=0

    (x^2-10x+25)+8\sqrt{21}-25=0

    (x-5)^2+8\sqrt{21}-25=0

    (x-5)^2=25-8\sqrt{21}

    x-5=\sqrt{25-8\sqrt{21}}

    x=5\pm\sqrt{25-8\sqrt{21}}

    Quadratic Formula

    for ax^2+bx+c=0

    then x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    so for x^2-10x+8\sqrt{21}=0

    then x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)8\sqrt{21}}}{2\times1}
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  5. #5
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    don't forget to throw away any negative values you arrive at!

    Area can't be negative.
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    That's what I tried to do, but I'm confused about how to deal with the sqrt21 under the sqrt sign.
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    Quote Originally Posted by Rinnie View Post
    That's what I tried to do, but I'm confused about how to deal with the sqrt21 under the sqrt sign.
    You may need to make an approximation for \sqrt{21} = 4.583 and continue you calculations.
    Last edited by pickslides; May 11th 2009 at 05:43 PM. Reason: typo
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    Using the quadratic forumula, I end up with 100 - 146.something in under the square root sign which makes it a negative though.
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    This would suggest your equation is not correct.

    Quote Originally Posted by pickslides View Post
    what is your equation? how did you arrive at it?
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  10. #10
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    Quote Originally Posted by Rinnie View Post
    Two noncongruent triangles have equal area. One triangle has sides 5, 5, and 4. The other has sides 5, 5, and x. Find the value of x.

    I managed to get a quadratic equation, but I have no idea how to solve it. Any tips here?
    If a triangle has sides of length a, b, c, and C is the angle opposite side c, then the area of the triangle is \tfrac12ab\sin C. If a = b = 5 then the area is 12.5\sin C. The only way that two different triangles could have that same area is if \sin C is the same for both triangles. But that would mean that the two angles are supplementary: C_2 = \pi - C_1 (or C_2 = 180 - C_1 if you prefer degrees to radians).

    Now use the cosine rule. You are told that one of the two triangles (say the one with angle C_1) has the third side equal to 4. The cosine rule says that \cos C_1 = \frac{5^2+5^2-4^2}{2\times5\times5} = \frac{34}{50}. For the other triangle, with third side x and angle C_2, the formula says \cos C_2 = \frac{5^2+5^2-x^2}{2\times5\times5} = \frac{50-x^2}{50}. But C_1 and C_2 are supplementary, and so \cos C_2 = -\cos C_1. Therefore \frac{50-x^2}{50} = -\frac{34}{50}. That's your quadratic equation. Now all you have to do is to solve it.
    Last edited by Opalg; May 13th 2009 at 08:33 AM. Reason: Usual silly arithmetic error
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