Constructing a Quadrilateral

Question

I am given three sides of length 4.5 and one more side of length 4.2 (dimensionless entities). Also the area is given as 19.575. Now I have been given the task to construct a quadrilateral with these figures. I don't know whether it is trivial or complex. Since the type of quadrilateral is not specified, I'm concerned whether such a polygon can be constructed (I tried considering trapezium), and if possible, whether it is unique! And when unique, how to construct. I need help with this. Any online source for construction is also welcome. The main goal is to know angles between the sides. Thanks in advance.

Answer 1

Before breaking out your straightedge and compasses, you might want to read the story of the Emperor's new clothes. You need to check that the solution, like the clothes, is really there. Read on to discover the naked truth.

Bounds of Decency

Given a set of side lengths for any polygon, the maximum possible area is obtained by setting up the angles so the polygon is inscribed in a circle. With a quadrilateral having three congruent sides that would, of course, be an isosceles trapezoid. Two of the three congruent sides are the legs of the trapezoid, the remaining congruent side and the fourth side are the bases.

Thus consider a trapezoid with bases $4.2$ and $4.5$ and both legs $4.5$. Its altitude is then

$\sqrt{4.5^2-[(4.5-4.2)/2]^2}=\sqrt{20.2275}$

The area, which is the maximum possible area for our quadrilateral, is then half the sum of bases times this altitude:

$S_{max}=4.35\sqrt{20.2275}$

$=\color{blue}{19.564...<19.575}$

The Emperor, in fact, is wearing no clothes. The construction is impossible.

Answer 2

I am going to assume that the quadrilateral is constructible.

The first thing to do is check out Heron's formula from here.

Draw a main diagonal and assume its length is $x$. Then you have two triangles. You can compute the area of both triangles using Heron's formula, as a function of $x$. Since the quadrilateral's area is given, you know the area of the two triangles formed by the main diagonal.

Therefore, you can solve for the length of the main diagonal. That is, you can solve for $x$. Once this is done, you can solve for two of the angles via the Law of Cosines.

Then, rinse and repeat, to get the length of the other main diagonal, and consequently the other two angles.

Addendum
Never having done this, it is unclear to me whether the computation of each diagonal will yield a unique value. From my perspective, that is actually irrelevant because when you identify each of the 4 angles that correspond to a pair of (possible diagonal lengths), you have the constraint that the 4 angles must sum to $360^{\circ}.$