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October 2nd, 2014, 02:08 AM   #1
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Arrow Quadrilateral

In a convex quadrilateral ABCD, sides AB and BC each have length 2, while side CD has length 2√3 and DA has length 2√5. Let M, N - mid diagonals, MN has length √2. What is the area of quadrilateral ABCD?
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October 2nd, 2014, 03:45 AM   #2
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First of all, you need to draw a clear picture and upload it.
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October 2nd, 2014, 04:29 AM   #3
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October 2nd, 2014, 06:26 AM   #4
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I am not sure so correct me if I am wrong.
Side AC is common in the two triangles of the quadrilateral.
So, substitute AC to be x. Then apply Heron's formula and write the equations so obtained for both the triangles. You have got 2 equations in variable x, so solve for x. Then apply Heron's formula again and find the numerical area of both the triangles and then add them to find the area of quadrilateral ABCD.
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October 2nd, 2014, 03:26 PM   #5
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That doesn't work. You can't find x without using the length of MN.
ConvexQuad.PNG
With some help from WolframAlpha,
I got angle(ABC) = 135°, area(ABC) = √2 and area(ACD) = 2 + 3√2.
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