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 `alpha` between sides 5 and 6. Then the corresponding diagonal c may be found by the Cosine law:

`c^2 = 5^2+6^2-2*5*6*cos(alpha)=61-60*cos(alpha).`

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

`1/2*5*6*sin(alpha) + 1/4* sqrt((169-c^2)(c^2-25))`

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

`15sin(alpha) + 1/4* sqrt((108+60cos(alpha))(36-60cos(alpha))).`

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