If you know the Diagonal and Area of a Rectangle, can you find the sides of the rectangle? If you know the Diagonal and Area of a Rectangle, can you find the sides of the rectangle? 
I was doing strange math yesterday, and I can across the realization that two different rectangles can’t have the same area and same diagonal. But I haven’t been able to solve a equation that shows this. I also haven’t be able to prove myself wrong? So maybe higher math gods will help?
$$\text{Diagonal}=\sqrt{x^2+y^2}$$
$$\text{Area}=xy$$
 A: Let $d=\sqrt{x^2+y^2}$ be the diagonal and $a=xy$ be the area. Assuming $x\ge y$, we note that
$$\sqrt{d^2+2a}=\sqrt{x^2+2xy+y^2}=x+y$$
$$\sqrt{d^2-2a}=\sqrt{x^2-2xy+y^2}=x-y$$
Thus, by solving a linear system,
$$\frac{\sqrt{d^2+2a}+\sqrt{d^2-2a}}2=x$$
$$\frac{\sqrt{d^2+2a}-\sqrt{d^2-2a}}2=y$$
A: In addition to the algebraic proofs, you can also demonstrate this with a geometric construction.
Take a given rectangle as shown in the figure below. The diagonal of length $d$ cuts the rectangle into two right triangles.
Each right triangle has some altitude measured from its right angle to the hypotenuse. Since the two triangles are congruent, they have the same altitude, which is labeled $h$ in the figure.

The area of each triangle is $\frac12 hd,$ and the area of the entire rectangle is $hd.$ Any other rectangle with the same length diagonal and the same area,
that is, diagonal $d$ and area $hd,$ must also be cut into two right triangles by its diagonal, each right triangle having hypotenuse $d$ and height $h$.
If the heights are greater than $h$ the area will be too large and if they are less than $h$ the area will be too small.
A fact about right triangles is that for a given hypotenuse $AB$, the right-angled vertex of the triangle always lies on the circle that has the segment $AB$ as its diameter.
So for a given diagonal of length $d,$ a rectangle with that diagonal must have two vertices at the ends of the diagonal and the other two lying on the circle with that diameter as shown in the figure below.
To find the other two vertices, we construct two lines parallel to the diameter at a distance $h$ from the diameter. A right triangle with the given hypotenuse and height $h$ from that hypotenuse must have its right-angle vertex at one of the intersections of one of those lines with the circle.
Choose one such intersection and the rectangle is determined as shown in the figure below.

The lengths of the sides of this rectangle are uniquely determined by the distance from the intersection point to the two ends of the diameter.
So they are uniquely determined by the given area and given length of the diagonal from which the figure was constructed.
Alternatively, you can choose one of the intersections that is not used in the figure above, but that just gives you a congruent rectangle in a different orientation.
A: Sure:
You have $xy = A$ a known constant.
And $\sqrt{x^2 +y^2} = d$ a known constant.
So just substitute.
$x = \frac Ay$ (assuming $y\ne 0$ which if the area is positive must be so... or we could do $y = \frac Ax$.... it doesn't matter.
And $\sqrt{(\frac Ay)^2 + y^2} = d$.
So $(\frac Ay)^2 + y^2 = d^2$
So $A^2 + y^4 =d^2 y^2$
$y^4 - d^2y^2 +A^2 =0$.  Use the quadratic formula:
$y^2 = \frac {d^2 \pm \sqrt{d^4-4A^2}}2$  We know $y$ is positive
So $y =\sqrt{\frac {d^2 \pm \sqrt{d^4-4A^2}}2}$ and $x =\frac {A}{\sqrt{\frac {d^2 \pm \sqrt{d^4-4A^2}}2}}=\frac {A}{\sqrt{\frac {d^2 \pm \sqrt{d^4-4A^2}}2}}\frac {\sqrt{\frac {d^2 \mp \sqrt{d^4-4^2A}}2}}{\sqrt{\frac {d^2 \mp \sqrt{d^4-4A^2}}2}}=\frac {A}{\sqrt{\frac {d^4-(d^4-4A^2)}4}}\sqrt{\frac {d^2 \mp \sqrt{d^4-4A^2}}2}=\frac AA\sqrt{\frac {d^2 \mp \sqrt{d^4-4A^2}}2}=\sqrt{\frac {d^2 \mp \sqrt{d^4-4A^2}}2}$
....
So if we take the rectangle with sides $3$ and $4$ and area $A=12$ and diagonal $d=5$ and pretended we didn't know the sides we'd have.
$y = \sqrt{\frac {5^2+\sqrt{5^4-4*12^2}}2}=\sqrt{\frac {25+\sqrt{(25+2*12)(25-2*12)}}2}=\sqrt{\frac{25+\sqrt{49*1}}2}=\sqrt{\frac {25+7}2}=\sqrt{16}=4$
And $x = \sqrt{\frac {5^2-\sqrt{5^4-4*12^2}}2}=\sqrt{\frac {25-\sqrt{(25+2*12)(25-2*12)}}2}=\sqrt{\frac{25-\sqrt{49*1}}2}=\sqrt{\frac {25-7}2}=\sqrt{9}=3$
A: With $d = $ Diagonal and $a = $ Area your equations are $x^2 + y^2 = d^2$ and $xy = a$. Then $y = a/x$, and the first equation becomes $x^2 + \frac{a^2}{x^2} = d^2$
A: Squaring the given information, you know $x^2+y^2=A$ and $x^2y^2=B$, say. Then $x^2$ and $y^2$ are the zeros of the quadratic polynomial $(t-x^2)(t-y^2)=t^2-At+b$, which you can solve by the quadratic formula; taking the positive square roots of those zeros yields $x$ and $y$.
