Let ABCD be an isosceles trapezoid with AB parallel to CD, angle ADC equal to angle BCD and AD = DC = CB. Let M be the midpoint of BC and N be the midpoint of CD.
If AB=MN=1 then how long is CD?
If we bisect the trapesium along the axes of symmetry and reassemble the pieces we obtain a rectangle.
Let AB = MN= 2x , AD = DC =CB = 2y and h be the height of the trapesium, then the sides of this rectangle are h and x+y with diagonal length 2 MN = 4 x, Also the rectangle generated by the line BC has sides h and y-x with diagonal length 2y = BC.
This gives two pythagorean equations. Eliminating h and substituting x = 1/2 gives
y = ( sqrt(17) -1 ) / 4
Hence CD = 2 y = ( sqrt(17) -1 ) / 2.
(= 1.56155....)
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Posted by goFish
on 2006-01-02 13:19:19 |
Geometrical solution
