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$a,b,c$ 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 $a,b,c$

enter image description here

$$y=a.\sin P +b \sin Q$$

$$|EF|=a \cos P=b \cos Q$$

$$x=a.\cos P +\sqrt{c^2-a^2 \sin^2 P}$$


$$y=a.\sin P +b \sin Q=a.\sin P +b \sqrt{1-\frac{a^2 \cos^2 P}{b^2}}$$

$$y=a.\sin P +b \sin Q=a.\sin P + \sqrt{b^2-a^2 \cos^2 P}$$

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 $a,b,c$

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  • $\begingroup$ another +1 for showing your work.from georgia $\endgroup$ – dato datuashvili Apr 5 '13 at 6:42
  • $\begingroup$ $a,b,c,d$ are length right?is $m,n$ given? $\endgroup$ – dato datuashvili Apr 5 '13 at 6:48
  • $\begingroup$ if m,n is known, then d is known. but if m,n are not known, it is imposible to find the area as there is unlimit possible. $\endgroup$ – chenbai Apr 5 '13 at 8:56
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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.

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  • $\begingroup$ Thanks a lot for answer and the graph. Yellow area does not follow the rule. Red and green lengths must be on opposite corner. Another important thing that $d=\sqrt {b^2+c^2-a^2}$. Thus you can draw another circle for $d$. I still suspect that $x.y$ can be constant. I need to check more some formulas. $\endgroup$ – Mathlover Apr 8 '13 at 10:39
  • $\begingroup$ @Mathlover, when m or n is not fixed,the d is not effect the result as the last point is always one circle if we have a circle for d. so it is impossible to find the area only with a,b,c $\endgroup$ – chenbai Apr 10 '13 at 7:37

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