Geometry Geometry Math Forum

 October 2nd, 2014, 02:08 AM #1 Newbie   Joined: Oct 2014 From: Ukriane Posts: 6 Thanks: 0 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?
 October 2nd, 2014, 03:45 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra First of all, you need to draw a clear picture and upload it.
 October 2nd, 2014, 04:29 AM #3 Newbie   Joined: Oct 2014 From: Ukriane Posts: 6 Thanks: 0
 October 2nd, 2014, 06:26 AM #4 Senior Member   Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230 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.
 October 2nd, 2014, 03:26 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 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. Thanks from VLad9pa

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