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.

The way I solve this problem is different from the previous educator. I tend to visualize the shape and use logic to solve the problem, which is what I did here; there may be other answers as to area, for other shapes.

I used the lengths provided for the sides and drew a quadrilateral with a width of 4. The height of the shorter side is 6, and the longer side, 9. This leaves the side length 5 for a slanted side connecting the tops of the short and long sides.

What this gives me is a quadrilateral composed of two shapes: a rectangle with width 4 and height 6, topped by a right triangle with sides 3, 4, and 5. My two shapes then have areas of 24 square units (rectangle), and 6 square units (triangle), for a combined area of 30 square units.

As noted above, to determine the area of a quadrilateral we generally need to know something about the angles.

If the quadrilateral is cyclic (i.e. can be inscribed in a circle) then we can use Brahmagupta's formula. (This is an extension of Hero's (Heron's) formula for the area of a triangle.)

The formula is `A=sqrt((s-a)(s-b)(s-c)(s-d)) ` where the side lengths are a,b,c, and d and s is the semiperimeter (one-half of the perimeter.)

Here we have a=5,b=6,c=4,d=9 and s=12.

Thus, if the quadrilateral is cyclic, the area is:

`A=sqrt((12-5)(12-6)(12-4)(12-9))=sqrt(1008)=12sqrt(7)~~31.75 `

This is true if the quadrilateral is a trapezoid with bases 6 and 9 each perpendicular to the leg of length 4. Without this information the problem has a range of possibilities; we were not given that the figure was a trapezoid let alone a trapezoid with bases perpendicular to a leg.