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Math Help - How to solve a shaded region in a weird shape?

  1. #1
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    How to solve a shaded region in a weird shape?

    How to solve a shaded region in a weird shape?-weird.png
    (Ah, I don't know how to upload it so it can show on this website instead of you having to click on it, sorry, so the picture is the one that says 'weird.png')
    Sorry, the left side is 3, and the right side is a 5
    The base is 6, and the base of the shaded is 'x'.

    What I was confuse with was how to start off?
    The problem was: Express the area of the shaded region as a function of 'x'.
    By the way, at the right bottom cornor is a right angle.

    I know that x has to be greater than 6 and less than 6.
    Though at first I thought I could've done something like 'x * 3'
    Since it makes up most of the area there (but only makes up the rectangle)
    So the slanted triangle wasn't included too.
    Then I knew that the highest height was '5' so I did the area of a triangle
    First I subtracted 3 (from the height on the left) from 5 (from the height on the right), to get the height of that triangle.
    So I did the triangle formula: (1/2) base * height
    (0.5)(x*2) = (0.5)(2x) = x

    So I thought it was x * 3 to get the rectangle + x (from the area of the triangle)
    So I got (x*3) + x, but this answer was wrong (I got the answer book)
    So can anyone help me out here?
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    Re: How to solve a shaded region in a weird shape?

    the shape is a trapezoid ...

    note the upper triangles are similar, and therefore their sides are proportional.

    \frac{y}{x} = \frac{2}{6}

    solve for y in terms of x, then find the area of the shaded trapezoid in terms of x.
    Attached Thumbnails Attached Thumbnails How to solve a shaded region in a weird shape?-trapezoid.jpg  
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    Re: How to solve a shaded region in a weird shape?

    Quote Originally Posted by skeeter View Post
    the shape is a trapezoid ...

    note the upper triangles are similar, and therefore their sides are proportional.

    \frac{y}{x} = \frac{2}{6}

    solve for y in terms of x, then find the area of the shaded trapezoid in terms of x.
    Oh... I see now!
    Darn, I didn't really get the shape in my head xD.
    So since they the same shape, but the shaded region is just a smaller version of it.
    y/x = 2/6
    y = (1/3)x

    Then the area of the triangle = (1/2)(b*h)
    (1/2)(x*y)
    (1/2)(x*(1/3)x)
    (1/2)x*(1/6)x
    Area of the rectangle = x*3
    So (x*3) + ((1/2)x *(1/6)x))
    3x + (1/2)x*(1/6)x
    3x + (1/12)x^2
    x (1/12x +3)?

    Darn, I somehow got the wrong answer
    Sorry
    What did I do wrong?
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    Re: How to solve a shaded region in a weird shape?

    one more time ... find the area of the shaded trapezoid
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    Re: How to solve a shaded region in a weird shape?

    Quote Originally Posted by skeeter View Post
    one more time ... find the area of the shaded trapezoid
    Oh....
    So the equation of a trapezoid is A = (1/2) * (b1 + b2) * h right?
    So A = (1/2) (6 * x) * 5
    A = (6 * (1/2)(x)) * 5
    (6 * (1/2)(x)) => 6 = 12/2 => 12/2 + 1/2 = 13/2x
    A = (13/2)x * 5
    A = (75/2)x?
    Sorry, I'm confused, so I am suppose to use the trapezoid formula right?
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    Re: How to solve a shaded region in a weird shape?

    A = \frac{x}{2}\left[3 + (y+3)\right]

    and as you already figured out, y = \frac{x}{3}
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    Re: How to solve a shaded region in a weird shape?

    Quote Originally Posted by skeeter View Post
    A = \frac{x}{2}\left[3 + (y+3)\right]

    and as you already figured out, y = \frac{x}{3}
    Is that the area of the shaded region?
    Cause in the answer book, it says
    (1/6)x(x+18)
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    Re: How to solve a shaded region in a weird shape?

    Quote Originally Posted by Chaim View Post
    Is that the area of the shaded region?
    Cause in the answer book, it says
    (1/6)x(x+18)
    \frac{x}{2}(y+6)

    \frac{x}{2}\left(\frac{x}{3} + 6\right)

    it's the same thing ... expand both expressions and see for yourself.
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    Re: How to solve a shaded region in a weird shape?

    Quote Originally Posted by skeeter View Post
    \frac{x}{2}(y+6)

    \frac{x}{2}\left(\frac{x}{3} + 6\right)

    it's the same thing ... expand both expressions and see for yourself.
    Oh.. I see, thanks!
    Sorry, I was lost with the [ and the ( xD
    Thank you very much
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