Problem: The area of a trapezoid is equal to 2 and the sum of his diagonals is equal to 4. Find the trapezoid height.
[QUESTION]: I find a result that implies that the height of the Triangle is not uniquely defined, any help discussing this result or with other solutions is appreciated
Attempt: So using the notation of the figure I have:
$$a(\square ABCD)=2,\: \: \: AC+BD=d_1+d_2=4, \: \: h=?$$
So I make the following construction mirroring the trapezoid twice, once vertically and once horizontally:
Here I have put both trapezoid together sharing the $\overline{BC}$ segment.
Clearly $a(\triangle DB\hat{A})=a(\square ABCD)=\Big(\frac{b_1+b_2}{2}\Big)h=2$
Using Heron's Formula:
$$a(\triangle DB\hat{A})=\sqrt{s(s-(b_2+b_1))(s-d_1)(s-d_2)}$$
whre $s=\frac{b_1+b_2+d_1+d_2}{2}$
Using $B=(b_2+b_1)$ for short :
$$a(\triangle DB\hat{A})=\frac{1}{4}\sqrt{(B+d_1+d_2)(-(B)+d_1+d_2)(B-d_1+d_2)(B+d_1-d_2)}$$
Using the beginning relations :
$$a(\triangle DB\hat{A})=\frac{1}{4}\sqrt{(B+4)(-(B)+4)(B-(d_1-d_2))(B+(d_1-d_2))}$$ $$a(\triangle DB\hat{A})=\frac{1}{4}\sqrt{(-B^2+16)(B^2-(d_1-d_2)^2))}$$ $$a(\triangle DB\hat{A})=\frac{1}{4}\sqrt{-B^4-B^2(d_1-d_2)^2+16B^2-16(d_1-d_2)^2)}$$
Now replacing $a(\triangle DB\hat{A})=2$
$$64=-B^4+B^2(16-(d_1-d_2)^2)-16(d_1-d_2)^2$$ $$64=-B^4+B^2(16-(4-2d_2)^2)-16(4-2d_2)^2$$
I can solve this Polynomial in Wolfram (where I changed $B\rightarrow y$ and $d_2 \rightarrow x$), obtaining many solutions
$$B = \pm\Big(\sqrt{2} \sqrt{-d_2^2+\sqrt{(d_2-2)^2 (d_2^2-4d_2-20)}+4 d_2}\Big)$$ $$B = \pm\Big(\sqrt{2} \sqrt{-d_2^2-\sqrt{(d_2-2)^2 (d_2^2-4d_2-20)}+4 d_2}\Big)$$
This yields that I can't find a unique value of $B$ and hence of $h$, which sort of make sense because I can find many triangles $\triangle DB\hat{A}$ that match the problem conditions of Area and sum of diagonals. This is kind of as if one have the Base and the height of a triangle but one cannot specify the side lengths.
I don't Know if there is another thing I can use, any help is appreciated.
There is a way to find $h$, though quite tedious.
By Heron's formula, $$ 2=\sqrt{s(s-\hat{A}B)(s-BD)(s-\hat{A}D)},$$ where $ s=\frac{\hat{A}B+BD+\hat{A}D}{2}=\frac{4+\frac{4}{h}}{2}=2+\frac{2}{h}.$
\begin{eqnarray} 2&=&\sqrt{s(s-\hat{A}B)(s-BD)(s-\hat{A}D)} \\&=& \sqrt{\left(2+\frac{2}{h}\right)\left(2+\frac{2}{h}-\hat{A}B\right)\left(2+\frac{2}{h}-BD\right)\left(2+\frac{2}{h}-\frac{4}{h}\right)} \\&=&\sqrt{\left(2+\frac{2}{h}\right)\left(\left(2+\frac{2}{h}\right)^2-\left(2+\frac{2}{h}\right)(\hat{A}B+BD)+\hat{A}BBD\right)\left(2-\frac{2}{h}\right)} \\&=& \sqrt{\left(4-\frac{4}{h^2}\right)\left(\left(2+\frac{2}{h}\right)^2-4\left(2+\frac{2}{h}\right)+\hat{A}BBD\right)} \end{eqnarray}
Let $k=\hat{A}BBD$. Back to Heron:
\begin{align} 4 &= &\left(4-\frac{4}{h^2}\right)\left(\left(2+\frac{2}{h}\right)^2-4\left(2+\frac{2}{h}\right)+k\right) \\ 4h^4&=&\left(4h^2-4\right)\left(\left(2h+2\right)^2-4\left(2h^2+2h\right)+kh^2\right) \\ &=&\left(4h^2-4\right)\left(-4h^2+4+kh^2\right) \\ &=&4(k-4)h^4-4(k-4)h^2+16h^2-16 \end{align} $$(4k-20)h^4+(32-4k)h^2-16=0 \tag{*}\label{*}$$
The discriminant $b^2-4ac\geq 0$ as $h^2$ is real.
\begin{align} (32-4k)^2-4(4k-20)(-16)&\geq& 0 \\ 1024-256k+16k^2+256k-1280 &\geq& 0 \\ 16k^2-256&\geq& 0 \end{align} Thus, $k \geq 4 $ or $ k \leq -4 $. Since $k$ is positive, $k \geq 4$.
$\hat{A}BBD=k$ and $\hat{A}B + BD =4$. Hence we now introduce quadratic equation $$x^2-4x+k=0. $$ Again, $b^2-4ac \geq 0$ since $\hat{A}B$ and $BD$ are real. Then \begin{align} 16-4k \geq 0 \\ k \leq 4 \end{align} Then $4 \leq k \leq 4$, which means that $k=4$.
Therefore, substituting $4$ for $k$ in $\eqref{*}$ brings us $h=\sqrt{2}$.
Appendix
If $h$ can be found, then the statement that $h$ cannot be determined is not true. The problem in the argument is that some clues are neglected such as the use of discriminant. Actually, $d_2$ can be determined if we find the discriminant of the (quadratic) Polynomial with respect to $B^2$. We will know that $d_2=2$ by some geometric interpretations, and hence that $B=2\sqrt2$. $B$ can be determined, so can $h$.