What is the area of the ABCD rectangle?

Question

$(1) \quad \dfrac{256}{17}$

$(2) \quad \dfrac{39}{4}$

$(3) \quad \dfrac{483}{8}$

$(4) \quad \dfrac{52}{4}$

$(5) \quad \dfrac{492}{18}$

Answer 1

Let $AE=a$, $BE=b$, $AF=c$, $DF=d$.

Now

\begin{align} a^2+c^2 &= 3^2 \tag{1} \\ d^2+(a+b)^2 &= 4^2 \tag{2} \\ b^2+(c+d)^2 &= 5^2 \tag{3} \\ a^2+c^2+d^2+(a+b)^2 &= b^2+(c+d)^2 \tag{$3^2+4^2=5^2$} \\ a(a+b) &= cd \tag{4} \\ d^2+\left( \frac{cd}{a} \right)^2 &= 4^2 \\ d^2(a^2+c^2) &= 4^2 a^2 \\ 3^2d^2 &= 4^2a^2 \\ d &= \frac{4a}{3} \\ c &= \sqrt{3^2-a^2} \\ a+b &= \frac{4c}{3} \\ (a+b)(c+d) &= \frac{4}{3} \sqrt{3^2-a^2} \left( \frac{4a}{3}+\sqrt{3^2-a^2} \right) \\ \end{align}

Unless $a+b=c+d$, that is a square, you cannot have a proper choice.

For the square case,

$$(a+b)(c+d)=\frac{256}{17}$$

I'm leaving the missing steps for above result as an exercise.

Answer 2

A unique triangle cannot be constructed so no area can be associated with it.

Shown here are two cases of rectangles (red and blue borders). The circles have radii $(5,4)$ for side lengths and a side of tangent length $3$ units between them.

EDIT1:

Area is not invariant. To calculate when hypotenuse makes an angle $u$ to vertical,

$$ (3 \cos u + 4 \sin u)(3 \sin u + +4 \cos u)=(12+12.5 \sin2u) $$

It has maximum value 24.5 when hypotenuse makes $45^{\circ}$ to vertical, and minimum 12 at time of vertical/horizontal positions of the shorter sides.