$a,b,c$ are given parameters . I would like to find Area of (ABCD) rectangular.
I can find $d$ from $a,b,c$.
$$(x-m)^2+(y-n)^2=a^2$$
$$(x-m)^2+n^2=b^2$$
$$m^2+(y-n)^2=c^2$$
$$m^2+n^2=d^2$$
$$m^2+n^2+(x-m)^2+(y-n)^2=a^2+d^2=b^2+c^2$$
$$d=\sqrt {b^2+c^2-a^2}$$
Let's define
$\angle AEB =\alpha$,
$\angle DEC =\beta$ ,$\angle AED =\gamma$ , $\angle BEC =\phi$
$$x^2=a^2+c^2-2ac \cos (\alpha)$$
$$x^2=b^2+d^2-2bd \cos (\beta)$$
$$y^2=c^2+d^2-2cd \cos (\gamma)$$
$$y^2=a^2+b^2-2ab \cos (\phi)$$
$$b^2+d^2-2bd \cos (\beta)=a^2+c^2-2ac \cos (\alpha)$$
$$a^2+b^2-2ab \cos (\phi)=c^2+d^2-2cd \cos (\gamma)$$
And also we know that
$$\alpha + \beta + \phi + \gamma = 2 \pi $$
$$\cos (\alpha + \beta + \phi + \gamma)= \cos (2 \pi)=1$$
Area of $ABCD =\frac{1}{2} [ac \sin (\alpha) + bd \sin (\beta) + cd \sin (\gamma))+ ab \sin (\phi)]=xy$
I am stuck to solve the equations and find the area by given $a,b,c$, Is it possible to find area of ABCD rectangular via given 3 parameters $a,b,c$ ?
Thanks for hints and answers.
UPDATE: Nov, 14th 2014:
I proved that Area of ABCD does not depend on only $a,b,c$
$$y=a.\sin P +b \sin Q$$
$$|EF|=a \cos P=b \cos Q$$
$$x=a.\cos P +\sqrt{c^2-a^2 \sin^2 P}$$
$$y=a.\sin P +b \sin Q=a.\sin P +b \sqrt{1-\frac{a^2 \cos^2 P}{b^2}}$$
$$y=a.\sin P +b \sin Q=a.\sin P + \sqrt{b^2-a^2 \cos^2 P}$$
Area of $ABCD=x.y=(a.\cos P +\sqrt{c^2-a^2 \sin^2 P})(a.\sin P + \sqrt{b^2-a^2 \cos^2 P})$
The formula shows that The Area also depends on an angle not only $a,b,c$
Best Answer
look at the above picture, you can see orange and yellow rectangles are all satisfy a,b,c if m,n is not fixed. and there are many such rectangles so you can't find the area. but you can find max area by a,b,c which might be another exercise you can do which is very hard to find out the final result.
for m,n is fixed, $d=\sqrt{m^2+n^2}$ .
BTW, if one of m and n is known,then the area is fixed.