Diagonals of quadrilateral
Is there any way to find out diagonals of quadrilateral if I only k开发者_如何学编程now the four sides - no angles?
I understand I could calculate it with the law of cosines:
p = sqrt( a^2 + b^2 - 2*a*b*cos(α) ) q = sqrt( b^2 + c^2 - 2*b*c*cos(β) ) // or whatever you name these angles
but I don't know the angles! So I'm kind of stuck here.
I don't believe that's possible, at least not in the general case. I believe least for parallelograms you can have multiple configurations of angles possible using the same length sides, and the different angles would end up giving you different lengths for the diagonals.
Even for non-parallelograms it's possible to have multiple configurations for the same lengths, and these would give you different diagonal lengths.
You really need to know an angle in order to calculate the diagonal.
In general, a quadrilateral is not uniquely determined by the lengths of its sides. For instance, if all sides are equal, it could be a square, or any of a family of diamond shapes with opposite angles equal. All of these shapes will have different diagonals, so no, this is not possible.
If you knew one angle, then it would be possible.
No, there is not a way to compute this without knowing the angles.
Consider taking four matches. Put them in a square. Then squeeze two corners together. The other corners will move from each other, but the sides remain the same. This process is the same if the sides aren't equal.
As it turns out: you cannot know the diagonal length unless you know the angles. If it is a square or a rectangle (angles 90 degrees) it's easy: Pythagoras. Otherwise, there's no way to tell unless you get at least one angle.
For parallelograms, the rule is that the sum of the squares of the diagonals equals the sum of the squares of the sides.
This provides one method for proving the equation for the length of a triangle median - see https://www.cut-the-knot.org/triangle/LengthOfMedian.shtml. The proof was published by C. P. Lawes, 'Proof Without Words: The Length of a Triangle Median via the Parallelogram Law', Math. Mag. issue 86(2013), page 146.
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