Posted by Pre Calc on Monday, October 5, 2009 at 1:43pm.
There may be simpler ways of doing, but this is what came to mind for me.
In fact, the problem boils down to finding the height h, and the offset of the two parallel sides, k.
Let ABCD be the vertices of the quadrilateral where AB and DC are parallel sides.
AB=4 centimetres, and
DC=10 centimetres.
Drop a perpendicular from A to the long side DC to intersect at E.
Assume for the moment E is between D and C, so that DEC is a straight line.
Let
k=distance DE, and
h=distance between the two parallel sides = length AE
Consider each of the oblique sides as the hypothenuse of a right triangle,
for ADE,
AE²+DE²=AD²
h²+k²=8² ..... (1)
For the side BC,
(10-k-4)²+h²=BC²
36-12k+k²+h²=144....(2)
Subtract (1) from (2)
36-12k-8²=144
k=-44/12=-11/3
Meaning that the point E is situated on the extension of the line CD.
Substitute k into (1) gives
h=sqrt(8²-(11/3)²)
=sqrt(455)/3
=7.110243
Can you take it from here?
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