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Math Help - Express the area of the trapezoid in terms of A and B.

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    Express the area of the trapezoid in terms of A and B.

    A trapezoid is divided into four triangles by its diagonals. Suppose the top and bottom triangles have areas of A and B. Express the AREA of the trapezoid in terms of A and B.
    Here is a picture of the problem:

    http://i216.photobucket.com/albums/c...hchallenge.jpg

    I know the solution already, but just want to see if there's a much simpler way.
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    Quote Originally Posted by mathwizard View Post
    A trapezoid is divided into four triangles by its diagonals. Suppose the top and bottom triangles have areas of A and B. Express the AREA of the trapezoid in terms of A and B.
    Here is a picture of the problem:

    http://i216.photobucket.com/albums/c...hchallenge.jpg

    I know the solution already, but just want to see if there's a much simpler way.
    DISCLAIMER: If this is wrong, just shoot me. I spent way too much time here.

    First, draw a line through the intersection of the diagonals and perpendicular to both parallel bases. This will give us the heights of the two triangles whose areas we know.

    Label the base of triangle A be b_1 and the base of triangle B be b_2

    The height of triangle A will be h_1 and the height of triangle B will be h_2

    Now, the areas of the two triangles:

    A=\frac{1}{2}b_1h_1

    B=\frac{1}{2}b_2h_2

    The area of the trapezoid is given as: T=\frac{1}{2}h(b_1+b_2)

    Substituting, we get,

    T=\frac{1}{2}(h_1+h_2)(b_1+b_2)

    \triangle{A}\sim\triangle {B} by AA postulate

    That means:

    \frac{b_1}{b_2}=\frac{h_1}{h_2}

    * \boxed{h_1b_2=h_2b_1}\Longrightarrow Remember \ \ this!

    T=\frac{1}{2}h_1+h_2)(b_1+b_2)

    T=\frac{1}{2}(h_1b_1+h_2b_2+h_1b_2+h_2b_1)

    Since we know from the * above,

    T=\frac{1}{2}h_1b_1 + \frac{1}{2}h_2b_2 +\frac{1}{2} \underbrace{(h_1b_2 + h_2b_1)}_{\text{equal to each other}}

    T=A+B+h_1b_2

    Now,

    AB=(\frac{1}{2}b_1h_1)(\frac{1}{2}h_2b_2)

    AB=\frac{1}{4} \underbrace{(b_1h_2)(b_2h_1)}_{\text{equal to each other}}

    AB=\frac{1}{4}(h_1b_2)^2

    4AB=(h_1b_2)^2

    2\sqrt{AB}=h_1b_2

    Substituting back into our last T equation we finally get:

    T=A+B+2\sqrt{AB}
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