Let ABHF be a trapezoid such that
1) sides AB and FH are parallel,
2) angle BAF is 90 degrees, and
3) side FH is larger than side AB.
Let G be a point on side FH such that line segment BG
is perpendicular to FH. Let line segment CDE be parallel
to AB with points C, D, and E on line segments AF, BG,
and BH respectively.
Let x = [ABDC], y = [CDGF], z = [BDE], and w = [DEHG].
Determine w in terms of x, y, and z.
Note: [PQ...UV] denotes the area of polygon PQ...UV.
Consider the parallel sides to be horizontal, for the discussion of height and base in the triangles below.
As the location of line CDE varies, the area of triangle BDE increases in proportion as the square of its height BD, as the base, DE is increasing in the same proportion. In the same way, consider the area of triangle BGH, which is z+w:
(z+w)/z = (GB/DB)^2
But on the rectangle side of the diagram:
(y+x)/x = GB/DB
Thus
(z+w)/z = ((y+x)/x)^2
z+w = z * ((y+x)/x)^2
w = z * (((y+x)/x)^2 - 1)
w = z * ((y^2+x^2+2*x*y)/x^2 - 1)
w = z * ((y^2+2*x*y)/x^2)
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Posted by Charlie
on 2010-03-27 16:00:37 |