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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.


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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is it a square (or rectangle)?...shouldn't CD = AB? – newbie Aug 18 '11 at 10:39
That's the point, it doesn't say so in the text. Only that it has four corners. But since only A and B is specifically said to be 90 degrees, D, and C is not. The pic is an illustration, it was not in the book. By Pythagorean theorem DA is 38.78m. – Algific Aug 18 '11 at 10:41
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. – kahen Aug 18 '11 at 10:55
up vote 2 down vote accepted

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. $$CD = \sqrt{(AD-BC)^{2} + AB^2} \approx 39.5851 $$

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Spot on, thanks. – Algific Aug 18 '11 at 10:56

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