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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)

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Some Thoughts re: Solution | Comment 3 of 4 |
(In reply to Solution by Bractals)


To prove the following is easy.
"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."
Let WXYZ be an trapezoid with side WX
parallel with side YZ and diagonals WY
and XZ intersecting at point P.
   [WYX] = [WZX]
because they both have the same base and
their altitudes are equal. Therefore,
   [XPY] = [WYX] - [WPX] = [WZX] - [WPX] = [ZPW].
To prove 
"For any convex quadrilateral we have
   [WPX][YPZ] = [XPY][ZPW]."
Let WXYZ be any convex quadrilateral instead
of a trapezoid.
   [WPX] = (1/2)|PW||PX|*sin(WPX)    (1)  
   [XPY] = (1/2)|PX||PY|*sin(WPX)    (2) 
   [YPZ] = (1/2)|PY||PZ|*sin(WPX)    (3) 
   [ZPW] = (1/2)|PZ||PW|*sin(WPX)    (4)
because sin(XPY) = sin(YPZ) = sin(ZPW)
                 = sin(WPX). Therefore,
   [WPX][YPZ] = (1/4)|PW||PX||PY||PZ|*
                     [sin(WPX)]^2
   [XPY][ZPW] = (1/4)|PX||PY||PZ||PW|*
                     [sin(WPX)]^2
 

  Posted by Bractals on 2011-02-10 18:33:34
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