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    #2
    Let's go through this step by step...can you find the area of $ \displaystyle \triangle ABC$ ?

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    Quote Originally Posted by MarkFL View Post
    Let's go through this step by step...can you find the area of $ \displaystyle \triangle ABC$ ?
    3200cm^2

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    #4
    Quote Originally Posted by Yazan975 View Post
    3200cm^2
    That's not correct...how did you arrive at that answer?

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    Quote Originally Posted by MarkFL View Post
    That's not correct...how did you arrive at that answer?
    1600cm^2?

    For 3200 I multiplied Base and Height but didn't divide by 2

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    #6
    Quote Originally Posted by Yazan975 View Post
    1600cm^2?

    For 3200 I multiplied Base and Height but didn't divide by 2
    Okay, good...so what must the area of $ \displaystyle \triangle AED$ be?

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    Quote Originally Posted by MarkFL View Post
    Okay, good...so what must the area of $ \displaystyle \triangle AED$ be?
    xy/2 cm^2

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    #8
    Quote Originally Posted by Yazan975 View Post
    xy/2 cm^2
    Yes, that's correct in terms of \(x\) and \(y\), but we should also be able to assign a numeric value to its area based on the fact that the areas of the two shaded regions are the same.

    This will give us an equation...can you state it?

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    Quote Originally Posted by MarkFL View Post
    Yes, that's correct in terms of \(x\) and \(y\), but we should also be able to assign a numeric value to its area based on the fact that the areas of the two shaded regions are the same.

    This will give us an equation...can you state it?
    xy/2 cm^2 = 800
    xy = 1600

    Is that right?

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    #10
    Quote Originally Posted by Yazan975 View Post
    xy/2 cm^2 = 800
    xy = 1600

    Is that right?
    Excellent!

    Okay, now the next thing I notice is that within $ \displaystyle \triangle ABC$ there are two similar triangles, with $ \displaystyle \triangle AED$ being the smaller of the two. This means we may state:

    $ \displaystyle \frac{y}{x}=\frac{40}{x+10}$

    Do you see where this comes from?

    What I would do here is solve both equations we now have for \(y\), and equate the two results to get an equation in \(x\)...can you state this equation?

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