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Let me first write out the task.

A square property ABCD, Angle A = B 90 degrees. AB = 39m, BC = 32m, and the diagonal BD is 55m. How long would a fence be to run around the property? I drew it up in GeoGebra.

Scetch

The problem for me is to find DC, I can find the other sides using Pythagorean theorem. I can find DC using the cosine rule, but since that hasn't been introduced yet in the book. I'm curious how one can find DC without using Sine rule or the cosine rule.

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    Your drawing is *extremely* deceptive. It suggests that the angles at C and D are 90 degrees as well which is impossible with the given lenghts.2011-08-18

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Dont use square in your question statement, it is just a polygon.

Based on given information, you can check $AD \neq BC $

$AD = \sqrt{BD^{2} - AB^{2} } = 4\sqrt{94}>BC = 32$

so, if you drew a line from C paralleled to AB, you can find a new triangle with diagonal CD. $CD = \sqrt{(AD-BC)^{2} + AB^2} \approx 39.5851 $