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Thread: Area of Trapezoid

  1. #1
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    Talking Area of Trapezoid

    The plane area shown in the figure consists of an isosceles trapezoid(non-parallel sides equal) and a segment of a circle. If the non-parallel sides are tangent to the segment at points A and B, find the area of the composite figure.


    Where should I start in this problem?
    Attached Thumbnails Attached Thumbnails Area of Trapezoid-drawing1.jpg  
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  2. #2
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    and what is that $\displaystyle 60^o$ ?
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  3. #3
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    I think it is needed to compute for the r?
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  4. #4
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    Quote Originally Posted by jasonlewiz View Post
    I think it is needed to compute for the r?
    you are not sure what is that $\displaystyle 60^o$ for???
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  5. #5
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    assume your $\displaystyle 60^o$ is for $\displaystyle \angle DBF$ as shown in my attached figure.

    I have worked out a pretty complicated, perhaps stupid, method...

    produce all the lines and points as shown in figure, such that
    AO = WO = BO = radius of a circle. O is centre of the circle.
    AOZ, BOX and WOY are all straight lines.
    AC//WY//BD and they are all perpendicular to both AB and EF
    since AE and BF are tangents, $\displaystyle \angle EAO$ = $\displaystyle \angle FBO$ = $\displaystyle 90^o$hence $\displaystyle \angle CAZ$ = $\displaystyle \angle DBX$ = $\displaystyle 30^o$, and
    $\displaystyle \angle BAZ$ = $\displaystyle \angle ABX$ = $\displaystyle 60^o$
    so we have a equilateral triangle ABO, with AO = BO = AB = 3
    you can prove that WOY is bisector of angles AOB and XOZ, therefore
    $\displaystyle \angle YOZ$ = $\displaystyle 30^o$
    Given that WOY = 5 and assumed that WO = 3, hence OY = 2

    In triangle YOZ,
    $\displaystyle cos30^o = \frac{OY}{OZ}$
    $\displaystyle \frac{\sqrt{3}}{2} = \frac{2}{OZ}$
    OZ = $\displaystyle \frac{4}{\sqrt{3}}$
    therefore AZ = $\displaystyle 3 + \frac{4}{\sqrt{3}}$

    In triangle CAZ,
    $\displaystyle cos30^o = \frac{AC}{AZ}$
    $\displaystyle (cos30^o)(AZ) = AC$
    AC = $\displaystyle (\frac{\sqrt{3}}{2})(3 + \frac{4}{\sqrt{3}})$
    AC = $\displaystyle \frac{3\sqrt{3}}{2} + 2$

    In triangle EAC,
    $\displaystyle tan60^o = \frac{EC}{AC}$
    ($\displaystyle tan60^o)(AC) = EC$
    EC = $\displaystyle (\sqrt{3})(\frac{3\sqrt{3}}{2} + 2)$
    EC = $\displaystyle \frac{9}{2} + 2\sqrt{3}$

    The area of sector AWBO
    $\displaystyle = (\frac{1}{6})(\pi)(3^2)$
    $\displaystyle = \frac{3\pi}{2}$
    and the area of triangle ABO
    $\displaystyle = (\frac{1}{2})(3)(3)(sin60^o)$
    $\displaystyle = (\frac{9}{2})(\frac{\sqrt{3}}{2})$
    $\displaystyle = \frac{9\sqrt{3}}{4}$
    Therefore the area of region AWB (green in figure)
    = area of sector AWBO - area of triangle ABO
    = 0.8153

    area of trapezium AEFB
    $\displaystyle = \frac{(2EC + CD + AB)(AC)}{2}$
    $\displaystyle = \frac{(6 + 2(\frac{9}{2} + 2\sqrt{3}))(\frac{3\sqrt{3}}{2} + 2)}{2}$
    $\displaystyle = (15 + 4\sqrt{3})(\frac{3\sqrt{3}}{4} + 1)$
    .....in the end you will get this area = 50.4138

    So the area of overall figure = 50.4138 + 0.8153 = 51.2291 ($\displaystyle cm^2 $??)
    Attached Thumbnails Attached Thumbnails Area of Trapezoid-temp.bmp  
    Last edited by ukorov; Nov 22nd 2009 at 02:25 AM.
    Thanks from frameup
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  6. #6
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    hey

    how come u assume WO= 3 ?? is that reasonable??
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  7. #7
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    how come u assume WO=3?? and OY=2 is u need to find for their value like WO= x then OY= 5-x?
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  8. #8
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    is it there is a special triagle 30 by 60 so ur WO must be 1.5squareroot of 3 or 2.598
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