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Math Help - Triangles - Should Be Easy

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    Triangles - Should Be Easy

    This might seem extremely easy but I just can't figure it out. Suppose I have a triangle ABC, where the line BC has length 4 and is perpendicular to AC which has length 9. Now suppose there is a rectangle BDEC with width 3, adjacent to the triangle on the BC side. Now we draw a line AD, and take the point at which that line intersects BC, call it F. Find the area of the triangle BFA.

    So if we consider the triangle ABC, since BC = 4 and AC = 9, and 16+81=97 so AB = 97^{1/2}. The area of ABC is 4*9/2 = 18. The area of whole figure, ABDE, is 18+12=30. The area of the triangle ABD is 3*(9+3)/2 = 18.

    For all that, I still can't pin down what I need. I need to find a way to determine the length of BF. This is an SAT question and shouldn't appeal to more than basic notions of geometry like area of a triangle, the properties of 30-60-90 triangles, Pythagorean theorem, and such.
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    Quote Originally Posted by ragnar View Post
    This might seem extremely easy but I just can't figure it out. Suppose I have a triangle ABC, where the line BC has length 4 and is perpendicular to AC which has length 9. Now suppose there is a rectangle BDEC with width 3, adjacent to the triangle on the BC side. Now we draw a line AD, and take the point at which that line intersects BC, call it F. Find the area of the triangle BFA.
    DE/EA = 4/12 = 1/3 = FC/CA = FC/9

    FC = 3

    area of BCA = 18

    area of FCA = 27/2

    area of BFA = area of BCA - area of FCA
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