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Find the area (Posted on 2011-02-09) Difficulty: 3 of 5
The diagonals of the trapezoid ABCD intersect at P.
The area of the triangle ABP is 216 and the area of CDP is 150.

What is the area of the trapezoid ABCD?

See The Solution Submitted by Ady TZIDON    
Rating: 3.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 4

The diagonals of a trapezoid divide it
into four triangles. Two of the triangles
( those having a side in common with the
lateral  sides of the trapezoid ) have
equal areas. Therefore, AB and CD are
parallel.
Let [UV...YZ] denote the area of polygon
UV...YZ.
From the above we have
   [BPC] = [DPA]                      (1)
For any convex quadrilateral we have
   [APB][CPD] = [BPC][DPA]            (2)
Combining (1) and (2)
   [BPC]^2 = [APB][CPD] = 216*150
           = 180^2
Therefore,
   [ABCD] = [APB] + [BPC] + [CPD] + [DPA]
          =  216  +  180  +  150  +  180
          =  726
 

  Posted by Bractals on 2011-02-09 19:02:24
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