The Notes: What is the area of a quadrilateral with sides 5, 6, 4 and 9.

Friday, August 19, 2011

What is the area of a quadrilateral with sides 5, 6, 4 and 9.

Hello!

A quadrilateral, to the contrast with a triangle, usually cannot be uniquely determined by the lengths of its sides. And its area may be different, too.

I suppose the sides are go in the given order. Consider an angle $\displaystyle\alpha$ between sides 5 and 6. Then the corresponding diagonal c may be found by the Cosine law:

$\displaystyle{{c}}^{{2}}={{5}}^{{2}}+{{6}}^{{2}}-{2}\cdot{5}\cdot{6}\cdot{\cos{{\left(\alpha\right)}}}={61}-{60}\cdot{\cos{{\left(\alpha\right)}}}.$

This diagonal also forms a triangle with the sides 4 and 9, so it cannot be greater than 4+9=13 and cannot be less than 9-4=5.

The area of a quadrilateral is the sum of the areas of these two triangles. It is

$\displaystyle\frac{{1}}{{2}}\cdot{5}\cdot{6}\cdot{\sin{{\left(\alpha\right)}}}+\frac{{1}}{{4}}\cdot\sqrt{{{\left({169}-{{c}}^{{2}}\right)}{\left({{c}}^{{2}}-{25}\right)}}}$

(Heron's formula is used for the second case). It is

$\displaystyle{15}{\sin{{\left(\alpha\right)}}}+\frac{{1}}{{4}}\cdot\sqrt{{{\left({108}+{60}{\cos{{\left(\alpha\right)}}}\right)}{\left({36}-{60}{\cos{{\left(\alpha\right)}}}\right)}}}.$

This function is not a constant, please look at its graph by the link attached. There one can see limits for $\displaystyle\alpha$ and for the area.

The Notes

Friday, August 19, 2011

What is the area of a quadrilateral with sides 5, 6, 4 and 9.

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